传递函数的值（？）而不是函数指针？

问题描述

很抱歉，如果这之前被问过，但我找不到它。

Sorry if this has been asked before, but I was unable to find it.

因此，我试图教育自己关于模板和新的C ++ 11功能（主要是lambdas，我总是喜欢在其他语言）。

So im trying to educate myself about templates and the new C++11 features (mainly lambdas, something I always liked in other languages).

但在我的测试中，我来到一个我不知道它的工作，我试图理解它是如何工作，但不能计算出来。

But in my tests I came to something I had no idea it worked, and I'm trying to understand how it works but cant figure it out..

以下代码：

template <class Func>
void Test( Func callback ) {
    callback( 3 );
}

void Callback( int i ) {
    std::cout << i << std::endl;
}

int main( int argc, char** argv ) {
    Test( &Callback ); // this I was expecting to work, compiler will see its a pointer to a function
    Test( Callback ); // this also works, but how?!
    return 0;
}

如果我理解模板的工作原理，因为第一次调用 Test（& Callback）; 我期望工作，因为编译器将看到模板接收到一个函数地址，并将假设参数应该是一个指针。

If I understand how templates work, basically they're a scheme for the compiler to know what to build, so the first call Test( &Callback ); I was expecting to work because the compiler will see the template receives a function address and will assume the arguments should be a pointer.

但是第二次调用是什么？假设模板是什么？

But what is the second call? What is the template assuming it is? A copy of a functio (if that even makes any sense)?

推荐答案

函数可隐式转换为指向自身的指针;这种转换几乎无处不在。 Test（Callback）与 Test（& Callback）完全相同。没有什么区别。在这两种情况下， Func 被推导为 void（*）（int）。

A function is implicitly convertible to a pointer to itself; this conversion happens pretty much everywhere. Test(Callback) is exactly the same as Test(&Callback). There is no difference. In both cases, Func is deduced to be void(*)(int).

函数指针很奇怪。您可以在

Function pointers are weird. You can find out more about them in "Why do all these crazy function pointer definitions all work?"

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