1736 lines
66 KiB
C++
1736 lines
66 KiB
C++
// btree.cpp
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/**
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* Copyright (C) 2008 10gen Inc.
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*
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* This program is free software: you can redistribute it and/or modify
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* it under the terms of the GNU Affero General Public License, version 3,
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* as published by the Free Software Foundation.
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*
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* This program is distributed in the hope that it will be useful,
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* but WITHOUT ANY WARRANTY; without even the implied warranty of
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* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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* GNU Affero General Public License for more details.
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*
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* You should have received a copy of the GNU Affero General Public License
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* along with this program. If not, see <http://www.gnu.org/licenses/>.
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*/
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#include "pch.h"
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#include "db.h"
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#include "btree.h"
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#include "pdfile.h"
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#include "json.h"
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#include "clientcursor.h"
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#include "client.h"
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#include "dbhelpers.h"
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#include "curop-inl.h"
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#include "stats/counters.h"
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namespace mongo {
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#if !defined(_DURABLE) || !defined(_DEBUG)
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#define VERIFYTHISLOC dassert( thisLoc.btree() == this );
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#else
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// with _DURABLE, this assert wouldn't work without getting fancier as there are multiple mmap views for _DEBUG mode...
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#define VERIFYTHISLOC
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#endif
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/**
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* give us a writable version of the btree bucket (declares write intent).
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* note it is likely more efficient to declare write intent on something smaller when you can.
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*/
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BtreeBucket* DiskLoc::btreemod() const {
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assert( _a != -1 );
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BtreeBucket *b = const_cast< BtreeBucket * >( btree() );
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return static_cast< BtreeBucket* >( dur::writingPtr( b, BucketSize ) );
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}
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_KeyNode& _KeyNode::writing() const {
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return *dur::writing( const_cast< _KeyNode* >( this ) );
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}
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KeyNode::KeyNode(const BucketBasics& bb, const _KeyNode &k) :
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prevChildBucket(k.prevChildBucket),
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recordLoc(k.recordLoc), key(bb.data+k.keyDataOfs())
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{ }
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// largest key size we allow. note we very much need to support bigger keys (somehow) in the future.
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static const int KeyMax = BucketSize / 10;
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// We define this value as the maximum number of bytes such that, if we have
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// fewer than this many bytes, we must be able to either merge with or receive
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// keys from any neighboring node. If our utilization goes below this value we
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// know we can bring up the utilization with a simple operation. Ignoring the
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// 90/10 split policy which is sometimes employed and our 'unused' nodes, this
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// is a lower bound on bucket utilization for non root buckets.
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//
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// Note that the exact value here depends on the implementation of
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// rebalancedSeparatorPos(). The conditions for lowWaterMark - 1 are as
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// follows: We know we cannot merge with the neighbor, so the total data size
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// for us, the neighbor, and the separator must be at least
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// BtreeBucket::bodySize() + 1. We must be able to accept one key of any
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// allowed size, so our size plus storage for that additional key must be
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// <= BtreeBucket::bodySize() / 2. This way, with the extra key we'll have a
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// new bucket data size < half the total data size and by the implementation
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// of rebalancedSeparatorPos() the key must be added.
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static const int lowWaterMark = BtreeBucket::bodySize() / 2 - KeyMax - sizeof( _KeyNode ) + 1;
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static const int split_debug = 0;
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static const int insert_debug = 0;
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extern int otherTraceLevel;
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/**
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* this error is ok/benign when doing a background indexing -- that logic in pdfile checks explicitly
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* for the 10287 error code.
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*/
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static void alreadyInIndex() {
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// we don't use massert() here as that does logging and this is 'benign' - see catches in _indexRecord()
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throw MsgAssertionException(10287, "btree: key+recloc already in index");
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}
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/* BucketBasics --------------------------------------------------- */
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string BtreeBucket::bucketSummary() const {
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stringstream ss;
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ss << " Bucket info:" << endl;
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ss << " n: " << n << endl;
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ss << " parent: " << parent.toString() << endl;
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ss << " nextChild: " << parent.toString() << endl;
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ss << " flags:" << flags << endl;
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ss << " emptySize: " << emptySize << " topSize: " << topSize << endl;
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return ss.str();
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}
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int BucketBasics::Size() const {
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assert( _wasSize == BucketSize );
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return BucketSize;
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}
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void BucketBasics::_shape(int level, stringstream& ss) const {
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for ( int i = 0; i < level; i++ ) ss << ' ';
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ss << "*\n";
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for ( int i = 0; i < n; i++ )
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if ( !k(i).prevChildBucket.isNull() )
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k(i).prevChildBucket.btree()->_shape(level+1,ss);
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if ( !nextChild.isNull() )
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nextChild.btree()->_shape(level+1,ss);
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}
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int bt_fv=0;
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int bt_dmp=0;
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void BtreeBucket::dumpTree(const DiskLoc &thisLoc, const BSONObj &order) const {
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bt_dmp=1;
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fullValidate(thisLoc, order);
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bt_dmp=0;
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}
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int BtreeBucket::fullValidate(const DiskLoc& thisLoc, const BSONObj &order, int *unusedCount, bool strict) const {
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{
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bool f = false;
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assert( f = true );
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massert( 10281 , "assert is misdefined", f);
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}
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killCurrentOp.checkForInterrupt();
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assertValid(order, true);
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if ( bt_dmp ) {
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out() << thisLoc.toString() << ' ';
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((BtreeBucket *) this)->dump();
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}
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// keycount
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int kc = 0;
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for ( int i = 0; i < n; i++ ) {
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const _KeyNode& kn = k(i);
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if ( kn.isUsed() ) {
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kc++;
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} else {
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if ( unusedCount ) {
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++( *unusedCount );
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}
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}
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if ( !kn.prevChildBucket.isNull() ) {
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DiskLoc left = kn.prevChildBucket;
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const BtreeBucket *b = left.btree();
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if ( strict ) {
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assert( b->parent == thisLoc );
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} else {
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wassert( b->parent == thisLoc );
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}
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kc += b->fullValidate(kn.prevChildBucket, order, unusedCount, strict);
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}
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}
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if ( !nextChild.isNull() ) {
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const BtreeBucket *b = nextChild.btree();
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if ( strict ) {
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assert( b->parent == thisLoc );
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} else {
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wassert( b->parent == thisLoc );
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}
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kc += b->fullValidate(nextChild, order, unusedCount, strict);
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}
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return kc;
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}
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int nDumped = 0;
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void BucketBasics::assertValid(const Ordering &order, bool force) const {
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if ( !debug && !force )
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return;
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wassert( n >= 0 && n < Size() );
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wassert( emptySize >= 0 && emptySize < BucketSize );
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wassert( topSize >= n && topSize <= BucketSize );
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// this is very slow so don't do often
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{
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static int _k;
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if( ++_k % 128 )
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return;
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}
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DEV {
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// slow:
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for ( int i = 0; i < n-1; i++ ) {
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BSONObj k1 = keyNode(i).key;
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BSONObj k2 = keyNode(i+1).key;
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int z = k1.woCompare(k2, order); //OK
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if ( z > 0 ) {
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out() << "ERROR: btree key order corrupt. Keys:" << endl;
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if ( ++nDumped < 5 ) {
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for ( int j = 0; j < n; j++ ) {
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out() << " " << keyNode(j).key.toString() << endl;
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}
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((BtreeBucket *) this)->dump();
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}
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wassert(false);
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break;
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}
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else if ( z == 0 ) {
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if ( !(k(i).recordLoc < k(i+1).recordLoc) ) {
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out() << "ERROR: btree key order corrupt (recordloc's wrong). Keys:" << endl;
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out() << " k(" << i << "):" << keyNode(i).key.toString() << " RL:" << k(i).recordLoc.toString() << endl;
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out() << " k(" << i+1 << "):" << keyNode(i+1).key.toString() << " RL:" << k(i+1).recordLoc.toString() << endl;
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wassert( k(i).recordLoc < k(i+1).recordLoc );
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}
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}
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}
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}
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else {
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//faster:
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if ( n > 1 ) {
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BSONObj k1 = keyNode(0).key;
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BSONObj k2 = keyNode(n-1).key;
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int z = k1.woCompare(k2, order);
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//wassert( z <= 0 );
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if ( z > 0 ) {
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problem() << "btree keys out of order" << '\n';
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ONCE {
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((BtreeBucket *) this)->dump();
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}
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assert(false);
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}
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}
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}
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}
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inline void BucketBasics::markUnused(int keypos) {
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assert( keypos >= 0 && keypos < n );
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k(keypos).setUnused();
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}
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inline int BucketBasics::totalDataSize() const {
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return (int) (Size() - (data-(char*)this));
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}
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void BucketBasics::init() {
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parent.Null();
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nextChild.Null();
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_wasSize = BucketSize;
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_reserved1 = 0;
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flags = Packed;
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n = 0;
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emptySize = totalDataSize();
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topSize = 0;
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reserved = 0;
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}
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/** see _alloc */
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inline void BucketBasics::_unalloc(int bytes) {
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topSize -= bytes;
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emptySize += bytes;
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}
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/**
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* we allocate space from the end of the buffer for data.
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* the keynodes grow from the front.
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*/
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inline int BucketBasics::_alloc(int bytes) {
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topSize += bytes;
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emptySize -= bytes;
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int ofs = totalDataSize() - topSize;
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assert( ofs > 0 );
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return ofs;
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}
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void BucketBasics::_delKeyAtPos(int keypos, bool mayEmpty) {
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assert( keypos >= 0 && keypos <= n );
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assert( childForPos(keypos).isNull() );
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// TODO audit cases where nextChild is null
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assert( ( mayEmpty && n > 0 ) || n > 1 || nextChild.isNull() );
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emptySize += sizeof(_KeyNode);
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n--;
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for ( int j = keypos; j < n; j++ )
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k(j) = k(j+1);
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setNotPacked();
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}
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/**
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* pull rightmost key from the bucket. this version requires its right child to be null so it
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* does not bother returning that value.
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*/
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void BucketBasics::popBack(DiskLoc& recLoc, BSONObj& key) {
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massert( 10282 , "n==0 in btree popBack()", n > 0 );
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assert( k(n-1).isUsed() ); // no unused skipping in this function at this point - btreebuilder doesn't require that
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KeyNode kn = keyNode(n-1);
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recLoc = kn.recordLoc;
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key = kn.key;
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int keysize = kn.key.objsize();
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massert( 10283 , "rchild not null in btree popBack()", nextChild.isNull());
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// weirdly, we also put the rightmost down pointer in nextchild, even when bucket isn't full.
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nextChild = kn.prevChildBucket;
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n--;
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emptySize += sizeof(_KeyNode);
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_unalloc(keysize);
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}
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/** add a key. must be > all existing. be careful to set next ptr right. */
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bool BucketBasics::_pushBack(const DiskLoc recordLoc, const BSONObj& key, const Ordering &order, const DiskLoc prevChild) {
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int bytesNeeded = key.objsize() + sizeof(_KeyNode);
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if ( bytesNeeded > emptySize )
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return false;
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assert( bytesNeeded <= emptySize );
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assert( n == 0 || keyNode(n-1).key.woCompare(key, order) <= 0 );
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emptySize -= sizeof(_KeyNode);
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_KeyNode& kn = k(n++);
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kn.prevChildBucket = prevChild;
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kn.recordLoc = recordLoc;
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kn.setKeyDataOfs( (short) _alloc(key.objsize()) );
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char *p = dataAt(kn.keyDataOfs());
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memcpy(p, key.objdata(), key.objsize());
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return true;
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}
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/**
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* insert a key in a bucket with no complexity -- no splits required
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* @return false if a split is required.
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*/
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bool BucketBasics::basicInsert(const DiskLoc thisLoc, int &keypos, const DiskLoc recordLoc, const BSONObj& key, const Ordering &order) const {
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assert( keypos >= 0 && keypos <= n );
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int bytesNeeded = key.objsize() + sizeof(_KeyNode);
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if ( bytesNeeded > emptySize ) {
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_pack(thisLoc, order, keypos);
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if ( bytesNeeded > emptySize )
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return false;
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}
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BucketBasics *b;
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{
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const char *p = (const char *) &k(keypos);
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const char *q = (const char *) &k(n+1);
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// declare that we will write to [k(keypos),k(n)]
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// todo: this writes a medium amount to the journal. we may want to add a verb "shift" to the redo log so
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// we can log a very small amount.
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b = (BucketBasics*) dur::writingAtOffset((void *) this, p-(char*)this, q-p);
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// e.g. n==3, keypos==2
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// 1 4 9
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// ->
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// 1 4 _ 9
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for ( int j = n; j > keypos; j-- ) // make room
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b->k(j) = b->k(j-1);
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}
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dur::declareWriteIntent(&b->emptySize, 12);
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b->emptySize -= sizeof(_KeyNode);
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b->n++;
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_KeyNode& kn = b->k(keypos);
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kn.prevChildBucket.Null();
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kn.recordLoc = recordLoc;
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kn.setKeyDataOfs((short) b->_alloc(key.objsize()) );
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char *p = b->dataAt(kn.keyDataOfs());
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dur::declareWriteIntent(p, key.objsize());
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memcpy(p, key.objdata(), key.objsize());
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return true;
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}
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/** with this implementation, refPos == 0 disregards effect of refPos */
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bool BucketBasics::mayDropKey( int index, int refPos ) const {
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return index > 0 && ( index != refPos ) && k( index ).isUnused() && k( index ).prevChildBucket.isNull();
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}
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int BucketBasics::packedDataSize( int refPos ) const {
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if ( flags & Packed ) {
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return BucketSize - emptySize - headerSize();
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}
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int size = 0;
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for( int j = 0; j < n; ++j ) {
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if ( mayDropKey( j, refPos ) ) {
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continue;
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}
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size += keyNode( j ).key.objsize() + sizeof( _KeyNode );
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}
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return size;
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}
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/**
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* when we delete things we just leave empty space until the node is
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* full and then we repack it.
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*/
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void BucketBasics::_pack(const DiskLoc thisLoc, const Ordering &order, int &refPos) const {
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if ( flags & Packed )
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return;
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thisLoc.btreemod()->_packReadyForMod(order, refPos);
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}
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void BucketBasics::_packReadyForMod( const Ordering &order, int &refPos ) {
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if ( flags & Packed )
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return;
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int tdz = totalDataSize();
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char temp[BucketSize];
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int ofs = tdz;
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topSize = 0;
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int i = 0;
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for ( int j = 0; j < n; j++ ) {
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if( mayDropKey( j, refPos ) ) {
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continue; // key is unused and has no children - drop it
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}
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if( i != j ) {
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if ( refPos == j ) {
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refPos = i; // i < j so j will never be refPos again
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}
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k( i ) = k( j );
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}
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short ofsold = k(i).keyDataOfs();
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int sz = keyNode(i).key.objsize();
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ofs -= sz;
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topSize += sz;
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memcpy(temp+ofs, dataAt(ofsold), sz);
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k(i).setKeyDataOfsSavingUse( ofs );
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++i;
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}
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if ( refPos == n ) {
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refPos = i;
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}
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n = i;
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int dataUsed = tdz - ofs;
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memcpy(data + ofs, temp + ofs, dataUsed);
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emptySize = tdz - dataUsed - n * sizeof(_KeyNode);
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assert( emptySize >= 0 );
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setPacked();
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assertValid( order );
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}
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inline void BucketBasics::truncateTo(int N, const Ordering &order, int &refPos) {
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n = N;
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setNotPacked();
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_packReadyForMod( order, refPos );
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}
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/**
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* In the standard btree algorithm, we would split based on the
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* existing keys _and_ the new key. But that's more work to
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* implement, so we split the existing keys and then add the new key.
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*
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* There are several published heuristic algorithms for doing splits,
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* but basically what you want are (1) even balancing between the two
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* sides and (2) a small split key so the parent can have a larger
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* branching factor.
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*
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* We just have a simple algorithm right now: if a key includes the
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* halfway point (or 10% way point) in terms of bytes, split on that key;
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* otherwise split on the key immediately to the left of the halfway
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* point.
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*
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* This function is expected to be called on a packed bucket.
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*/
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int BucketBasics::splitPos( int keypos ) const {
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assert( n > 2 );
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int split = 0;
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int rightSize = 0;
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// when splitting a btree node, if the new key is greater than all the other keys, we should not do an even split, but a 90/10 split.
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// see SERVER-983
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int rightSizeLimit = ( topSize + sizeof( _KeyNode ) * n ) / ( keypos == n ? 10 : 2 );
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for( int i = n - 1; i > -1; --i ) {
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rightSize += keyNode( i ).key.objsize() + sizeof( _KeyNode );
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if ( rightSize > rightSizeLimit ) {
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split = i;
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break;
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}
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}
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// safeguards - we must not create an empty bucket
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if ( split < 1 ) {
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split = 1;
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} else if ( split > n - 2 ) {
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split = n - 2;
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}
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return split;
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}
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void BucketBasics::reserveKeysFront( int nAdd ) {
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assert( emptySize >= int( sizeof( _KeyNode ) * nAdd ) );
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emptySize -= sizeof( _KeyNode ) * nAdd;
|
|
for( int i = n - 1; i > -1; --i ) {
|
|
k( i + nAdd ) = k( i );
|
|
}
|
|
n += nAdd;
|
|
}
|
|
|
|
void BucketBasics::setKey( int i, const DiskLoc recordLoc, const BSONObj &key, const DiskLoc prevChildBucket ) {
|
|
_KeyNode &kn = k( i );
|
|
kn.recordLoc = recordLoc;
|
|
kn.prevChildBucket = prevChildBucket;
|
|
short ofs = (short) _alloc( key.objsize() );
|
|
kn.setKeyDataOfs( ofs );
|
|
char *p = dataAt( ofs );
|
|
memcpy( p, key.objdata(), key.objsize() );
|
|
}
|
|
|
|
void BucketBasics::dropFront( int nDrop, const Ordering &order, int &refpos ) {
|
|
for( int i = nDrop; i < n; ++i ) {
|
|
k( i - nDrop ) = k( i );
|
|
}
|
|
n -= nDrop;
|
|
setNotPacked();
|
|
_packReadyForMod( order, refpos );
|
|
}
|
|
|
|
/* - BtreeBucket --------------------------------------------------- */
|
|
|
|
/** @return largest key in the subtree. */
|
|
void BtreeBucket::findLargestKey(const DiskLoc& thisLoc, DiskLoc& largestLoc, int& largestKey) {
|
|
DiskLoc loc = thisLoc;
|
|
while ( 1 ) {
|
|
const BtreeBucket *b = loc.btree();
|
|
if ( !b->nextChild.isNull() ) {
|
|
loc = b->nextChild;
|
|
continue;
|
|
}
|
|
|
|
assert(b->n>0);
|
|
largestLoc = loc;
|
|
largestKey = b->n-1;
|
|
|
|
break;
|
|
}
|
|
}
|
|
|
|
/**
|
|
* NOTE Currently the Ordering implementation assumes a compound index will
|
|
* not have more keys than an unsigned variable has bits. The same
|
|
* assumption is used in the implementation below with respect to the 'mask'
|
|
* variable.
|
|
*/
|
|
int BtreeBucket::customBSONCmp( const BSONObj &l, const BSONObj &rBegin, int rBeginLen, bool rSup, const vector< const BSONElement * > &rEnd, const vector< bool > &rEndInclusive, const Ordering &o, int direction ) {
|
|
BSONObjIterator ll( l );
|
|
BSONObjIterator rr( rBegin );
|
|
vector< const BSONElement * >::const_iterator rr2 = rEnd.begin();
|
|
vector< bool >::const_iterator inc = rEndInclusive.begin();
|
|
unsigned mask = 1;
|
|
for( int i = 0; i < rBeginLen; ++i, mask <<= 1 ) {
|
|
BSONElement lll = ll.next();
|
|
BSONElement rrr = rr.next();
|
|
++rr2;
|
|
++inc;
|
|
|
|
int x = lll.woCompare( rrr, false );
|
|
if ( o.descending( mask ) )
|
|
x = -x;
|
|
if ( x != 0 )
|
|
return x;
|
|
}
|
|
if ( rSup ) {
|
|
return -direction;
|
|
}
|
|
for( ; ll.more(); mask <<= 1 ) {
|
|
BSONElement lll = ll.next();
|
|
BSONElement rrr = **rr2;
|
|
++rr2;
|
|
int x = lll.woCompare( rrr, false );
|
|
if ( o.descending( mask ) )
|
|
x = -x;
|
|
if ( x != 0 )
|
|
return x;
|
|
if ( !*inc ) {
|
|
return -direction;
|
|
}
|
|
++inc;
|
|
}
|
|
return 0;
|
|
}
|
|
|
|
bool BtreeBucket::exists(const IndexDetails& idx, const DiskLoc &thisLoc, const BSONObj& key, const Ordering& order) const {
|
|
int pos;
|
|
bool found;
|
|
DiskLoc b = locate(idx, thisLoc, key, order, pos, found, minDiskLoc);
|
|
|
|
// skip unused keys
|
|
while ( 1 ) {
|
|
if( b.isNull() )
|
|
break;
|
|
const BtreeBucket *bucket = b.btree();
|
|
const _KeyNode& kn = bucket->k(pos);
|
|
if ( kn.isUsed() )
|
|
return bucket->keyAt(pos).woEqual(key);
|
|
b = bucket->advance(b, pos, 1, "BtreeBucket::exists");
|
|
}
|
|
return false;
|
|
}
|
|
|
|
/**
|
|
* @param self - don't complain about ourself already being in the index case.
|
|
* @return true = there is a duplicate.
|
|
*/
|
|
bool BtreeBucket::wouldCreateDup(
|
|
const IndexDetails& idx, const DiskLoc &thisLoc,
|
|
const BSONObj& key, const Ordering& order,
|
|
const DiskLoc &self) const
|
|
{
|
|
int pos;
|
|
bool found;
|
|
DiskLoc b = locate(idx, thisLoc, key, order, pos, found, minDiskLoc);
|
|
|
|
while ( !b.isNull() ) {
|
|
// we skip unused keys
|
|
const BtreeBucket *bucket = b.btree();
|
|
const _KeyNode& kn = bucket->k(pos);
|
|
if ( kn.isUsed() ) {
|
|
if( bucket->keyAt(pos).woEqual(key) )
|
|
return kn.recordLoc != self;
|
|
break;
|
|
}
|
|
b = bucket->advance(b, pos, 1, "BtreeBucket::dupCheck");
|
|
}
|
|
|
|
return false;
|
|
}
|
|
|
|
string BtreeBucket::dupKeyError( const IndexDetails& idx , const BSONObj& key ){
|
|
stringstream ss;
|
|
ss << "E11000 duplicate key error ";
|
|
ss << "index: " << idx.indexNamespace() << " ";
|
|
ss << "dup key: " << key;
|
|
return ss.str();
|
|
}
|
|
|
|
/**
|
|
* Find a key withing this btree bucket.
|
|
*
|
|
* When duplicate keys are allowed, we use the DiskLoc of the record as if it were part of the
|
|
* key. That assures that even when there are many duplicates (e.g., 1 million) for a key,
|
|
* our performance is still good.
|
|
*
|
|
* assertIfDup: if the key exists (ignoring the recordLoc), uassert
|
|
*
|
|
* pos: for existing keys k0...kn-1.
|
|
* returns # it goes BEFORE. so key[pos-1] < key < key[pos]
|
|
* returns n if it goes after the last existing key.
|
|
* note result might be an Unused location!
|
|
*/
|
|
char foo;
|
|
bool BtreeBucket::find(const IndexDetails& idx, const BSONObj& key, const DiskLoc &recordLoc, const Ordering &order, int& pos, bool assertIfDup) const {
|
|
#if defined(_EXPERIMENT1)
|
|
{
|
|
char *z = (char *) this;
|
|
int i = 0;
|
|
while( 1 ) {
|
|
i += 4096;
|
|
if( i >= BucketSize )
|
|
break;
|
|
foo += z[i];
|
|
}
|
|
}
|
|
#endif
|
|
|
|
globalIndexCounters.btree( (char*)this );
|
|
|
|
// binary search for this key
|
|
bool dupsChecked = false;
|
|
int l=0;
|
|
int h=n-1;
|
|
while ( l <= h ) {
|
|
int m = (l+h)/2;
|
|
KeyNode M = keyNode(m);
|
|
int x = key.woCompare(M.key, order);
|
|
if ( x == 0 ) {
|
|
if( assertIfDup ) {
|
|
if( k(m).isUnused() ) {
|
|
// ok that key is there if unused. but we need to check that there aren't other
|
|
// entries for the key then. as it is very rare that we get here, we don't put any
|
|
// coding effort in here to make this particularly fast
|
|
if( !dupsChecked ) {
|
|
dupsChecked = true;
|
|
if( idx.head.btree()->exists(idx, idx.head, key, order) ) {
|
|
if( idx.head.btree()->wouldCreateDup(idx, idx.head, key, order, recordLoc) )
|
|
uasserted( ASSERT_ID_DUPKEY , dupKeyError( idx , key ) );
|
|
else
|
|
alreadyInIndex();
|
|
}
|
|
}
|
|
}
|
|
else {
|
|
if( M.recordLoc == recordLoc )
|
|
alreadyInIndex();
|
|
uasserted( ASSERT_ID_DUPKEY , dupKeyError( idx , key ) );
|
|
}
|
|
}
|
|
|
|
// dup keys allowed. use recordLoc as if it is part of the key
|
|
DiskLoc unusedRL = M.recordLoc;
|
|
unusedRL.GETOFS() &= ~1; // so we can test equality without the used bit messing us up
|
|
x = recordLoc.compare(unusedRL);
|
|
}
|
|
if ( x < 0 ) // key < M.key
|
|
h = m-1;
|
|
else if ( x > 0 )
|
|
l = m+1;
|
|
else {
|
|
// found it.
|
|
pos = m;
|
|
return true;
|
|
}
|
|
}
|
|
// not found
|
|
pos = l;
|
|
if ( pos != n ) {
|
|
BSONObj keyatpos = keyNode(pos).key;
|
|
wassert( key.woCompare(keyatpos, order) <= 0 );
|
|
if ( pos > 0 ) {
|
|
wassert( keyNode(pos-1).key.woCompare(key, order) <= 0 );
|
|
}
|
|
}
|
|
|
|
return false;
|
|
}
|
|
|
|
void BtreeBucket::delBucket(const DiskLoc thisLoc, const IndexDetails& id) {
|
|
ClientCursor::informAboutToDeleteBucket(thisLoc); // slow...
|
|
assert( !isHead() );
|
|
|
|
const BtreeBucket *p = parent.btree();
|
|
int parentIdx = indexInParent( thisLoc );
|
|
p->childForPos( parentIdx ).writing().Null();
|
|
deallocBucket( thisLoc, id );
|
|
}
|
|
|
|
void BtreeBucket::deallocBucket(const DiskLoc thisLoc, const IndexDetails &id) {
|
|
#if 0
|
|
// as a temporary defensive measure, we zap the whole bucket, AND don't truly delete
|
|
// it (meaning it is ineligible for reuse).
|
|
memset(this, 0, Size());
|
|
#else
|
|
// defensive:
|
|
n = -1;
|
|
parent.Null();
|
|
string ns = id.indexNamespace();
|
|
theDataFileMgr._deleteRecord(nsdetails(ns.c_str()), ns.c_str(), thisLoc.rec(), thisLoc);
|
|
#endif
|
|
}
|
|
|
|
/** note: may delete the entire bucket! this invalid upon return sometimes. */
|
|
void BtreeBucket::delKeyAtPos( const DiskLoc thisLoc, IndexDetails& id, int p, const Ordering &order) {
|
|
assert(n>0);
|
|
DiskLoc left = childForPos(p);
|
|
|
|
if ( n == 1 ) {
|
|
if ( left.isNull() && nextChild.isNull() ) {
|
|
if ( isHead() ) {
|
|
_delKeyAtPos(p); // we don't delete the top bucket ever
|
|
} else {
|
|
delBucket(thisLoc, id);
|
|
}
|
|
return;
|
|
}
|
|
markUnused(p);
|
|
return;
|
|
}
|
|
|
|
if ( left.isNull() ) {
|
|
_delKeyAtPos(p);
|
|
balanceWithNeighbors( thisLoc, id, order );
|
|
} else {
|
|
markUnused(p);
|
|
}
|
|
}
|
|
|
|
void BtreeBucket::replaceWithNextChild( const DiskLoc thisLoc, IndexDetails &id ) {
|
|
assert( n == 0 && !nextChild.isNull() );
|
|
if ( parent.isNull() ) {
|
|
assert( id.head == thisLoc );
|
|
id.head.writing() = nextChild;
|
|
} else {
|
|
parent.btree()->childForPos( indexInParent( thisLoc ) ).writing() = nextChild;
|
|
}
|
|
nextChild.btree()->parent.writing() = parent;
|
|
ClientCursor::informAboutToDeleteBucket( thisLoc );
|
|
deallocBucket( thisLoc, id );
|
|
}
|
|
|
|
bool BtreeBucket::mayMergeChildren( const DiskLoc &thisLoc, int leftIndex ) const {
|
|
assert( leftIndex >= 0 && leftIndex < n );
|
|
DiskLoc leftNodeLoc = childForPos( leftIndex );
|
|
DiskLoc rightNodeLoc = childForPos( leftIndex + 1 );
|
|
if ( leftNodeLoc.isNull() || rightNodeLoc.isNull() ) {
|
|
// TODO if this situation is possible in long term implementation, maybe we should compact somehow anyway
|
|
return false;
|
|
}
|
|
int pos = 0;
|
|
{
|
|
const BtreeBucket *l = leftNodeLoc.btree();
|
|
const BtreeBucket *r = rightNodeLoc.btree();
|
|
if ( ( headerSize() + l->packedDataSize( pos ) + r->packedDataSize( pos ) + keyNode( leftIndex ).key.objsize() + sizeof(_KeyNode) > unsigned( BucketSize ) ) ) {
|
|
return false;
|
|
}
|
|
}
|
|
return true;
|
|
}
|
|
|
|
/**
|
|
* This implementation must respect the meaning and value of lowWaterMark.
|
|
* Also see comments in splitPos().
|
|
*/
|
|
int BtreeBucket::rebalancedSeparatorPos( const DiskLoc &thisLoc, int leftIndex ) const {
|
|
int split = -1;
|
|
int rightSize = 0;
|
|
const BtreeBucket *l = childForPos( leftIndex ).btree();
|
|
const BtreeBucket *r = childForPos( leftIndex + 1 ).btree();
|
|
|
|
int KNS = sizeof( _KeyNode );
|
|
int rightSizeLimit = ( l->topSize + l->n * KNS + keyNode( leftIndex ).key.objsize() + KNS + r->topSize + r->n * KNS ) / 2;
|
|
// This constraint should be ensured by only calling this function
|
|
// if we go below the low water mark.
|
|
assert( rightSizeLimit < BtreeBucket::bodySize() );
|
|
for( int i = r->n - 1; i > -1; --i ) {
|
|
rightSize += r->keyNode( i ).key.objsize() + KNS;
|
|
if ( rightSize > rightSizeLimit ) {
|
|
split = l->n + 1 + i;
|
|
break;
|
|
}
|
|
}
|
|
if ( split == -1 ) {
|
|
rightSize += keyNode( leftIndex ).key.objsize() + KNS;
|
|
if ( rightSize > rightSizeLimit ) {
|
|
split = l->n;
|
|
}
|
|
}
|
|
if ( split == -1 ) {
|
|
for( int i = l->n - 1; i > -1; --i ) {
|
|
rightSize += l->keyNode( i ).key.objsize() + KNS;
|
|
if ( rightSize > rightSizeLimit ) {
|
|
split = i;
|
|
break;
|
|
}
|
|
}
|
|
}
|
|
// safeguards - we must not create an empty bucket
|
|
if ( split < 1 ) {
|
|
split = 1;
|
|
} else if ( split > l->n + 1 + r->n - 2 ) {
|
|
split = l->n + 1 + r->n - 2;
|
|
}
|
|
|
|
return split;
|
|
}
|
|
|
|
void BtreeBucket::doMergeChildren( const DiskLoc thisLoc, int leftIndex, IndexDetails &id, const Ordering &order ) {
|
|
DiskLoc leftNodeLoc = childForPos( leftIndex );
|
|
DiskLoc rightNodeLoc = childForPos( leftIndex + 1 );
|
|
BtreeBucket *l = leftNodeLoc.btreemod();
|
|
BtreeBucket *r = rightNodeLoc.btreemod();
|
|
int pos = 0;
|
|
l->_packReadyForMod( order, pos );
|
|
r->_packReadyForMod( order, pos ); // pack r in case there are droppable keys
|
|
|
|
int oldLNum = l->n;
|
|
{
|
|
KeyNode kn = keyNode( leftIndex );
|
|
l->pushBack( kn.recordLoc, kn.key, order, l->nextChild ); // left child's right child becomes old parent key's left child
|
|
}
|
|
for( int i = 0; i < r->n; ++i ) {
|
|
KeyNode kn = r->keyNode( i );
|
|
l->pushBack( kn.recordLoc, kn.key, order, kn.prevChildBucket );
|
|
}
|
|
l->nextChild = r->nextChild;
|
|
l->fixParentPtrs( leftNodeLoc, oldLNum );
|
|
r->delBucket( rightNodeLoc, id );
|
|
childForPos( leftIndex + 1 ) = leftNodeLoc;
|
|
childForPos( leftIndex ) = DiskLoc();
|
|
_delKeyAtPos( leftIndex, true );
|
|
if ( n == 0 ) {
|
|
// will trash this and thisLoc
|
|
replaceWithNextChild( thisLoc, id );
|
|
} else {
|
|
// balance recursively - maybe we should do this even when n == 0?
|
|
balanceWithNeighbors( thisLoc, id, order );
|
|
}
|
|
}
|
|
|
|
int BtreeBucket::indexInParent( const DiskLoc &thisLoc ) const {
|
|
assert( !parent.isNull() );
|
|
const BtreeBucket *p = parent.btree();
|
|
if ( p->nextChild == thisLoc ) {
|
|
return p->n;
|
|
} else {
|
|
for( int i = 0; i < p->n; ++i ) {
|
|
if ( p->k( i ).prevChildBucket == thisLoc ) {
|
|
return i;
|
|
}
|
|
}
|
|
}
|
|
out() << "ERROR: can't find ref to child bucket.\n";
|
|
out() << "child: " << thisLoc << "\n";
|
|
dump();
|
|
out() << "Parent: " << parent << "\n";
|
|
p->dump();
|
|
assert(false);
|
|
return -1; // just to compile
|
|
}
|
|
|
|
bool BtreeBucket::tryBalanceChildren( const DiskLoc thisLoc, int leftIndex, IndexDetails &id, const Ordering &order ) const {
|
|
// If we can merge, then we must merge rather than balance to preserve
|
|
// bucket utilization constraints.
|
|
if ( mayMergeChildren( thisLoc, leftIndex ) ) {
|
|
return false;
|
|
}
|
|
thisLoc.btreemod()->doBalanceChildren( thisLoc, leftIndex, id, order );
|
|
return true;
|
|
}
|
|
|
|
void BtreeBucket::doBalanceLeftToRight( const DiskLoc thisLoc, int leftIndex, int split,
|
|
BtreeBucket *l, const DiskLoc lchild,
|
|
BtreeBucket *r, const DiskLoc rchild,
|
|
IndexDetails &id, const Ordering &order ) {
|
|
// TODO maybe do some audits the same way pushBack() does?
|
|
int rAdd = l->n - split;
|
|
r->reserveKeysFront( rAdd );
|
|
for( int i = split + 1, j = 0; i < l->n; ++i, ++j ) {
|
|
KeyNode kn = l->keyNode( i );
|
|
r->setKey( j, kn.recordLoc, kn.key, kn.prevChildBucket );
|
|
}
|
|
{
|
|
KeyNode kn = keyNode( leftIndex );
|
|
r->setKey( rAdd - 1, kn.recordLoc, kn.key, l->nextChild ); // left child's right child becomes old parent key's left child
|
|
}
|
|
r->fixParentPtrs( rchild, 0, rAdd - 1 );
|
|
{
|
|
KeyNode kn = l->keyNode( split );
|
|
l->nextChild = kn.prevChildBucket;
|
|
setInternalKey( thisLoc, leftIndex, kn.recordLoc, kn.key, order, lchild, rchild, id );
|
|
}
|
|
int zeropos = 0;
|
|
l->truncateTo( split, order, zeropos );
|
|
}
|
|
|
|
void BtreeBucket::doBalanceRightToLeft( const DiskLoc thisLoc, int leftIndex, int split,
|
|
BtreeBucket *l, const DiskLoc lchild,
|
|
BtreeBucket *r, const DiskLoc rchild,
|
|
IndexDetails &id, const Ordering &order ) {
|
|
int lN = l->n;
|
|
{
|
|
KeyNode kn = keyNode( leftIndex );
|
|
l->pushBack( kn.recordLoc, kn.key, order, l->nextChild ); // left child's right child becomes old parent key's left child
|
|
}
|
|
for( int i = 0; i < split - lN - 1; ++i ) {
|
|
KeyNode kn = r->keyNode( i );
|
|
l->pushBack( kn.recordLoc, kn.key, order, kn.prevChildBucket );
|
|
}
|
|
{
|
|
KeyNode kn = r->keyNode( split - lN - 1 );
|
|
l->nextChild = kn.prevChildBucket;
|
|
l->fixParentPtrs( lchild, lN + 1, l->n );
|
|
setInternalKey( thisLoc, leftIndex, kn.recordLoc, kn.key, order, lchild, rchild, id );
|
|
}
|
|
int zeropos = 0;
|
|
r->dropFront( split - lN, order, zeropos );
|
|
}
|
|
|
|
void BtreeBucket::doBalanceChildren( DiskLoc thisLoc, int leftIndex, IndexDetails &id, const Ordering &order ) {
|
|
DiskLoc lchild = childForPos( leftIndex );
|
|
DiskLoc rchild = childForPos( leftIndex + 1 );
|
|
int zeropos = 0;
|
|
BtreeBucket *l = lchild.btreemod();
|
|
l->_packReadyForMod( order, zeropos );
|
|
BtreeBucket *r = rchild.btreemod();
|
|
r->_packReadyForMod( order, zeropos );
|
|
int split = rebalancedSeparatorPos( thisLoc, leftIndex );
|
|
|
|
// By definition, if we are below the low water mark and cannot merge
|
|
// then we must actively balance.
|
|
assert( split != l->n );
|
|
if ( split < l->n ) {
|
|
doBalanceLeftToRight( thisLoc, leftIndex, split, l, lchild, r, rchild, id, order );
|
|
} else {
|
|
doBalanceRightToLeft( thisLoc, leftIndex, split, l, lchild, r, rchild, id, order );
|
|
}
|
|
}
|
|
|
|
void BtreeBucket::balanceWithNeighbors( const DiskLoc thisLoc, IndexDetails &id, const Ordering &order ) const {
|
|
if ( parent.isNull() ) { // we are root, there are no neighbors
|
|
return;
|
|
}
|
|
|
|
if ( packedDataSize( 0 ) >= lowWaterMark ) {
|
|
return;
|
|
}
|
|
|
|
const BtreeBucket *p = parent.btree();
|
|
int parentIdx = indexInParent( thisLoc );
|
|
|
|
// TODO will missing neighbor case be possible long term? Should we try to merge/balance somehow in that case if so?
|
|
bool mayBalanceRight = ( ( parentIdx < p->n ) && !p->childForPos( parentIdx + 1 ).isNull() );
|
|
bool mayBalanceLeft = ( ( parentIdx > 0 ) && !p->childForPos( parentIdx - 1 ).isNull() );
|
|
|
|
// Balance if possible on one side - we merge only if absolutely necessary
|
|
// to preserve btree bucket utilization constraints since that's a more
|
|
// heavy duty operation (especially if we must re-split later).
|
|
if ( mayBalanceRight &&
|
|
p->tryBalanceChildren( parent, parentIdx, id, order ) ) {
|
|
return;
|
|
}
|
|
if ( mayBalanceLeft &&
|
|
p->tryBalanceChildren( parent, parentIdx - 1, id, order ) ) {
|
|
return;
|
|
}
|
|
|
|
BtreeBucket *pm = parent.btreemod();
|
|
if ( mayBalanceRight ) {
|
|
pm->doMergeChildren( parent, parentIdx, id, order );
|
|
} else if ( mayBalanceLeft ) {
|
|
pm->doMergeChildren( parent, parentIdx - 1, id, order );
|
|
}
|
|
}
|
|
|
|
/** remove a key from the index */
|
|
bool BtreeBucket::unindex(const DiskLoc thisLoc, IndexDetails& id, const BSONObj& key, const DiskLoc recordLoc ) const {
|
|
if ( key.objsize() > KeyMax ) {
|
|
OCCASIONALLY problem() << "unindex: key too large to index, skipping " << id.indexNamespace() << /* ' ' << key.toString() << */ endl;
|
|
return false;
|
|
}
|
|
|
|
int pos;
|
|
bool found;
|
|
DiskLoc loc = locate(id, thisLoc, key, Ordering::make(id.keyPattern()), pos, found, recordLoc, 1);
|
|
if ( found ) {
|
|
loc.btreemod()->delKeyAtPos(loc, id, pos, Ordering::make(id.keyPattern()));
|
|
return true;
|
|
}
|
|
return false;
|
|
}
|
|
|
|
BtreeBucket* BtreeBucket::allocTemp() {
|
|
BtreeBucket *b = (BtreeBucket*) malloc(BucketSize);
|
|
b->init();
|
|
return b;
|
|
}
|
|
|
|
inline void BtreeBucket::fix(const DiskLoc thisLoc, const DiskLoc child) {
|
|
if ( !child.isNull() ) {
|
|
if ( insert_debug )
|
|
out() << " " << child.toString() << ".parent=" << thisLoc.toString() << endl;
|
|
child.btree()->parent.writing() = thisLoc;
|
|
}
|
|
}
|
|
|
|
/** this sucks. maybe get rid of parent ptrs. */
|
|
void BtreeBucket::fixParentPtrs(const DiskLoc thisLoc, int firstIndex, int lastIndex) const {
|
|
VERIFYTHISLOC
|
|
if ( lastIndex == -1 ) {
|
|
lastIndex = n;
|
|
}
|
|
for ( int i = firstIndex; i <= lastIndex; i++ ) {
|
|
fix(thisLoc, childForPos(i));
|
|
}
|
|
}
|
|
|
|
void BtreeBucket::setInternalKey( const DiskLoc thisLoc, int keypos,
|
|
const DiskLoc recordLoc, const BSONObj &key, const Ordering &order,
|
|
const DiskLoc lchild, const DiskLoc rchild, IndexDetails &idx ) {
|
|
childForPos( keypos ).Null();
|
|
// This may leave the bucket empty (n == 0) which is ok only as a
|
|
// transient state. In the instant case, the implementation of
|
|
// insertHere behaves correctly when n == 0 and as a side effect
|
|
// increments n.
|
|
_delKeyAtPos( keypos, true );
|
|
|
|
// just set temporarily, required to pass validation in insertHere()
|
|
childForPos( keypos ) = lchild;
|
|
|
|
insertHere( thisLoc, keypos, recordLoc, key, order, lchild, rchild, idx );
|
|
}
|
|
|
|
/**
|
|
* insert a key in this bucket, splitting if necessary.
|
|
* @keypos - where to insert the key i3n range 0..n. 0=make leftmost, n=make rightmost.
|
|
* NOTE this function may free some data, and as a result the value passed for keypos may
|
|
* be invalid after calling insertHere()
|
|
*/
|
|
void BtreeBucket::insertHere( const DiskLoc thisLoc, int keypos,
|
|
const DiskLoc recordLoc, const BSONObj& key, const Ordering& order,
|
|
const DiskLoc lchild, const DiskLoc rchild, IndexDetails& idx) const
|
|
{
|
|
if ( insert_debug )
|
|
out() << " " << thisLoc.toString() << ".insertHere " << key.toString() << '/' << recordLoc.toString() << ' '
|
|
<< lchild.toString() << ' ' << rchild.toString() << " keypos:" << keypos << endl;
|
|
|
|
DiskLoc oldLoc = thisLoc;
|
|
|
|
if ( !basicInsert(thisLoc, keypos, recordLoc, key, order) ) {
|
|
thisLoc.btreemod()->split(thisLoc, keypos, recordLoc, key, order, lchild, rchild, idx);
|
|
return;
|
|
}
|
|
|
|
{
|
|
const _KeyNode *_kn = &k(keypos);
|
|
_KeyNode *kn = (_KeyNode *) dur::alreadyDeclared((_KeyNode*) _kn); // already declared intent in basicInsert()
|
|
if ( keypos+1 == n ) { // last key
|
|
if ( nextChild != lchild ) {
|
|
out() << "ERROR nextChild != lchild" << endl;
|
|
out() << " thisLoc: " << thisLoc.toString() << ' ' << idx.indexNamespace() << endl;
|
|
out() << " keyPos: " << keypos << " n:" << n << endl;
|
|
out() << " nextChild: " << nextChild.toString() << " lchild: " << lchild.toString() << endl;
|
|
out() << " recordLoc: " << recordLoc.toString() << " rchild: " << rchild.toString() << endl;
|
|
out() << " key: " << key.toString() << endl;
|
|
dump();
|
|
assert(false);
|
|
}
|
|
kn->prevChildBucket = nextChild;
|
|
assert( kn->prevChildBucket == lchild );
|
|
nextChild.writing() = rchild;
|
|
if ( !rchild.isNull() )
|
|
rchild.btree()->parent.writing() = thisLoc;
|
|
}
|
|
else {
|
|
kn->prevChildBucket = lchild;
|
|
if ( k(keypos+1).prevChildBucket != lchild ) {
|
|
out() << "ERROR k(keypos+1).prevChildBucket != lchild" << endl;
|
|
out() << " thisLoc: " << thisLoc.toString() << ' ' << idx.indexNamespace() << endl;
|
|
out() << " keyPos: " << keypos << " n:" << n << endl;
|
|
out() << " k(keypos+1).pcb: " << k(keypos+1).prevChildBucket.toString() << " lchild: " << lchild.toString() << endl;
|
|
out() << " recordLoc: " << recordLoc.toString() << " rchild: " << rchild.toString() << endl;
|
|
out() << " key: " << key.toString() << endl;
|
|
dump();
|
|
assert(false);
|
|
}
|
|
const DiskLoc *pc = &k(keypos+1).prevChildBucket;
|
|
*dur::alreadyDeclared((DiskLoc*) pc) = rchild; // declared in basicInsert()
|
|
if ( !rchild.isNull() )
|
|
rchild.btree()->parent.writing() = thisLoc;
|
|
}
|
|
return;
|
|
}
|
|
}
|
|
|
|
void BtreeBucket::split(const DiskLoc thisLoc, int keypos, const DiskLoc recordLoc, const BSONObj& key, const Ordering& order, const DiskLoc lchild, const DiskLoc rchild, IndexDetails& idx)
|
|
{
|
|
if ( split_debug )
|
|
out() << " " << thisLoc.toString() << ".split" << endl;
|
|
|
|
int split = splitPos( keypos );
|
|
DiskLoc rLoc = addBucket(idx);
|
|
BtreeBucket *r = rLoc.btreemod();
|
|
if ( split_debug )
|
|
out() << " split:" << split << ' ' << keyNode(split).key.toString() << " n:" << n << endl;
|
|
for ( int i = split+1; i < n; i++ ) {
|
|
KeyNode kn = keyNode(i);
|
|
r->pushBack(kn.recordLoc, kn.key, order, kn.prevChildBucket);
|
|
}
|
|
r->nextChild = nextChild;
|
|
r->assertValid( order );
|
|
|
|
if ( split_debug )
|
|
out() << " new rLoc:" << rLoc.toString() << endl;
|
|
r = 0;
|
|
rLoc.btree()->fixParentPtrs(rLoc);
|
|
|
|
{
|
|
KeyNode splitkey = keyNode(split);
|
|
nextChild = splitkey.prevChildBucket; // splitkey key gets promoted, its children will be thisLoc (l) and rLoc (r)
|
|
if ( split_debug ) {
|
|
out() << " splitkey key:" << splitkey.key.toString() << endl;
|
|
}
|
|
|
|
// promote splitkey to a parent node
|
|
if ( parent.isNull() ) {
|
|
// make a new parent if we were the root
|
|
DiskLoc L = addBucket(idx);
|
|
BtreeBucket *p = L.btreemod();
|
|
p->pushBack(splitkey.recordLoc, splitkey.key, order, thisLoc);
|
|
p->nextChild = rLoc;
|
|
p->assertValid( order );
|
|
parent = idx.head.writing() = L;
|
|
if ( split_debug )
|
|
out() << " we were root, making new root:" << hex << parent.getOfs() << dec << endl;
|
|
rLoc.btree()->parent.writing() = parent;
|
|
}
|
|
else {
|
|
// set this before calling _insert - if it splits it will do fixParent() logic and change the value.
|
|
rLoc.btree()->parent.writing() = parent;
|
|
if ( split_debug )
|
|
out() << " promoting splitkey key " << splitkey.key.toString() << endl;
|
|
parent.btree()->_insert(parent, splitkey.recordLoc, splitkey.key, order, /*dupsallowed*/true, thisLoc, rLoc, idx);
|
|
}
|
|
}
|
|
|
|
int newpos = keypos;
|
|
// note this may trash splitkey.key. thus we had to promote it before finishing up here.
|
|
truncateTo(split, order, newpos); // note this may trash splitkey.key. thus we had to promote it before finishing up here.
|
|
|
|
// add our new key, there is room now
|
|
{
|
|
if ( keypos <= split ) {
|
|
if ( split_debug )
|
|
out() << " keypos<split, insertHere() the new key" << endl;
|
|
insertHere(thisLoc, newpos, recordLoc, key, order, lchild, rchild, idx);
|
|
} else {
|
|
int kp = keypos-split-1;
|
|
assert(kp>=0);
|
|
rLoc.btree()->insertHere(rLoc, kp, recordLoc, key, order, lchild, rchild, idx);
|
|
}
|
|
}
|
|
|
|
if ( split_debug )
|
|
out() << " split end " << hex << thisLoc.getOfs() << dec << endl;
|
|
}
|
|
|
|
/** start a new index off, empty */
|
|
DiskLoc BtreeBucket::addBucket(const IndexDetails& id) {
|
|
string ns = id.indexNamespace();
|
|
DiskLoc loc = theDataFileMgr.insert(ns.c_str(), 0, BucketSize, true);
|
|
BtreeBucket *b = loc.btreemod();
|
|
b->init();
|
|
return loc;
|
|
}
|
|
|
|
void BtreeBucket::renameIndexNamespace(const char *oldNs, const char *newNs) {
|
|
renameNamespace( oldNs, newNs );
|
|
}
|
|
|
|
const DiskLoc BtreeBucket::getHead(const DiskLoc& thisLoc) const {
|
|
DiskLoc p = thisLoc;
|
|
while ( !p.btree()->isHead() )
|
|
p = p.btree()->parent;
|
|
return p;
|
|
}
|
|
|
|
DiskLoc BtreeBucket::advance(const DiskLoc& thisLoc, int& keyOfs, int direction, const char *caller) const {
|
|
if ( keyOfs < 0 || keyOfs >= n ) {
|
|
out() << "ASSERT failure BtreeBucket::advance, caller: " << caller << endl;
|
|
out() << " thisLoc: " << thisLoc.toString() << endl;
|
|
out() << " keyOfs: " << keyOfs << " n:" << n << " direction: " << direction << endl;
|
|
out() << bucketSummary() << endl;
|
|
assert(false);
|
|
}
|
|
int adj = direction < 0 ? 1 : 0;
|
|
int ko = keyOfs + direction;
|
|
DiskLoc nextDown = childForPos(ko+adj);
|
|
if ( !nextDown.isNull() ) {
|
|
while ( 1 ) {
|
|
keyOfs = direction>0 ? 0 : nextDown.btree()->n - 1;
|
|
DiskLoc loc = nextDown.btree()->childForPos(keyOfs + adj);
|
|
if ( loc.isNull() )
|
|
break;
|
|
nextDown = loc;
|
|
}
|
|
return nextDown;
|
|
}
|
|
|
|
if ( ko < n && ko >= 0 ) {
|
|
keyOfs = ko;
|
|
return thisLoc;
|
|
}
|
|
|
|
// end of bucket. traverse back up.
|
|
DiskLoc childLoc = thisLoc;
|
|
DiskLoc ancestor = parent;
|
|
while ( 1 ) {
|
|
if ( ancestor.isNull() )
|
|
break;
|
|
const BtreeBucket *an = ancestor.btree();
|
|
for ( int i = 0; i < an->n; i++ ) {
|
|
if ( an->childForPos(i+adj) == childLoc ) {
|
|
keyOfs = i;
|
|
return ancestor;
|
|
}
|
|
}
|
|
assert( direction<0 || an->nextChild == childLoc );
|
|
// parent exhausted also, keep going up
|
|
childLoc = ancestor;
|
|
ancestor = an->parent;
|
|
}
|
|
|
|
return DiskLoc();
|
|
}
|
|
|
|
DiskLoc BtreeBucket::locate(const IndexDetails& idx, const DiskLoc& thisLoc, const BSONObj& key, const Ordering &order, int& pos, bool& found, const DiskLoc &recordLoc, int direction) const {
|
|
int p;
|
|
found = find(idx, key, recordLoc, order, p, /*assertIfDup*/ false);
|
|
if ( found ) {
|
|
pos = p;
|
|
return thisLoc;
|
|
}
|
|
|
|
DiskLoc child = childForPos(p);
|
|
|
|
if ( !child.isNull() ) {
|
|
DiskLoc l = child.btree()->locate(idx, child, key, order, pos, found, recordLoc, direction);
|
|
if ( !l.isNull() )
|
|
return l;
|
|
}
|
|
|
|
pos = p;
|
|
if ( direction < 0 )
|
|
return --pos == -1 ? DiskLoc() /*theend*/ : thisLoc;
|
|
else
|
|
return pos == n ? DiskLoc() /*theend*/ : thisLoc;
|
|
}
|
|
|
|
bool BtreeBucket::customFind( int l, int h, const BSONObj &keyBegin, int keyBeginLen, bool afterKey, const vector< const BSONElement * > &keyEnd, const vector< bool > &keyEndInclusive, const Ordering &order, int direction, DiskLoc &thisLoc, int &keyOfs, pair< DiskLoc, int > &bestParent ) const {
|
|
while( 1 ) {
|
|
if ( l + 1 == h ) {
|
|
keyOfs = ( direction > 0 ) ? h : l;
|
|
DiskLoc next = thisLoc.btree()->k( h ).prevChildBucket;
|
|
if ( !next.isNull() ) {
|
|
bestParent = make_pair( thisLoc, keyOfs );
|
|
thisLoc = next;
|
|
return true;
|
|
} else {
|
|
return false;
|
|
}
|
|
}
|
|
int m = l + ( h - l ) / 2;
|
|
int cmp = customBSONCmp( thisLoc.btree()->keyNode( m ).key, keyBegin, keyBeginLen, afterKey, keyEnd, keyEndInclusive, order, direction );
|
|
if ( cmp < 0 ) {
|
|
l = m;
|
|
} else if ( cmp > 0 ) {
|
|
h = m;
|
|
} else {
|
|
if ( direction < 0 ) {
|
|
l = m;
|
|
} else {
|
|
h = m;
|
|
}
|
|
}
|
|
}
|
|
}
|
|
|
|
/**
|
|
* find smallest/biggest value greater-equal/less-equal than specified
|
|
* starting thisLoc + keyOfs will be strictly less than/strictly greater than keyBegin/keyBeginLen/keyEnd
|
|
* All the direction checks below allowed me to refactor the code, but possibly separate forward and reverse implementations would be more efficient
|
|
*/
|
|
void BtreeBucket::advanceTo(DiskLoc &thisLoc, int &keyOfs, const BSONObj &keyBegin, int keyBeginLen, bool afterKey, const vector< const BSONElement * > &keyEnd, const vector< bool > &keyEndInclusive, const Ordering &order, int direction ) const {
|
|
int l,h;
|
|
bool dontGoUp;
|
|
if ( direction > 0 ) {
|
|
l = keyOfs;
|
|
h = n - 1;
|
|
dontGoUp = ( customBSONCmp( keyNode( h ).key, keyBegin, keyBeginLen, afterKey, keyEnd, keyEndInclusive, order, direction ) >= 0 );
|
|
} else {
|
|
l = 0;
|
|
h = keyOfs;
|
|
dontGoUp = ( customBSONCmp( keyNode( l ).key, keyBegin, keyBeginLen, afterKey, keyEnd, keyEndInclusive, order, direction ) <= 0 );
|
|
}
|
|
pair< DiskLoc, int > bestParent;
|
|
if ( dontGoUp ) {
|
|
// this comparison result assures h > l
|
|
if ( !customFind( l, h, keyBegin, keyBeginLen, afterKey, keyEnd, keyEndInclusive, order, direction, thisLoc, keyOfs, bestParent ) ) {
|
|
return;
|
|
}
|
|
} else {
|
|
// go up parents until rightmost/leftmost node is >=/<= target or at top
|
|
while( !thisLoc.btree()->parent.isNull() ) {
|
|
thisLoc = thisLoc.btree()->parent;
|
|
if ( direction > 0 ) {
|
|
if ( customBSONCmp( thisLoc.btree()->keyNode( thisLoc.btree()->n - 1 ).key, keyBegin, keyBeginLen, afterKey, keyEnd, keyEndInclusive, order, direction ) >= 0 ) {
|
|
break;
|
|
}
|
|
} else {
|
|
if ( customBSONCmp( thisLoc.btree()->keyNode( 0 ).key, keyBegin, keyBeginLen, afterKey, keyEnd, keyEndInclusive, order, direction ) <= 0 ) {
|
|
break;
|
|
}
|
|
}
|
|
}
|
|
}
|
|
customLocate( thisLoc, keyOfs, keyBegin, keyBeginLen, afterKey, keyEnd, keyEndInclusive, order, direction, bestParent );
|
|
}
|
|
|
|
void BtreeBucket::customLocate(DiskLoc &thisLoc, int &keyOfs, const BSONObj &keyBegin, int keyBeginLen, bool afterKey, const vector< const BSONElement * > &keyEnd, const vector< bool > &keyEndInclusive, const Ordering &order, int direction, pair< DiskLoc, int > &bestParent ) const {
|
|
if ( thisLoc.btree()->n == 0 ) {
|
|
thisLoc = DiskLoc();
|
|
return;
|
|
}
|
|
// go down until find smallest/biggest >=/<= target
|
|
while( 1 ) {
|
|
int l = 0;
|
|
int h = thisLoc.btree()->n - 1;
|
|
// leftmost/rightmost key may possibly be >=/<= search key
|
|
bool firstCheck;
|
|
if ( direction > 0 ) {
|
|
firstCheck = ( customBSONCmp( thisLoc.btree()->keyNode( 0 ).key, keyBegin, keyBeginLen, afterKey, keyEnd, keyEndInclusive, order, direction ) >= 0 );
|
|
} else {
|
|
firstCheck = ( customBSONCmp( thisLoc.btree()->keyNode( h ).key, keyBegin, keyBeginLen, afterKey, keyEnd, keyEndInclusive, order, direction ) <= 0 );
|
|
}
|
|
if ( firstCheck ) {
|
|
DiskLoc next;
|
|
if ( direction > 0 ) {
|
|
next = thisLoc.btree()->k( 0 ).prevChildBucket;
|
|
keyOfs = 0;
|
|
} else {
|
|
next = thisLoc.btree()->nextChild;
|
|
keyOfs = h;
|
|
}
|
|
if ( !next.isNull() ) {
|
|
bestParent = pair< DiskLoc, int >( thisLoc, keyOfs );
|
|
thisLoc = next;
|
|
continue;
|
|
} else {
|
|
return;
|
|
}
|
|
}
|
|
bool secondCheck;
|
|
if ( direction > 0 ) {
|
|
secondCheck = ( customBSONCmp( thisLoc.btree()->keyNode( h ).key, keyBegin, keyBeginLen, afterKey, keyEnd, keyEndInclusive, order, direction ) < 0 );
|
|
} else {
|
|
secondCheck = ( customBSONCmp( thisLoc.btree()->keyNode( 0 ).key, keyBegin, keyBeginLen, afterKey, keyEnd, keyEndInclusive, order, direction ) > 0 );
|
|
}
|
|
if ( secondCheck ) {
|
|
DiskLoc next;
|
|
if ( direction > 0 ) {
|
|
next = thisLoc.btree()->nextChild;
|
|
} else {
|
|
next = thisLoc.btree()->k( 0 ).prevChildBucket;
|
|
}
|
|
if ( next.isNull() ) {
|
|
// if bestParent is null, we've hit the end and thisLoc gets set to DiskLoc()
|
|
thisLoc = bestParent.first;
|
|
keyOfs = bestParent.second;
|
|
return;
|
|
} else {
|
|
thisLoc = next;
|
|
continue;
|
|
}
|
|
}
|
|
if ( !customFind( l, h, keyBegin, keyBeginLen, afterKey, keyEnd, keyEndInclusive, order, direction, thisLoc, keyOfs, bestParent ) ) {
|
|
return;
|
|
}
|
|
}
|
|
}
|
|
|
|
|
|
/** @thisLoc disk location of *this */
|
|
int BtreeBucket::_insert(const DiskLoc thisLoc, const DiskLoc recordLoc,
|
|
const BSONObj& key, const Ordering &order, bool dupsAllowed,
|
|
const DiskLoc lChild, const DiskLoc rChild, IndexDetails& idx) const {
|
|
if ( key.objsize() > KeyMax ) {
|
|
problem() << "ERROR: key too large len:" << key.objsize() << " max:" << KeyMax << ' ' << key.objsize() << ' ' << idx.indexNamespace() << endl;
|
|
return 2;
|
|
}
|
|
assert( key.objsize() > 0 );
|
|
|
|
int pos;
|
|
bool found = find(idx, key, recordLoc, order, pos, !dupsAllowed);
|
|
if ( insert_debug ) {
|
|
out() << " " << thisLoc.toString() << '.' << "_insert " <<
|
|
key.toString() << '/' << recordLoc.toString() <<
|
|
" l:" << lChild.toString() << " r:" << rChild.toString() << endl;
|
|
out() << " found:" << found << " pos:" << pos << " n:" << n << endl;
|
|
}
|
|
|
|
if ( found ) {
|
|
const _KeyNode& kn = k(pos);
|
|
if ( kn.isUnused() ) {
|
|
log(4) << "btree _insert: reusing unused key" << endl;
|
|
massert( 10285 , "_insert: reuse key but lchild is not null", lChild.isNull());
|
|
massert( 10286 , "_insert: reuse key but rchild is not null", rChild.isNull());
|
|
kn.writing().setUsed();
|
|
return 0;
|
|
}
|
|
|
|
DEV {
|
|
log() << "_insert(): key already exists in index (ok for background:true)\n";
|
|
log() << " " << idx.indexNamespace() << " thisLoc:" << thisLoc.toString() << '\n';
|
|
log() << " " << key.toString() << '\n';
|
|
log() << " " << "recordLoc:" << recordLoc.toString() << " pos:" << pos << endl;
|
|
log() << " old l r: " << childForPos(pos).toString() << ' ' << childForPos(pos+1).toString() << endl;
|
|
log() << " new l r: " << lChild.toString() << ' ' << rChild.toString() << endl;
|
|
}
|
|
alreadyInIndex();
|
|
}
|
|
|
|
DEBUGGING out() << "TEMP: key: " << key.toString() << endl;
|
|
DiskLoc child = childForPos(pos);
|
|
if ( insert_debug )
|
|
out() << " getChild(" << pos << "): " << child.toString() << endl;
|
|
if ( child.isNull() || !rChild.isNull() /* means an 'internal' insert */ ) {
|
|
insertHere(thisLoc, pos, recordLoc, key, order, lChild, rChild, idx);
|
|
return 0;
|
|
}
|
|
|
|
return child.btree()->bt_insert(child, recordLoc, key, order, dupsAllowed, idx, /*toplevel*/false);
|
|
}
|
|
|
|
void BtreeBucket::dump() const {
|
|
out() << "DUMP btreebucket n:" << n;
|
|
out() << " parent:" << hex << parent.getOfs() << dec;
|
|
for ( int i = 0; i < n; i++ ) {
|
|
out() << '\n';
|
|
KeyNode k = keyNode(i);
|
|
out() << '\t' << i << '\t' << k.key.toString() << "\tleft:" << hex <<
|
|
k.prevChildBucket.getOfs() << "\tRecLoc:" << k.recordLoc.toString() << dec;
|
|
if ( this->k(i).isUnused() )
|
|
out() << " UNUSED";
|
|
}
|
|
out() << " right:" << hex << nextChild.getOfs() << dec << endl;
|
|
}
|
|
|
|
/** todo: meaning of return code unclear clean up */
|
|
int BtreeBucket::bt_insert(const DiskLoc thisLoc, const DiskLoc recordLoc,
|
|
const BSONObj& key, const Ordering &order, bool dupsAllowed,
|
|
IndexDetails& idx, bool toplevel) const
|
|
{
|
|
if ( toplevel ) {
|
|
if ( key.objsize() > KeyMax ) {
|
|
problem() << "Btree::insert: key too large to index, skipping " << idx.indexNamespace() << ' ' << key.objsize() << ' ' << key.toString() << endl;
|
|
return 3;
|
|
}
|
|
}
|
|
|
|
int x = _insert(thisLoc, recordLoc, key, order, dupsAllowed, DiskLoc(), DiskLoc(), idx);
|
|
assertValid( order );
|
|
|
|
return x;
|
|
}
|
|
|
|
void BtreeBucket::shape(stringstream& ss) const {
|
|
_shape(0, ss);
|
|
}
|
|
|
|
int BtreeBucket::getLowWaterMark() {
|
|
return lowWaterMark;
|
|
}
|
|
|
|
int BtreeBucket::getKeyMax() {
|
|
return KeyMax;
|
|
}
|
|
|
|
DiskLoc BtreeBucket::findSingle( const IndexDetails& indexdetails , const DiskLoc& thisLoc, const BSONObj& key ) const {
|
|
int pos;
|
|
bool found;
|
|
// TODO: is it really ok here that the order is a default?
|
|
Ordering o = Ordering::make(BSONObj());
|
|
DiskLoc bucket = locate( indexdetails , indexdetails.head , key , o , pos , found , minDiskLoc );
|
|
if ( bucket.isNull() )
|
|
return bucket;
|
|
|
|
const BtreeBucket *b = bucket.btree();
|
|
while ( 1 ){
|
|
const _KeyNode& knraw = b->k(pos);
|
|
if ( knraw.isUsed() )
|
|
break;
|
|
bucket = b->advance( bucket , pos , 1 , "findSingle" );
|
|
if ( bucket.isNull() )
|
|
return bucket;
|
|
b = bucket.btree();
|
|
}
|
|
KeyNode kn = b->keyNode( pos );
|
|
if ( key.woCompare( kn.key ) != 0 )
|
|
return DiskLoc();
|
|
return kn.recordLoc;
|
|
}
|
|
|
|
} // namespace mongo
|
|
|
|
#include "db.h"
|
|
#include "dbhelpers.h"
|
|
|
|
namespace mongo {
|
|
|
|
void BtreeBucket::a_test(IndexDetails& id) {
|
|
BtreeBucket *b = id.head.btreemod();
|
|
|
|
// record locs for testing
|
|
DiskLoc A(1, 20);
|
|
DiskLoc B(1, 30);
|
|
DiskLoc C(1, 40);
|
|
|
|
DiskLoc rl;
|
|
BSONObj key = fromjson("{x:9}");
|
|
BSONObj orderObj = fromjson("{}");
|
|
Ordering order = Ordering::make(orderObj);
|
|
|
|
b->bt_insert(id.head, A, key, order, true, id);
|
|
A.GETOFS() += 2;
|
|
b->bt_insert(id.head, A, key, order, true, id);
|
|
A.GETOFS() += 2;
|
|
b->bt_insert(id.head, A, key, order, true, id);
|
|
A.GETOFS() += 2;
|
|
b->bt_insert(id.head, A, key, order, true, id);
|
|
A.GETOFS() += 2;
|
|
assert( b->k(0).isUsed() );
|
|
// b->k(0).setUnused();
|
|
b->k(1).setUnused();
|
|
b->k(2).setUnused();
|
|
b->k(3).setUnused();
|
|
|
|
b->dumpTree(id.head, orderObj);
|
|
|
|
/* b->bt_insert(id.head, B, key, order, false, id);
|
|
b->k(1).setUnused();
|
|
|
|
b->dumpTree(id.head, order);
|
|
|
|
b->bt_insert(id.head, A, key, order, false, id);
|
|
|
|
b->dumpTree(id.head, order);
|
|
*/
|
|
|
|
// this should assert. does it? (it might "accidentally" though, not asserting proves a problem, asserting proves nothing)
|
|
b->bt_insert(id.head, C, key, order, false, id);
|
|
|
|
// b->dumpTree(id.head, order);
|
|
}
|
|
|
|
/* --- BtreeBuilder --- */
|
|
|
|
BtreeBuilder::BtreeBuilder(bool _dupsAllowed, IndexDetails& _idx) :
|
|
dupsAllowed(_dupsAllowed),
|
|
idx(_idx),
|
|
n(0),
|
|
order( idx.keyPattern() ),
|
|
ordering( Ordering::make(idx.keyPattern()) )
|
|
{
|
|
first = cur = BtreeBucket::addBucket(idx);
|
|
b = cur.btreemod();
|
|
committed = false;
|
|
}
|
|
|
|
void BtreeBuilder::newBucket() {
|
|
DiskLoc L = BtreeBucket::addBucket(idx);
|
|
b->tempNext() = L;
|
|
cur = L;
|
|
b = cur.btreemod();
|
|
}
|
|
|
|
void BtreeBuilder::addKey(BSONObj& key, DiskLoc loc) {
|
|
if( !dupsAllowed ) {
|
|
if( n > 0 ) {
|
|
int cmp = keyLast.woCompare(key, order);
|
|
massert( 10288 , "bad key order in BtreeBuilder - server internal error", cmp <= 0 );
|
|
if( cmp == 0 ) {
|
|
//if( !dupsAllowed )
|
|
uasserted( ASSERT_ID_DUPKEY , BtreeBucket::dupKeyError( idx , keyLast ) );
|
|
}
|
|
}
|
|
keyLast = key;
|
|
}
|
|
|
|
if ( ! b->_pushBack(loc, key, ordering, DiskLoc()) ){
|
|
// no room
|
|
if ( key.objsize() > KeyMax ) {
|
|
problem() << "Btree::insert: key too large to index, skipping " << idx.indexNamespace() << ' ' << key.objsize() << ' ' << key.toString() << endl;
|
|
}
|
|
else {
|
|
// bucket was full
|
|
newBucket();
|
|
b->pushBack(loc, key, ordering, DiskLoc());
|
|
}
|
|
}
|
|
n++;
|
|
}
|
|
|
|
void BtreeBuilder::buildNextLevel(DiskLoc loc) {
|
|
int levels = 1;
|
|
while( 1 ) {
|
|
if( loc.btree()->tempNext().isNull() ) {
|
|
// only 1 bucket at this level. we are done.
|
|
dur::writingDiskLoc(idx.head) = loc;
|
|
break;
|
|
}
|
|
levels++;
|
|
|
|
DiskLoc upLoc = BtreeBucket::addBucket(idx);
|
|
DiskLoc upStart = upLoc;
|
|
BtreeBucket *up = upLoc.btreemod();
|
|
|
|
DiskLoc xloc = loc;
|
|
while( !xloc.isNull() ) {
|
|
BtreeBucket *x = xloc.btreemod();
|
|
BSONObj k;
|
|
DiskLoc r;
|
|
x->popBack(r,k);
|
|
bool keepX = ( x->n != 0 );
|
|
DiskLoc keepLoc = keepX ? xloc : x->nextChild;
|
|
|
|
if ( ! up->_pushBack(r, k, ordering, keepLoc) ){
|
|
// current bucket full
|
|
DiskLoc n = BtreeBucket::addBucket(idx);
|
|
up->tempNext() = n;
|
|
upLoc = n;
|
|
up = upLoc.btreemod();
|
|
up->pushBack(r, k, ordering, keepLoc);
|
|
}
|
|
|
|
DiskLoc nextLoc = x->tempNext(); // get next in chain at current level
|
|
if ( keepX ) {
|
|
x->parent = upLoc;
|
|
} else {
|
|
if ( !x->nextChild.isNull() )
|
|
x->nextChild.btreemod()->parent = upLoc;
|
|
x->deallocBucket( xloc, idx );
|
|
}
|
|
xloc = nextLoc;
|
|
}
|
|
|
|
loc = upStart;
|
|
}
|
|
|
|
if( levels > 1 )
|
|
log(2) << "btree levels: " << levels << endl;
|
|
}
|
|
|
|
/** when all addKeys are done, we then build the higher levels of the tree */
|
|
void BtreeBuilder::commit() {
|
|
buildNextLevel(first);
|
|
committed = true;
|
|
}
|
|
|
|
BtreeBuilder::~BtreeBuilder() {
|
|
if( !committed ) {
|
|
log(2) << "Rolling back partially built index space" << endl;
|
|
DiskLoc x = first;
|
|
while( !x.isNull() ) {
|
|
DiskLoc next = x.btree()->tempNext();
|
|
string ns = idx.indexNamespace();
|
|
theDataFileMgr._deleteRecord(nsdetails(ns.c_str()), ns.c_str(), x.rec(), x);
|
|
x = next;
|
|
}
|
|
assert( idx.head.isNull() );
|
|
log(2) << "done rollback" << endl;
|
|
}
|
|
}
|
|
|
|
}
|