如何用--wrap选项正确包装函数? [英] How to wrap functions with the `--wrap` option correctly?

问题描述

gcc 6.3的手册页说:

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.

因此，我创建了一个简单的示例:

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;
}

并使用以下代码进行编译:

And compiled it with:

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

这给了我一个警告:

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

好吧.现在，我建议这样的输出:

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

foo:wrap foo
wrapfoo:wrap foo
realfoo:foo

相反，我得到了:

foo:foo
wrapfoo:wrap foo
realfoo:foo

我希望事情已经清楚了.我对警告感到困惑.通常，链接器应将__real函数链接到foo().此外，对foo()的调用应链接到__wrap_foo.但是输出显示，正在执行foo()代替.

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.

如何正确使用--wrap?

推荐答案

正如StoryTeller告诉我的那样，我忽略了上面已经发布的未定义引用"要求:

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

...对符号的任何未定义引用都将解析为"__wrap_symbol".任何对"__real_symbol"的未定义引用都将解析为符号.

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

要使用--wrap选项，我重新排列了代码示例，如下所示:

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;
}

然后编译:

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

运行./main之后的惊人输出:

foo:wrap foo
wrapfoo:wrap foo
realfoo:foo

技巧是(如果我错了，请纠正我)是在编译时未定义foo()和__real_foo()的引用. I. E.他们有** undefined reference，这是链接器将foo()链接到__wrap_foo()和__real_foo()链接到foo()的要求.

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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