什么是 `R(*pf)(void*, Args...)`，function 指向方法的指针？

Question

I saw this type here . I believe he's trying to create a variable pf for a member pointer type-erased (that's why there's void* there). I then noticed this type signature in similar such classes.

But according to isocpp a non-static member pointer type is defined like this: int (Fred::*)(char,float) (for some class Fred ) and a function pointer type is defined like this: int (*)(char,float)

Therefore one would create a member pointer variable mp like this: int (S::*mp)(int) = nullptr;

Maybe this void* represents this* and its another way to define a member pointer variable by defining a function pointer variable? Is this possible?

What is R(*pf)(void*, Args...) ?

Answer 1

It's the declaration of a function pointer. Nothing more than that.

Compatible functions take void* and Args... , and return R .

In the given example, the compatible function that's assigned to the pointer, is a lambda. The void* is the type-erased address of some callable f , and the Args... members are, well, the arguments that'll be passed to that callable. The callable's type is restored by capturing of type aliases inside the lambda (nice.).

Answer 2

R(*pf)(void*, Args...) is a function pointer (regular one, not pointer-to-member) to a function that returns R and has (void*, Args...) parameters, where Args... is a list of types (an expanded template parameter pack).

Maybe this void* represents this* and its another way to define a member pointer variable

Nah, there is no such feature in C++.

If you look at the code , the only things assigned to this pointer are lambdas, like this one:

pf = [](void* ptr, Args... args)->R{
  return blah;
};

I'm not sure why you expected pointers-to-members to be involved.