直接从指针转换为模板函数? [英] Directly casting from pointer to a template function?

问题描述

我试图获取一个指向函数模板实例的指针，并将其强制转换为void *:

I am trying to take a pointer to an instance of function template and cast it to void*:

#include <stdio.h>

void plainFunction(int *param) {}

template <typename T>
void templateFunction(T *param) {}

int main() {
    void *addr1=&plainFunction; //OK
    void *addr2=&templateFunction<int>; //Compile error
}

我收到以下错误(在Visual Studio 2008中)

I get the following error (in Visual Studio 2008)

main.cu(10) : error C2440: 'initializing' : cannot convert from 'void (__cdecl *)(T *)' to 'void *'
Context does not allow for disambiguation of overloaded function

为什么会这样?函数 templateFunction (对于具体类型 T = int )没有重载.可以推断出我要引用的函数实例.

Why is it happening? Function templateFunction (for concrete type T=int) is not overloaded. It is possible to deduct which instance of the function I am refering to.

如果我将错误的行替换为:

If I replace the erroneus line with:

void (*foo)(int*)=&templateFunction<int>;
void *addr2=foo;

它编译没有问题.

谢谢！

更新:

如詹姆士(谢谢你)所建议的那样，当普通指针 void * 被伪函数指针 void(*)()代替时，它使错误消失了:

When normal pointer void* is replaced by dummy function pointer void(*)(), as suggested by James (thank you), it makes the error go away:

void (*addr1)()=(void(*)())&plainFunction;
void (*addr2)()=(void(*)())(&templateFunction<int>);

但是，如果错误是由于将函数指针转换为普通指针而引起的，则在两种情况下编译器都应进行投诉.但是，事实并非如此，因此我继续假设它至少对于此编译器是正确的.如果我没记错的话，该标准只是说函数指针不必像普通指针那样被表示，但这并不禁止.

However, if the error was caused by casting a function-pointer to a normal pointer, the compiler should complain in both cases. It does not however, so I continue to assume that it is correct at least for this compiler. If I am not mistaken, the standard just says that function pointers don't have to be represented like normal pointers, but it does not forbid that.

推荐答案

从技术上来讲都是错误的:在C ++中，您不能将函数指针转换为 void * .

Both are technically wrong: in C++, you can't convert a function pointer to a void*.

指向函数的类型(如 void(*)(int *))与指向对象的类型(如 void * 此处).

Pointer-to-function types (like void (*)(int*) here) are a completely different class of types than pointer-to-object types (like void* here).

完全允许转换的Visual C ++(例如，在 void * addr1 =& plainFunction; 中)是一种语言扩展(使用/Za 标志进行编译，会禁用语言扩展，导致两条线都被拒绝.

Visual C++ allowing the conversion at all (e.g. in void* addr1 = &plainFunction;) is a language extension (compiling with the /Za flag, which disables language extensions, causes both lines to be rejected).

可以肯定，该错误有点令人误解，尽管其他一些编译器同样无济于事(Comeau报告错误:没有函数模板"templateFunction"的实例与所需类型相匹配").

The error is a bit misleading, for sure, though some other compilers are equally unhelpful (Comeau reports "error: no instance of function template "templateFunction" matches the required type").

这篇关于直接从指针转换为模板函数?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！