conradk conradk - 1 month ago 11
C Question

warning: comparison of unsigned expression >= 0 is always true

I have the following error when compiling a C file:

t_memmove.c: In function ‘ft_memmove’:
ft_memmove.c:19: warning: comparison of unsigned expression >= 0 is always true


Here's the full code, via
cat ft_memmove.c
:

#include "libft.h"
#include <string.h>

void *ft_memmove(void *s1, const void *s2, size_t n)
{
char *s1c;
char *s2c;
size_t i;

if (!s1 || !s2 || !n)
{
return s1;
}
i = 0;
s1c = (char *) s1;
s2c = (char *) s2;
if (s1c > s2c)
{
while (n - i >= 0) // this triggers the error
{
s1c[n - i] = s2c[n - i];
++i;
}
}
else
{
while (i < n)
{
s1c[i] = s2c[i];
++i;
}
}
return s1;
}


I do understand that size_t is unsigned and that both integers will be >= 0 because of that. But since I'm subtracting one from the other, I don't get it. Why does this error come up?

Answer

If you subtract two unsigned integers in C, the result will be interpreted as unsigned. It doesn't automatically treat it as signed just because you subtracted. One way to fix that is use n >= i instead of n - i >= 0.

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