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.

> if (cBoard[x-1][y-1] == NULL)
I'm guessing NULL should be 0 or '\0'

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?

No, the code should be something along the lines of:

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] == 0)
				{
					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;
}

NULL is defined as (void*)0. You cannot compare a pointer to a non-pointer, and you can't compare a pointer of one type to another type.
(Or you can, but you shouldn't.)

Originally Posted by Elysia

NULL is defined as (void*)0.

No it's not, but it could be.

If it could be defined as (void*)0, then it cannot not defined as (void*)0
It can be one, but not both!
But in any case, it seems to be defined as (void*)0 in this case.

Whatever the case, NULL == NULL

And that's all that matters.

I think we can agree to that

If it could be defined as (void*)0, then it cannot not defined as (void*)0

That NULL could be defined as (void*)0 does not imply that it is defined as (void*)0. C99 states that the macro NULL "expands to an implementation-defined null pointer constant" and that "an integer constant expression with the value 0, or such an expression cast to type void *, is called a null pointer constant".

It was a figure of speech. If something is X (not Y), then it cannot be of Y at the same time.
Or, in other words, if NULL is not defined to (void*)0 (X), then it cannot be defined to (void*)0 (Y).
It can be one of those possibilities, but not both!

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

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