How to wrap functions with the `--wrap` option correctly?

Question

The man page of gcc 6.3 says:

--wrap=symbol
           Use a wrapper function for symbol.  Any undefined reference to
           symbol will be resolved to \         


        

Answer 1

As StoryTeller told me, I ignored the "undefined reference" requirement which I already posted above:

... Any undefined reference to symbol will be resolved to "__wrap_symbol". Any undefined reference to "__real_symbol" will be resolved to symbol.

To use the --wrap option I rearranged my code example like this:

main.c:

#include <stdio.h>
extern int foo();
extern int __real_foo();

int __wrap_foo() {
    printf("wrap foo\n");
    return 0;
}

int main () {
    printf("foo:");foo();
    printf("wrapfoo:");__wrap_foo();
    printf("realfoo:");__real_foo();
    return 0;
}

foo.c:

#include <stdio.h>
int foo() {
    printf("foo\n");
    return 0;
}

Then compile:

gcc main.c foo.c -Wl,--wrap=foo -o main

And the the amazing output after running ./main:

foo:wrap foo
wrapfoo:wrap foo
realfoo:foo

The trick is (correct me if I am wrong) that the reference of foo() and __real_foo() is not defined at compile time. I. E. they have **undefined references" which is the requierement for the linker to link foo() to __wrap_foo() and __real_foo() to foo().