使用指针打印功能地址 [英] Printing address of function by using pointer

问题描述

我正在尝试使用下面代码中所示的 printf 语句两次打印函数指针地址...

I am trying to print a function pointer address two times using a printf statement shown in below code...

class B
{
public:
        int fun()
        {
        }
};

int main()
{
        int (B::*pb)()=&B::fun;
        printf("ptr:%x | %x\n",pb,pb);   //Output is ptr:8048730 | 0
}

当我将同一变量传递给 printf 它应该打印相同的值，但是得到结果后，我感到很惊讶。

when i am passing same variable to printf it should print the same value but after getting the result i am surprised.

任何人都可以解释这个原因或我做错了什么地方。

Can anyone explain the reason for this or somewhere i am doing wrong.

gcc版本4.8.2（GCC）

推荐答案

如果打开警告（将 -Wall 标志传递给编译器），编译器将告诉您您在做什么。错误：

If you turn on the warnings (pass the -Wall flag to the compiler), the compiler will tell you what you are doing wrong:

warning: format ‘%x’ expects argument of type ‘unsigned int’, but argument 2 has type ‘int (B::*)()’ [-Wformat]
warning: format ‘%x’ expects argument of type ‘unsigned int’, but argument 3 has type ‘int (B::*)()’ [-Wformat]

通常，我只能鼓励您打开警告。

In general, I can only encourage you to turn on warnings.

现在有以下情况（未定义的行为）：

Now a hack (undefined behavior) is the following:

std :: printf（ ptr：％ p |％p\n，（void *）funcptr，（void *）funcptr）;

这仍然给出以下警告：

警告：从'int转换（B :: *）（）'到'void *'[-Wpmf-conversions]

，但是程序会打印两次相同的地址如您所愿。仍然是未定义的行为，请参见如何格式化函数指针？通过遵循公认的答案（从那里被无耻地偷走的代码） ，则可以按以下方式打印地址：

but the program prints the same address twice as you wished. It is still undefined behavior, see How to format a function pointer? By following the accepted answer (code shamelessly stolen from there), one could print the address as follows:

#include <cstdio>

class B {
public:
        int fun() { return 0; }
};

int main()
{
    int (B::*funcptr)()=&B::fun;

    unsigned char *p = (unsigned char *)&funcptr;

    for (unsigned int i = 0; i < sizeof funcptr; ++i)
    {
        std::printf("%02x ", p[i]);
    }

    std::putchar('\n');

    return 0;
}

根据该答案，这是实现此目的的唯一合法方法。它不会发出任何警告。

According to that answer it is the only legal way to achieve this. It doesn't give any warning.

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