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

问题

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 "__wrap_symbol".  Any undefined
           reference to "__real_symbol" will be resolved to symbol.
           ...
           If you link other code with this file using --wrap malloc, then all
           calls to "malloc" will call the function "__wrap_malloc" instead.
           The call to "__real_malloc" in "__wrap_malloc" will call the real
           "malloc" function.

So I created a simple example:

#include <stdio.h>

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

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

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

And compiled it with:

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

This gave me a warning:

main.c:18:21: warning: implicit declaration of function ‘__real_foo’ [-Wimplicit-function-declaration]
  printf("realfoo:");__real_foo();
                     ^~~~~~~~~~

Well going on. Now I would suggest an output like this:

foo:wrap foo
wrapfoo:wrap foo
realfoo:foo

Instead I get this:

foo:foo
wrapfoo:wrap foo
realfoo:foo

I hope the thing is clear. I am confused about the warning. Normally the __real function should be linked by the linker to foo(). Furthermore a call to foo() should be linked to __wrap_foo. But the output showes, that foo() is being executed instead.

How to use --wrap correctly?

回答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().

来源：https://stackoverflow.com/questions/46444052/how-to-wrap-functions-with-the-wrap-option-correctly

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