C++ copy on member access - c++

Question

I was doing some experiments to see when copy is performed apart from copy elision, RVO, NRVO cases.
So I've written some code like this:
class X {
public:
X() { std::cout << "Default constructor" << std::endl; }
X(const X&) { std::cout << "Copy constructor" << std::endl; }
X(X&&) { std::cout << "Move constructor" << std::endl; }
X& operator=(const X) {
std::cout << "Assignment operator" << std::endl;
return *this;
}
X& operator=(X&&) {
std::cout << "Move assignment operator" << std::endl;
return *this;
}
~X() { std::cout << "Destructor" << std::endl; }
};
class Y {
private:
X x;
public:
const X& getX() const {
std::cout << "getX" << std::endl;
return x;
}
};
int main() {
Y y;
std::cout << "assign to ref" << std::endl;
const X& x1 = y.getX();
(void)x1;
std::cout << "assign to const" << std::endl;
const X x2 = y.getX();
return 0;
}
and I receive the following as output:
Default constructor
assign to ref
getX
assign to const
getX
Copy constructor
Destructor
Destructor
Both when compiled with gcc or clang with -O3 and tried -std=c++{11,14,17} all produced the same output.
Which surprised me was, I wasn't expecting any copy to be performed when using y.getX(); to a const variable. It is something I used frequently just to ease my access to that variable and its members in the following code, but I wasn't doing it over a const reference instead I was just using const hoping the compiler would regard it just as a renaming.
Does anyone knows why exactly is that copy performed? Only reason that comes to my mind is that it is to make code thread-safe. If there are multiple threads working with object y, then my assignment to const would not be that const after all. Since it would just reference the member x in object y. Which might be changed by other threads. But I am not sure whether that's the real intention or not.

Answer 1

To see the effect of RVO verses compiler forced use of NRVO, play with -fno-elide-constructors compiler switch on the following modified program below. With the usual options you get:
Default constructor 1
assign to ref
getX (with id: 1)
x1 (id:1)
assign to const
getX (with id: 1)
Copy constructor 2
x2 (id:2)
make_X copy
Default constructor 3
make_X (with id: 3)
x3 (id:3)
make_X ref
Default constructor 4
make_X (with id: 4)
x4 (id:4)
Destructor 4
Destructor 3
Destructor 2
Destructor 1
But with NRVO you get:
Default constructor 1
assign to ref
getX (with id: 1)
x1 (id:1)
assign to const
getX (with id: 1)
Copy constructor 2
x2 (id:2)
additional 1
Default constructor 3
make_X (with id: 3)
Move constructor 4
Destructor 3
Move constructor 5
Destructor 4
x3 (id:5)
additional 2
Default constructor 6
make_X (with id: 6)
Move constructor 7
Destructor 6
x4 (id:7)
Destructor 7
Destructor 5
Destructor 2
Destructor 1
Code example:
#include <iostream>
int global_id;
class X {
public:
X() : id(++global_id) {
std::cout << "Default constructor " << id << std::endl;
}
X(const X&) : id(++global_id) {
std::cout << "Copy constructor " << id << std::endl;
}
X(X&&) : id(++global_id) {
std::cout << "Move constructor " << id << std::endl;
}
X& operator=(const X&) {
std::cout << "Assignment operator " << id << std::endl;
return *this;
}
X& operator=(X&&) {
std::cout << "Move assignment operator " << id << std::endl;
return *this;
}
~X() {
std::cout << "Destructor " << id << std::endl;
}
int id;
};
class Y {
X x;
public:
const X& getX() const {
std::cout << "getX (with id: " << x.id << ')' << std::endl;
return x;
}
X make_X() const {
X extra;
std::cout << "make_X (with id: " << extra.id << ')' << std::endl;
return extra;
}
};
int main()
{
Y y;
std::cout << "assign to ref" << std::endl;
const X& x1 = y.getX();
std::cout << "x1 (id:" << x1.id << ")\n";
(void) x1;
std::cout << "assign to const" << std::endl;
const X x2 = y.getX();
std::cout << "x2 (id:" << x2.id << ")\n";
std::cout << "make_X copy" << std::endl;
const X x3 = y.make_X();
std::cout << "x3 (id:" << x3.id << ")\n";
std::cout << "make_X ref" << std::endl;
const X& x4 = y.make_X();
std::cout << "x4 (id:" << x4.id << ")\n";
return 0;
}
As you see, the RVO really only comes to play with local variables.