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- Why is the copy constructor not called? 4 answers
I have this piece of code. I was expecting d3
to be created with move constructor as well, as I passed a rvalue temporary.
#include <iostream>
using namespace std;
struct Data {
Data(): x(1)
{
cout << "constructor" << endl;
}
Data(const Data& original): x(2)
{
cout << "copy constructor" << endl;
}
Data(Data&& original): x(3)
{
cout << "move constructor" << endl;
}
int x;
};
int main() {
Data d1; // constructor
cout << "d1:" << d1.x << endl << endl;
Data d2(d1); // copy constructor
cout << "d2:" << d2.x << endl << endl;
Data d3(Data{}); // move constructor?
cout << "d3:" << d3.x << endl << endl;
Data d4(move(Data{})); // move constructor?
cout << "d4:" << d4.x << endl << endl;
return 0;
}
I saw output as:
constructor
d1:1
copy constructor
d2:2
constructor
d3:1
constructor
move constructor
d4:3
Although d4
is constructed with move constructor as I expected, I don't understand why d3.x
got value 1. It seems d3
is constructed by the default constructor?
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