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Home/ Questions/Q 8618147
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Editorial Team
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Editorial Team
Asked: June 12, 20262026-06-12T06:02:13+00:00 2026-06-12T06:02:13+00:00

#include <iostream> using namespace std; struct A { A() { cout << A() <<

  • 0
#include <iostream>

using namespace std;

struct A
{
    A()
    {
        cout << "A()" << endl;
    }

    ~A()
    {
        cout << "~A()" << endl;
    }

    A(A&&)
    {
        cout << "A(A&&)" << endl;
    }

    A& operator =(A&&)
    {
        cout << "A& operator =(A&&)" << endl;
        return *this;
    }
};

struct B
{
    // According to the C++11, the move ctor/assignment operator
    // should be implicitly declared and defined. The move ctor
    // /assignment operator should implicitly call class A's move
    // ctor/assignment operator to move member a.
    A a;
};

B f()
{
    B b;

    // The compiler knows b is a temporary object, so implicitly 
    // defined move ctor/assignment operator of class B should be
    // called here. Which will cause A's move ctor is called.
    return b; 
}

int main()
{
    f();
    return 0;
}

My expected output should be:

A()
A(A&&)
~A()
~A()

However, the actual output is: (The C++ compiler is: Visual Studio 2012)

A()
~A()
~A()

Is this a bug of VC++? or just my misunderstanding?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-12T06:02:15+00:00Added an answer on June 12, 2026 at 6:02 am

    According to this blog post, VC++ 2012 currently implements N2844 + DR1138, but not N3053. As a result, the compiler is not implicitly generating move constructors or assignment operators for you. If you add explicit default and move constructors to B then you will get the output you expect.

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