Question

I’m a C beginner, today I encountered a problem that puzzled me for hours and I subtracted the clauses where the problem occurs.

I compiled it with Archlinux(gcc).

#include <stdio.h>

#define SIZE 10

int main()
{
    char s[SIZE]; 
    int i;
    for (i = 0; i < SIZE; )
        s[i++] = 'm';
    s[i++] = '\n';
    s[i] = '\0';
    printf("%s/D\n", s, i);

    return 0;
}

It worked without errors.

The output is mmmmmmmmmm 11.

With one line removed. s[i++] = '\n';

#include <stdio.h>

#define SIZE 10

int main()
{
    char s[SIZE]; 
    int i;
    for (i = 0; i < SIZE; )
        s[i++] = 'm';
    s[i] = '\0';
    printf("%s %d\n", s, i);

    return 0;
}

The “i” became 0.
The output: mmmmmmmmmm 0

but once compiled with Cent OS(gcc).

The “i” didn’t become 0.

back to Archlinux. I entered another line. int a = i", to reference i;

#include <stdio.h>

#define SIZE 10

int main()
{
    char s[SIZE]; 
    int i;
    for (i = 0; i < SIZE; )
        s[i++] = 'm';
    s[i] = '\0';
    int a = i;
    printf("%s/D\n", s, i);

    return 0;
}

And this time “i” didn’t become 0.

I’m a newbie, someone please tell me what was happening?

If this is just some stupid mistake I made, please let me know and I’ll delete the post.

Thanks!

Answer 1

C arrays are zero-based so valid indices in your example are [0..SIZE-1]. At the end of your loop, i==SIZE. You then write to s[SIZE] which is one element beyond the end of your array. This has undefined consequences.

In your test, &s[SIZE] == &i so you write to i in both cases. In the first case, the ascii value of '\n' happens to be what you expected for i so you don’t notice the bug. In the second case you get luckier, reset i to 0 and spot the array overflow.

The fix is to exit your loop one iteration sooner, leaving space for the null terminator in your char array

for (i = 0; i < SIZE-1; )
//                  ^^