C ++ lambda two copy constructor calls

Question

I have the following code snippet.

#include <iostream> #include <functional> using namespace std; struct A { A() { cout << "A "; data = 1; } A(const A& a) { cout << "cA "; data = a.data; } ~A() { cout << " dA"; } int data; }; void f(A& a, function<void(A)> f) { cout << "("; f(a); cout << ")"; } int main() { A temp; auto fun = [](A a) {cout << a.data;}; f(temp, fun); } 

Output:

A (cA cA 1 dA dA) dA

Why is temp copied twice?

I am using Visual C ++ (vc140).

Answer 1

function<void(A)> has a call function operator with this signature: operator()(A) i.e. it takes its argument by value, so calling f(a) makes a copy.

lambda also takes its argument by value, so when it is called inside the function<void(A)> statement function<void(A)> , another copy is created.

If you define a move constructor for A , you should see that initializing the lambda argument (from the first copy made by function ) can be a move, not a copy, but only if the type has a move constructor. Otherwise, it must be copied.

Alternatively, if you use std::function<void(const A&)> , then the call operator will take its argument by reference not by value, therefore there will only be one copy to initialize the lambda argument.