Understanding void func(A())

Question

This is my code

#include <iostream>

class A {
public:
    int a = 0;
    A(int i = 0) : a(i) {}
};

void func(A())
{
    std::cout << "Hello" << std::endl;
}

int main()
{
    A(*p)() = NULL;
    func(p);
}

What confused me is that A() in void func(A()) is equal to A(*)() instead of A's constructor. How does this work?

Anything that can be parsed as a function declaration is a function declaration. Look up _"most vexing parse"_. Can't find a canonical duplicate but lots of similar. — Richard Critten, Nov 28 '19 at 13:44
@Richard [Most Vexing Parse](https://stackoverflow.com/questions/7007817/a-confusing-detail-about-the-most-vexing-parse) — Adrian Mole, Nov 28 '19 at 13:50
@Adrian-ReinstateMonica Thanks, I found that one - as this Q is so common was hoping for higher votes. — Richard Critten, Nov 28 '19 at 13:54
@RichardCritten I’m not sure this is related to the Most Vexing Parse, as this is specifically in the context of a function declaration where other parses couldn’t (?) apply. But perhaps I’m mistaken? — templatetypedef, Nov 28 '19 at 13:57

score 9 · Accepted Answer · answered Nov 28 '19 at 13:49

Let’s reason by analogy. If you define a function

void doSomething(A [137]) {

}

then C++ treats it as though you’d actually written

void doSomething(A *) {

}

In other words, there are some types where, if you use them as a parameter to a function, C++ will automatically replace them with a different type, the type you’d get by decaying the type.

In your case, A() is the type of a function that takes in no arguments and returns an A. If you have a C++ function that takes an A() as an argument, C++ will instead have the function take as input an A (*)(), a pointer to a function taking no arguments and returning an A. The reason for this is that you can’t have an object of type A() in C++, though you can have a pointer to an A().

Understanding void func(A())

1 Answers1