传递给模板函数时，lambda自动衰减为函数指针 [英] automatic decay of lambda to function pointer when passing to template function

问题描述

是否有一种方法可以使lambda衰减到指针，而无需显式转换为正确的签名?这样可以整理一些代码:

Is there a way to make a lambda decay to a pointer, without explicitly casting to the right signature? This would tidy some code:

template<typename T> T call(T(*func)()){ return func(); }
int ptr(){ return 0; }
int main(){
    auto ret1 = call(ptr);
    auto ret2 = call((int(*)())([]{ return 0; }));
    auto ret3 = call([]{ return 0; });  //won't compile
}

很明显，仅当lambda衰减到指针时，对call的调用才有效，但是我猜测只有在选择了正确的函数重载/模板之后，这种情况才会发生.不幸的是，我只能想到涉及模板​​的解决方案，以使λ具有任何签名衰减，所以我回到正题.

It's evident that a call to call works only if the lambda decays to a pointer, but I'm guessing that that can happen only after the right function overload/template is chosen. Unfortunately I can only think of solutions that involve templates to make a lambda with any signature decay, so I'm back to square one.

推荐答案

您可以将lambda更改为使用一元+运算符:+[]{ return 0; }

You can change your lambda to use the unary + operator: +[]{ return 0; }

之所以可行，是因为一元加号可以应用于指针，并且会触发对函数指针的隐式转换.

This works because unary plus can be applied to pointers, and will trigger the implicit conversion to function pointer.

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