C函数指针铸造空指针 [英] C function pointer casting to void pointer
问题描述
我试图运行下面的程序,但得到一些奇怪的错误:
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 */
}
下面的gfnPtr()是当用gcc编译64位的SUSE Linux崩溃。但它成功地调用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,如果我改变,因为下面给出的函数,它是成功的工作。
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));
}
请与问题的原因有所帮助。谢谢你。
Please help with the cause of problem. Thanks.
推荐答案
标准不允许投函数指针无效*
。您只可以转换为另一种函数指针类型。 6.3.2.3§8:
The standard does not allow to cast function pointers to void*
. You may only cast to another function pointer type. 6.3.2.3 §8:
一个指向包含一个类型的函数
可被转化为一个指针
另一种类型和背部的功能
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,必须使用指针调用函数之前转换回原始类型(从技术上说,到兼容类型定义的兼容的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).
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