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basic C++ question:

So I have this 2 dimensional Array (3X3)

and I want to make it so

"if there is an "X" in slot 2-2, return a "YES"

tried doing

if ( board[2][2] hahahaha 'X' )
{
cout << "Yes";
}
else
{
cout << "No";
}

and I keep on getting "No"s at every turn

can anybody help me out on this one?

what I am actually trying to do, is that it's a TIC TAC TOE game, and I need to stop the program once someone wins...

help me out yooo!!!

178340550

ah shit.. haven't done any C++ in over 10years

damn.. now I wanna know too
and I know it's a fuckin simple thing :disgust

post the code somewhere where it wont go hahahaha

Quote:

Originally Posted by skinnay

post the code somewhere where it wont go hahahaha

its just 2 "=" next to eachother

Quote:

Originally Posted by bdjuf

Because [2][2] is actually referencing the 'bottom right' element of a 3x3 array. Arrays in C/C++ are referenced from 0. If you're trying to check the MIDDLE square, you're checking the wrong spot... and if you reference [3][3], you're overflowing and begging for a h4x0r 2 r0x0r j00r b0x0rz. :winkwink:

Try this:

Code:

  if(board[1][1]hahahaha'X') {       // Is there an X in the middle square?

    cout << "Yes";

  } else {

    cout << "No";

  }

BTW - The best way to make a TTT game is to have the program 'learn'. I did the same thing back in 80s on an atari 400... it's possible to make a damn-near unbeatable computerized opponent.

A 'win' state checker:

Code:

char *checkwin(char board[2][2])  // Prototype might be a bit off, adjust accordingly ;)

{

  // Check horizontal

  for(int y=0;y<3;y++) {

    int xs, os;

    for(int x=0;x<3;x++) {

      xs += (board[x][y] = = 'X');

      os += (board[x][y] = = 'O');

    }

  }

  if(xs= =3) return("X Wins");

  if(os= =3) return("O Wins");



  // Check vertical

  for(int x=0;x<3;x++) {

    int xs, os;

    for(int y=0;y<3;y++) {

      xs += (board[x][y] = = 'X');

      os += (board[x][y] = = 'O');

    }

  }

  if(xs= =3) return("X Wins");

  if(os= =3) return("O Wins");



  // Check left-right diag

  for(int a=0,xs=0,os=0;a<3;a++) {

    xs += (board[a][a] = = 'X');

    os += (board[a][a] = = 'O');

  }

  if(xs= =3) return("X Wins");

  if(os= =3) return("O Wins");



  // Check right-left diag

  for(int a=0,xs=0,os=0;a<3;a++) {

    xs += (board[a][2-a] = = 'X');

    os += (board[a][2-a] = = 'O');

  }

  if(xs= =3) return("X Wins");

  if(os= =3) return("O Wins");



// No winnar.

  return;

}

There is almost certainly better ways of doing this (making a string of 3 characters in the array and matching that against a string of 'XXX' or 'OOO', for instance). This was just a 60 second slap together that'll do the job and demonstrate the logic cleanly. :2 cents:

Quote:

Originally Posted by bdjuf

Doing a assignment for cs class?

use the min max algo, i coded a tic tac toe game for school a couple years ago using this algo, and it was undefeatable..

http://personal.vsnl.com/erwin/tictactoe.htm

array element numbering starts with 0 not 1

Nerds :error

just kidding :pimp

Nerds :error

:1orglaugh

got it! thanx guys

now another question, how do I reset an array? make it all empty again?

Damn! I used to know C++, now I don't remember shit!

Quote:

Originally Posted by bdjuf

got it! thanx guys

now another question, how do I reset an array? make it all empty again?

Assuming it's a contiguous area of memory and a simple array:

Code:

memset((void *)array_ptr, number_of_elements * sizeof(array_type), 0);

Alternately, you can request it be zeroed while allocating:

Code:

array_ptr = (array_type *)calloc(number_of_elements, sizeof(array_type));

You're doing c++, so you can always just destroy the old object and create a new one, which should set all values to their prototypes as well... if not, you can do the memset() thing in the object constructor.

HTH. HAND. :thumbsup

PS - If you malloc()'d the array originally, make sure you free() it before doing a new malloc(), otherwise you'll leak ram. :2 cents:

everything works!!! thanx alot for your help guys