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?
wrap() is used to wrap a function inside other function. It means that the first calling function (a function which is calling another function in its body) is called and then the called function is being executed. If the calling function does not call the called function then the second function will not be executed.
A wrapper function is a subroutine (another word for a function) in a software library or a computer program whose main purpose is to call a second subroutine or a system call with little or no additional computation.
In programming languages such as JavaScript, a wrapper is a function that is intended to call one or more other functions, sometimes purely for convenience, and sometimes adapting them to do a slightly different task in the process.
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()
.
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