Question

If I have the following code, where I allocate my memory in the main function and then pass it to a function, which fills it for me like this:

main()
{
    char *bar = (char*) malloc(sizeof(char));

    while(1)
    {
        foo(bar);
        free(bar);
        bar = (char*) malloc(sizeof(char));
    }
}

foo(char *bar)
{
    int c;
    int i = 0;
    while((c = getchar()) != '\n' && c != EOF)
    {
        bar[i++] = c;
        bar = (char*) realloc(bar, sizeof(char) * (i+1));
    }
}

I get a segfault after a few inputs. If I do this however:

main()
{
    char *bar;

    while(1)
    {
        bar = foo();
        free(bar);
    }
}

char *foo()
{
    char *bar = (char*) malloc(sizeof(char));
    int c;
    int i = 0;
    while((c = getchar()) != '\n' && c != EOF)
    {
        bar[i++] = c;
        bar = (char*) realloc(bar, sizeof(char) * (i+1));
    }

    return bar;
}

, i.e. put the memory allocation to the function, everything seems to work fine. Why is that?

Answer 1

This is quite expected behaviour. realloc() may or may not “move” the pointer you pass into it to a new address. If that happens, it also free()’s the original address. Since C is pass-by-value and not pass-by-reference, the bar pointer in your main() function will still refer to the old address, and you’ll effectively do a double free (and the realloc()’d pointer will be lost).