Why does GCC not warn for unreachable code?

Question

I wonder why gcc (4.6.3) gives me no warning for the unreachable code in this example:

#include <stdio.h>

int status(void)
{
    static int first_time = 1;

    if (first_time) {
        return 1;   
        first_time = 0; /* never reached */
    } else {
        return 0;   
    }     
}

int main(int argc, const char *argv[])
{
    printf("first call %d\n", status());
    printf("second call %d\n", status());
    return 0;
}

Note, the purpose of the faulty status() function was to maintain a status. I had expected to get a warning for this with -Wall. I tried also -Wunreachable-code, -Wextra, -pedantic and -ansi (as it was discussed here). Yet, none of those give me a warning.

It appears gcc silently removes the static variable assignment.

In my opinion gcc options -Wall -Werror should throw an error.

Answer 1

gcc 4.4 will give you warning. In the later versions of gcc this feature (-Wunreachable-code) has been removed.

See here: http://gcc.gnu.org/ml/gcc-help/2011-05/msg00360.html

The -Wunreachable-code has been removed, because it was unstable: it relied on the optimizer, and so different versions of gcc would warn about different code. The compiler still accepts and ignores the command line option so that existing Makefiles are not broken. In some future release the option will be removed entirely.

Ian