有没有人用C ++封装函数的例子？ [英] Has anyone an example for wrapping a function in C++?

问题描述

我在网上搜索了很多，但是我找不到一个适用于g ++的例子，所有例子都使用gcc。
我收到的错误是：

  wrap_malloc.o：函数`__wrap_malloc（unsigned int）'：
 wrap_malloc.cc:(.text+0x20）：未定义引用`__real_malloc（unsigned int）'
 wrap_malloc.o：在函数`main'中：
 wrap_malloc.cc:(.text+0x37 ）：未定义的引用`__wrap_malloc'
 collect2：ld返回1退出状态
  

创建这个错误的代码如下（如果我使用gcc编译它，并将头文件从cstdio更改为stdio.h，则此代码有效）：

  #include< cstdio> 
 #include< cstdlib> 
 
 void * __ real_malloc（size_t）; 
 $ b无效* __ wrap_malloc（size_t c）{
 printf（我的malloc调用％d \ n，c）; 
 return __real_malloc（c）; 
} 
 
 int main（void）{
 void * ptr = malloc（12）; 
免费（ptr）; 
返回0; 
} 
  

这是我编译它的方式：

  wrap_malloc.o：wrap_malloc.cc 
 g ++ -c wrap_malloc.cc -o wrap_malloc.o 
 
 wrap_malloc：wrap_malloc.o 
 g ++ wrap_malloc.o -o wrap_malloc -Wl，wrap，malloc 
  

谢谢！

解决方案

当您使用C ++编译器时，所有名称都会被损坏。当你运行 nm wrap_malloc.o 时，这意味着什么，它应该给你这样的东西：

<$ p $ 00000000 b .bss
00000000 d .data
00000000 r .rdata
00000000 t .text
U __Z13__real_mallocj
00000000 T __Z13__wrap_mallocj
U _printf

这意味着您使用（ U ）a符号称为 __ Z13__real_mallocj 并且您在文本段（ T ）中定义了一个名为 __ Z13__wrap_mallocj 。但是您可能需要一个名为 __ real_malloc 的符号。为了达到这个目的，你必须说编译器： __ real_malloc 是一个C风格的函数，像这样：



  externCvoid * __ real_malloc（size_t）; 
 
 externCvoid * __ wrap_malloc（size_t c）{
 printf（我的malloc调用％d \ n，c）; 
 return __real_malloc（c）; 
 
  

现在输出 nm 是：



  00000000 b .bss 
 00000000 d .data 
 00000000 r .rdata 
 00000000 t .text 
 U ___real_malloc 
 00000000 T ___wrap_malloc 
 U _printf 
  

您可以看到名称 _printf 没有改变。这是因为在头文件中，许多函数已被声明为 externC已经存在。

注意：I在cygwin环境中，在Windows上执行了上述所有操作。这就是为什么在外部符号中有一个额外的前导下划线。

I have searched online a lot but I couldn't find an example that works with g++, all examples work with gcc. The error I keep getting is

wrap_malloc.o: In function `__wrap_malloc(unsigned int)':
wrap_malloc.cc:(.text+0x20): undefined reference to `__real_malloc(unsigned int)'
wrap_malloc.o: In function `main':
wrap_malloc.cc:(.text+0x37): undefined reference to `__wrap_malloc'
collect2: ld returned 1 exit status

The code that creates this error is the following (this code works if I compile it with gcc and change the headers from cstdio to stdio.h):

#include <cstdio>
#include <cstdlib>

void *__real_malloc(size_t);

void *__wrap_malloc(size_t c) {
  printf("My malloc called with %d\n", c);
  return __real_malloc(c);
}

int main(void) {
  void *ptr = malloc(12);
  free(ptr);
  return 0;
}

This is how I compile it:

wrap_malloc.o: wrap_malloc.cc
    g++ -c wrap_malloc.cc -o wrap_malloc.o

wrap_malloc: wrap_malloc.o
    g++ wrap_malloc.o -o wrap_malloc -Wl,--wrap,malloc

Thank you!

解决方案

When you use a C++ compiler, all names are mangled. What this means becomes clear when you run nm wrap_malloc.o, which should give you something like this:

00000000 b .bss
00000000 d .data
00000000 r .rdata
00000000 t .text
         U __Z13__real_mallocj
00000000 T __Z13__wrap_mallocj
         U _printf

This means that you use (U) a symbol called __Z13__real_mallocj and that you define a symbol in the text segment (T) called __Z13__wrap_mallocj. But you probably want a symbol called __real_malloc. To achieve this you have to say the compiler that __real_malloc is a C-style function, like this:

extern "C" void *__real_malloc(size_t);

extern "C" void *__wrap_malloc(size_t c) {
  printf("My malloc called with %d\n", c);
  return __real_malloc(c);
}

Now the output of nm is:

00000000 b .bss
00000000 d .data
00000000 r .rdata
00000000 t .text
         U ___real_malloc
00000000 T ___wrap_malloc
         U _printf

You can see that the name _printf hasn't changed. This is because in the header files, many functions are declared as extern "C" already.

Note: I did all of the above on Windows in the cygwin environment. That's why there is an additional leading underscore in the external symbols.

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