Why is $? always 0 after system() is called?

Question

I'm testing this tiny program under Linux:

// foo.c
#include <stdio.h>
#include <stdlib.h>

int main(int argc, char* argv[])
{
    int n = system(argv[1]);
    printf("%d\n", n);
    return n;
}

No matter what is fed into the command-line, an echo $? always prints 0, e.g.:

$ ./foo anything
sh: anything: not found
32512
$ echo $?
0

My question is: Why doesn't $? take the same value as n? I've also tested the program under Win32, and echo %errorlevel% gives the same value as n. Thanks!

Answer 1

If you print n in octal or hex, you'll discover that the low byte of it is always 0.

If you return WEXITSTATUS(n);, your program will exit with the status you are expecting.

Read man system and man wait carefully, and you'll understand.

Answer 2

Only lower 8 bits of the return value are recognized as the exit status, because the exit status is calculated by WEXITSTATUS macro, see SUSv4