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When all the target rowids have been found, the engine can sort them and then visit the inner table in a single pass, working along the length of the table just once, picking the rows in whatever order they happen to appear in this case (a, b, b, a, a, c, b, c, a, c) This may take fewer logical I/Os the diagram suggests that the first two rows in the inner table may be in the same block, which means they could be acquired with one consistent get, rather than two The mechanism may even result in fewer physical I/Os that one block with the first two rows may have been flushed from the buffer between the two visits that would be needed using the traditional mechanism.

On the down side, this strategy may mean that an order by clause could need a sort that would not otherwise be necessary Clearly, the choice of plan needs to be cost-based; however, it may also be affected by a cache-monitoring routine carried out by the CKPT (checkpoint) process as event 10299 is described by the comment Trace prefetch tracking decisions made by CKPT Looking at either picture, it is quite easy to see that a simple approximation to the cost of performing a nested loop could be described by this rationale: What is the cost of getting all the required rows from the first table How many rows will appear in the first table What is the typical cost of finding related rows once in the second table, given the newly available information I have from the current row in the first table.

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// Display the current scores and tiles in the bag or // in each player's rack. void PrintScores() { Console::Write("Current stats: "); if (bag->Count != 0) { Console::WriteLine("{0} tiles remaining in tile bag.", bag->Count); } else { Console::WriteLine("No tiles remaining in tile bag."); } for (int i = 0; i < nPlayer; i++) { Console::WriteLine("{0,-10} -- Score: {1,3} Number of tiles: {2} -- ", players[i]->Name, players[i]->Score, players[i]->TileCount); } } // Display the permanent gameboard (overload). void PrintBoard() { PrintBoard(gameBoard); } // Display the gameboard. This overload takes a board // as an argument, so it is possible to display the proposed // play before committing it to the permanent gameboard. void PrintBoard(array<Tile^, 2>^ board) { Console::WriteLine(); Console::Write(" "); for (int i = 0; i < BOARD_SIZE; i++) Console::Write(" {0:X1} ", i); Console::WriteLine(); for (int i = 0; i < BOARD_SIZE; i++) { Console::Write(" {0:X1} ", i); for (int j = 0; j < BOARD_SIZE; j++) {