CODEWITHSUNDEEP
c++c++11copy-constructormove-semanticsmove-constructor
fizzbuzz
fizzbuzz

Reputation: 157

Why is copy constructor called rather than move constructor?

I have the following code:

#include <bits/stdc++.h>

using namespace std;

class A {
public:
    A(const A& a) noexcept { cout << "copy constructor" << endl; }
    A& operator=(const A& a) noexcept { cout << "copy assignment operator" << endl; }
    A(A&& a) noexcept { cout << "move constructor" << endl; }
    A& operator=(A&& a) noexcept { cout << "move assignment operator" << endl; }
    A() { cout << "default constructor" << endl; }
};

vector<A> aList;

void AddData(const A&& a)
{
    aList.push_back(std::move(a));
}

int main()
{
    AddData(A());
    return 0;
}

The output is default constructor copy constructor. please tell me is the rvalue reference push_back(T&&)called? And when is copy constructor called?

Upvotes: 2

Views: 319

Answers (1)

jfMR
jfMR

Reputation: 24738

The issue is with the a parameter in AddData():

void AddData(const A&& a) // <-- const reference!!!
{
    aList.push_back(std::move(a)); // selects push_back(const A&)
}

The a parameter above is a const rvalue reference. You are marking with std::move() a const object.

Marking a const object with std::move() for moving has no effect when it comes to move semantics because you can't move from a const object (i.e., you need to alter the moved-from object, but it is const-qualified).

An rvalue reference doesn't bind to a const object, but a const lvalue reference does. As a result, the push_back(const A&) overload is selected instead of the push_back(A&&) one, and therefore the A object is copy constructed.

Solution

Use a non-const rvalue reference instead:

void AddData(A&& a) // <-- non-const reference
{
    aList.push_back(std::move(a)); // selects push_back(A&&)
}

Upvotes: 8

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