Thread: tic-tac-toe diagonal

Here is the code I have. Hopefully it's self explanatory:

Code:

// Game functions
#include "game.h"

bool check::win(char icon, char board[3][3])
{
	if(check::horizontal(icon, board))
		return true;
	else if(check::vertical(icon, board))
		return true;
	else if(check::diagonal(icon, board))
		return true;

	return false;
}

bool check::horizontal(char icon, char board[3][3])
{
	int counter = 0;

	for(int i = 0; i < 3; i ++)
	{
		for(int t = 0; t < 3; t ++)
		{
			if(board[i][t] == icon)
				counter ++;
		}

		if(counter == 3)
			return true;

		counter = 0;
	}

	return false;
}

bool check::vertical(char icon, char board[3][3])
{
	int counter = 0;

	for(int i = 0; i < 3; i ++)
	{
		for(int t = 0; t < 7; t += 3)
		{
			if(board[i][t] == icon)
				counter ++;
		}

		if(counter == 3)
			return true;

		counter = 0;
	}

	return false;
}

bool check::diagonal(char icon, char board[3][3])
{
	int counter = 0;

	return false;
}

I'm stumped on how I could use for loops to check for a diagonal win. Any suggestions?

Ahhh, that'll work. Thanks!

Code:

bool check::diagonal(char icon, char board[3][3])
{
        if[board[2][2] == icon)
        {
            if(board[1][1] == icon && board[3][1] == icon  || board[1][3] = icon && board[3][1]        )
            {
             return true; 
            }
            else 
            {
             return false;
            }
       }

          return false;
}

i don't have an other idea...
the other ideas don't work correctly because
it doesn't do when you have
[ ] [ ] [x]
[ ] [x ] [ ]
[x ] [ ] [ ]

Thanks for the help, guys. I ended up doing this:

Code:

bool check::diagonal(char icon, char board[3][3])
{
	int counter = 0;

	for(int i = 0; i < 3; i ++)
	{
		if(board[i][i] == icon)
			counter ++;
	}

	if(counter == 3)
		return true;

	counter = 0;

	int t;

	for(i = 0, t = 2; i > 3, t >= 0; i ++, t --)
	{
		if(board[t][i] == icon)
			counter ++; 
	}

	if(counter == 3)
		return true;

	return false;
}