Thread: What's Wrong With This?

Code:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int main()
{
	char cBoard[3][3] = {0};
	int x;
	int y;
	int z = 0;
	int iSquare;
	printf("Enter a square number <1-9>: ");
	scanf("%d", &iSquare);
	for (x=1; x <= 3; x++)
	{
		for (y=1; y <= 3; y++)
		{ 
			z++;
			if (iSquare == z)
			{
				if (cBoard[x-1][y-1] == NULL)
				{
					printf("\n\n%d is Empty!\n\n", iSquare);
				}
				else
				{	
					printf("looped %d times", z);
					printf("\n\n%d IS FULL!\n\n", iSquare);
				}
			}
		}
	}
return 0;
}

[mag1x@localhost ~]$ gcc wtf.c -o wtf
wtf.c: In function ‘main’:
wtf.c:21: warning: comparison between pointer and integer

I don't understand. Does that mean my program is wrong? Thanks.

I don't understand. Does that mean my program is wrong?

This is a warning so your program is compiled and you can execute it but except if you know precisely what you're doing and what's the reason of the warning, there are chances the execution does not produce what you expected. Try to compile with the -Werror flag so that warnings are considered as errors.

Concerning your code:

You use a char array:

Code:

char cBoard[3][3] = {0};

and you try to compare one of its element to a pointer:

Code:

cBoard[x-1][y-1] == NULL

Note: NULL is probably declared as (void*)0.
As a character is not a pointer, you don't have to compare those 2 different objects, hence the compiler warning. Replace NULL by 0.

Originally Posted by root4

This is a warning so your program is compiled and you can execute it but except if you know precisely what you're doing and what's the reason of the warning, there are chances the execution does not produce what you expected. Try to compile with the -Werror flag so that warnings are considered as errors.

Concerning your code:

You use a char array:

Code:

char cBoard[3][3] = {0};

and you try to compare one of its element to a pointer:

Code:

cBoard[x-1][y-1] == NULL

Note: NULL is probably declared as (void*)0.
As a character is not a pointer, you don't have to compare those 2 different objects, hence the compiler warning. Replace NULL by 0.

So if I was to replace:

Code:

char cBoard[3][3] = {0};

with:

Code:

char cBoard[3][3] = {NULL};

that would fix the problem? Or am I not understanding you properly?

Originally Posted by Elysia

NULL is defined as (void*)0.

No it's not, but it could be.

In a given implementations that is true but it doesn't hold over all implementations.

#define NULL (rand()? 0 : (void*)0)