CODEWITHSUNDEEP
c++c++11
xeonqq
xeonqq

Reputation: 63

understand the move constructor behavior in this example

I am learning the move semantics in C++11. I wrote a small program to test the behavior of move semantics. But it does not behave as what I expected, could someone explain me why?

#include<iostream>

using namespace std;


class Vector
{
public:
    Vector()
    {
    cout << "empty Ctor"<<endl;
    array = new int[10];
    size = 10;
    }

    Vector(int n)
    {
    array = new int[n];
    size = n;
    for (int i=0; i<size; ++i)
        array[i] = i;
    cout << "Ctor"<<endl;
    }

    Vector(const Vector& v):size(v.size)
    {
    array = new int[size];
    for (int i=0; i<size; ++i)
        array[i] = v.array[i];
    cout << "copy"<<endl;
    }

    Vector(Vector&& v):size(v.size)
    {
    array = v.array;
    v.array = nullptr;
    cout << "move"<<endl;

    }

    ~Vector()
    {
    delete array;
    }

private:
    int* array;
    int size;

};

int main() {
    Vector v(10); //print Ctor. (as expected)
    Vector v1(std::move(v)); //print move. (as expected)
    Vector v2(*(new Vector(2)));  //print Ctor, copy. (I expect Ctor, move)
    Vector v3(Vector(2)); //print only Ctor. (I expect Ctor, move)

}

So, why the print is not what I expected. Since I think both value passed to v2 and v3 are Rvalue. And for v3, why it print only Ctor without printing "move" or "copy"

Upvotes: 4

Views: 112

Answers (1)

TartanLlama
TartanLlama

Reputation: 65620

Vector v2(*(new Vector(2)));

new Vector(2) is an rvalue, but dereferencing it produces an lvalue, hence the copy rather than the move.

Vector v3(Vector(2));

Since the temporary Vector is unnecessary, the copy will be elided by the compiler. Your compiler may have a flag to disable copy elision so that you can see the additional move, e.g. -fno-elide-constructors in GCC and Clang.

Upvotes: 11

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