C 函数指针转换为 void 指针 [英] C function pointer casting to void pointer

问题描述

我正在尝试运行以下程序，但遇到一些奇怪的错误:

I am trying to run the following program but getting some strange errors:

文件 1.c:

typedef unsigned long (*FN_GET_VAL)(void);

FN_GET_VAL gfnPtr;

void setCallback(const void *fnPointer)
{
    gfnPtr = *((FN_GET_VAL*) (&fnPointer));
}

文件 2.c:

extern FN_GET_VAL gfnPtr;

unsigned long myfunc(void)
{
    return 0;
}

main()
{
   setCallback((void*)myfunc);
   gfnPtr(); /* Crashing as value was not properly 
                assigned in setCallback function */
}

这里 gfnPtr() 在使用 gcc 编译时在 64 位 suse linux 上崩溃.但它成功调用了 gfnPtr() VC6 和 SunOS.

Here the gfnPtr() is crashing on 64-Bit suse linux when compiled with gcc. But it successfully calling gfnPtr() VC6 and SunOS.

但是如果我按照下面给出的方法更改功能，它就会成功运行.

But if I change the function as given below, it is working successfully.

void setCallback(const void *fnPointer)
{
    int i; // put any statement here
    gfnPtr = *((FN_GET_VAL*) (&fnPointer));
}

请帮助找出问题的原因.谢谢.

Please help with the cause of problem. Thanks.

推荐答案

C 标准不允许将函数指针强制转换为 void*.您只能转换为另一种函数指针类型.在 C11 标准，6.3.2.3 §8 中:

The C standard does not allow to cast function pointers to void*. You may only cast to another function pointer type. In the C11 standard, 6.3.2.3 §8:

指向一种类型的函数的指针可以转换为指向 a 的指针另一种类型和返回的功能再次

A pointer to a function of one type may be converted to a pointer to a function of another type and back again

重要的是，您必须在使用指针调用函数之前转换回原始类型(从技术上讲，转换为兼容类型.6.2.7).

Importantly, you must cast back to the original type before using the pointer to call the function (technically, to a compatible type. Definition of "compatible" at 6.2.7).

请注意 POSIX 标准，许多(但不是全部)C 编译器由于使用它们的上下文也必须遵循该标准，它要求函数指针可以转换为 void*然后回来.这对于某些系统功能(例如 dlsym)是必需的.

Note that the POSIX standard, which many (but not all) C compilers have to follow too because of the context in which they are used, mandates that a function pointer can be converted to void* and back. This is necessary for some system functions (e.g. dlsym).

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