/* Since the last noteLocation(), our key may have moved around, and that old cached
information may thus be stale and wrong (although often it is right). We check
that here; if we have moved, we have to search back for where we were at.
i.e., after operations on the index, the BtreeCursor's cached location info may
be invalid. This function ensures validity, so you should call it before using
the cursor if other writers have used the database since the last noteLocation
call.
*/
void BtreeCursor::checkLocation() {
if ( eof() )
return;
if ( keyOfs >= 0 ) {
BtreeBucket *b = bucket.btree();
assert( !keyAtKeyOfs.isEmpty() );
// Note keyAt() returns an empty BSONObj if keyOfs is now out of range,
// which is possible as keys may have been deleted.
if ( b->keyAt(keyOfs).woEqual(keyAtKeyOfs) &&
b->k(keyOfs).recordLoc == locAtKeyOfs ) {
if ( !b->k(keyOfs).isUsed() ) {
/* we were deleted but still exist as an unused
marker key. advance.
*/
skipUnusedKeys();
}
return;
}
}
/* normally we don't get to here. when we do, old position is no longer
valid and we must refind where we left off (which is expensive)
*/
bool found;
/* TODO: Switch to keep indexdetails and do idx.head! */
bucket = indexDetails.head.btree()->locate(indexDetails, indexDetails.head, keyAtKeyOfs, order, keyOfs, found, locAtKeyOfs, direction);
RARELY log() << " key seems to have moved in the index, refinding. found:" << found << endl;
if ( found )
skipUnusedKeys();
}
/* Since the last noteLocation(), our key may have moved around, and that old cached
information may thus be stale and wrong (although often it is right). We check
that here; if we have moved, we have to search back for where we were at.
i.e., after operations on the index, the BtreeCursor's cached location info may
be invalid. This function ensures validity, so you should call it before using
the cursor if other writers have used the database since the last noteLocation
call.
*/
void BtreeCursor::checkLocation() {
if ( eof() )
return;
multikey = d->isMultikey(idxNo);
if ( keyOfs >= 0 ) {
BtreeBucket *b = bucket.btree();
assert( !keyAtKeyOfs.isEmpty() );
// Note keyAt() returns an empty BSONObj if keyOfs is now out of range,
// which is possible as keys may have been deleted.
int x = 0;
while( 1 ) {
if ( b->keyAt(keyOfs).woEqual(keyAtKeyOfs) &&
b->k(keyOfs).recordLoc == locAtKeyOfs ) {
if ( !b->k(keyOfs).isUsed() ) {
/* we were deleted but still exist as an unused
marker key. advance.
*/
skipUnusedKeys( false );
}
return;
}
/* we check one key earlier too, in case a key was just deleted. this is
important so that multi updates are reasonably fast.
*/
if( keyOfs == 0 || x++ )
break;
keyOfs--;
}
}
/* normally we don't get to here. when we do, old position is no longer
valid and we must refind where we left off (which is expensive)
*/
bool found;
/* TODO: Switch to keep indexdetails and do idx.head! */
bucket = indexDetails.head.btree()->locate(indexDetails, indexDetails.head, keyAtKeyOfs, _ordering, keyOfs, found, locAtKeyOfs, direction);
RARELY log() << " key seems to have moved in the index, refinding. found:" << found << endl;
if ( ! bucket.isNull() )
skipUnusedKeys( false );
}