passing Temporary variables to reference arg in Cons works. but not for functions in general. Why?

Question

Consider the following code.
Here, A a(B()) compiles even though the constructor is A(B& b); But print(B()) does not work. But print is also declared as print(B& b); Why this inconsistency?

#include <iostream>
using namespace std;

class B{
    public:
            char b;
};

class A {
    public:
            B b;
            A(B& b);
            A() { }
};

A::A(B& b) {
    this->b = b;
}

void print(B& b) { }

int main(){
    print(B());
    A a(B());
}

Answer 1

You should not be passing a non-constant reference to a temporary, so the print statement should not compile. If you modify that reference in print to const, it will work.

Answer 2

This:

A a(B());

isn't doing what you think. It is actually parsed as a function declaration. This is commonly referred to as the "most vexing parse" in C++ (there are many posts here about it, if you search for that phrase).

Answer 3

It compiles because it's not creating an instance of A. It's declaring a function named a that returns an A and receives one unnamed parameter of type pointer-to-function-returning-B. Since it's just a declaration, it compiles. If you're referred to a elsewhere in the code, you'd have seen additional problems. For why that's a function declaration instead of an object definition, the term to look up is most vexing parse.

Answer 4

Where you are trying to call the constructor you are actually declaring a function:

A a(B());

This declares a as a function returning A and taking as parameter a function pointer returning B and taking no parameters.

Actually trying to call the constructor results in an error, as expected:

A a = A(B());

tst.cpp:32: error: no matching function for call to ‘A::A(B)’