For a function that takes a const struct, does the compiler not optimize the function body?

Question

I have the following piece of code:

#include <stdio.h>

typedef struct {
    bool some_var;
} model_t;

const model_t model = {
    true
};

void bla(const model_t *m) {
    if (m->some_var) {
        printf("Some var is true!\n");
    }
    else {
        printf("Some var is false!\n");
    }
}

int main() {
    bla(&model);
}

I'd imagine that the compiler has all the information required to eliminate the else clause in the bla() function. The only code path that calls the function comes from main, and it takes in const model_t, so it should be able to figure out that that code path is not being used. However:

With GCC 12.2 we see that the second part is linked in.

If I inline the function this goes away though:

What am I missing here? And is there some way I can make the compiler do some smarter work? This happens in both C and C++ with -O3 and -Os.

Answer 1

The compiler does eliminate the else path in the inlined function in main. You're confusing the global function that is not called anyway and will be discarded by the linker eventually.

If you use the -fwhole-program flag to let the compiler know that no other file is going to be linked, that unused segment is discarded:

[See online]

Additionally, you use static or inline keywords to achieve something similar.

Answer 2

The compiler cannot optimize the else path away as the object file might be linked against any other code. This would be different if the function would be static or you use whole program optimization.