strtoull and long long arithmetic

Question

Can anyone explain the output of this program and how I can fix it?

unsigned long long ns = strtoull(\"123110724001300\", (char **)NULL, 10);
fprintf(stderr,         


        

Answer 1

Do you have <stdlib.h> included?

I can reproduce on MacOS X if I omit <stdlib.h>.

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

int main(void)
{
    unsigned long long ns = strtoll("123110724001300", (char **)NULL, 10);
    printf("%llu\n", ns);
    return(0);
}

Omit the header, I get your result. Include the header, I get the correct answer.

Both 32-bit and 64-bit compiles.

---

As noted in the comments, in the absence of a declaration for strtoll(), the compiler treats it as a function returning int.

To see more of what goes on, look at the hex outputs:

     123110724001300    0x00006FF7_F2F8DE14    Correct
18446744073490980372    0xFFFFFFFF_F2F8DE14    Incorrect

Manually inserted underscores...

Answer 2

I cannot explain the behavior. However, on 32 bit Windows XP with Cygwin gcc-4.3.2:

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

int main(void) {
    unsigned long long ns = strtoull("123110724001300", NULL, 10);
    fprintf(stderr, "%llu\n", ns);
    return 0;
}

prints

E:\Home> t.exe
123110724001300

Answer 3

Why not use strtoull if you want an unsigned long long?