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// Get the position of the start of the play on the board. bool GetPlayStartPosition(int% row, int% col) { // Input the row and column of the first letter. Console::Write( "Enter Location to Play as [row][col]: 00 (top left) to EE (bottom right): "); String^ locString = Console::ReadLine(); // Parse as a hex number. int x = Int32::Parse(locString, System::Globalization::NumberStyles::HexNumber); row = x / 16; col = x % 16; if (row > 14 || col > 14 || row < 0 || col < 0) { Console::WriteLine("I did not understand that input."); Console::WriteLine("The first digit is the row (0 to E);" " the second is the column (0 to E)."); throw gcnew Exception(); } // Check to see that this is an unoccupied space. if (gameBoard[row, col] != nullptr) { Console::WriteLine("Sorry, that space is occupied by the tile: {0}", gameBoard[row, col]); return false; } return true; } // Return true if the play is successful. // Return false if the play is invalid and needs to be restarted. bool GetTilesForPlay(int row, int col, PlayType playType, List<Tile^>^ workingTiles, array<Tile^, 2>^ workingBoard ) { // Get the desired tiles to play from the user. Console::WriteLine( "Enter letters to play (_<letter> to play a blank as <letter>): "); int code;

You will notice the two resp figures these show the cost of using parallel execution, but since the trace file came from a query where the degree of parallelism was one, this matches the serial cost.

You may decide you don t want to use Documents to Go. In that case, just select the View option when you go to open the attachment. The document won t have the same look, but you will be able to navigate through it quickly.

Looking back a few pages to the original example (prefetch_test_02.sql in the online code suite), you may start to question whether this formula is actually correct. Here s the (traditional) execution plan again: Execution Plan (9.2.0.6 autotrace) ------------------------------------------------------0 SELECT STATEMENT Optimizer=ALL_ROWS (Cost=324 Card=320 Bytes=11840) 1 0 NESTED LOOPS (Cost=324 Card=320 Bytes=11840) 2 1 TABLE ACCESS (FULL) OF 'DRIVER' (Cost=3 Card=320 Bytes=2560) 3 1 TABLE ACCESS (BY INDEX ROWID) OF 'TARGET' (Cost=2 Card=1 Bytes=29) 4 3 INDEX (UNIQUE SCAN) OF 'T_PK' (UNIQUE) (Cost=1 Card=1)

Using Sheet to Go or Slideshow to Go (MS Excel or PowerPoint)

do { code = Console::Read(); wchar_t character = safe_cast<wchar_t>(code); Letter letter = Letter::_; if (character == safe_cast<wchar_t>(Characters::NEWLINE)) { Console::ReadLine(); break; } if (character == '_') { letter = Letter::_; // If a blank is entered, read the next character. code = Console::Read(); character = safe_cast<wchar_t>(code); } else if (Char::IsLetter(character)) { if (Char::IsUpper(character)) { letter = safe_cast<Letter>(character - 'A' + 1); } else // character is a lowercase letter. { letter = safe_cast<Letter>(character - 'a' + 1); } } // See if the letter is in the player's hand. Tile^ tile = gcnew Tile(letter); if (letter == Letter::_) { tile->BlankValue = character; } Tile^ tileToRemove = nullptr; bool tileFound = false; for each (Tile^ t in workingTiles) { if (t->LetterValue == tile->LetterValue) { tileToRemove = t; tileFound = true; } }

Line 2 tells us that the cost of scanning the driver table was 3, with 320 rows returned Line 3 tells us that the cost of one trip to the target table was 2 Looking at line 4, we can infer that that cost was made up of 1 for hitting the index, plus 1 for hitting the table itself Applying the formula, then, we should see Total cost = 3 + (320 * 2) = 643 But the reported cost is 324 so what went wrong The answer lies in a combination of rounding errors and the way that intermediate results are reported I had CPU costing enabled for the example, with the following settings: begin dbms_statsset_system_stats('MBRC',8); dbms_statsset_system_stats('MREADTIM',20); dbms_statsset_system_stats('SREADTIM',10); dbms_stats.

Follow the same steps you did earlier when you opened the word processing document. Sheet to Go View Mode:

set_system_stats('CPUSPEED',500); end; / This means, in particular, that the CPU runs at 500 MHz (or 500 million Oracle operations per second) and that a single block read takes 10 milliseconds Put another way, 5,000,000 operations equate to one single block read If you look at the output from a full report on the plan_table (which will include the I/O costs and CPU costs as separate items), you will find the following figures (I have excluded many details, including the cardinality from this output): Execution Plan (920.

Opened with Sheet to Go (you can view formulas and edit).