## How std::move breaks return value optimization

Posted by Diego Assencio on 2017.08.07 under Programming (C/C++)

Consider the following program:

std::vector<int> random_numbers(const size_t n)
{
std::vector<int> numbers;

/* generate n random numbers, store them on numbers */

return numbers;
}

int main()
{
/* create an array with 100 random numbers */
std::vector<int> my_numbers = random_numbers(100);

...
}


Returning an std::vector<int> by value can make a developer nervous: will this cause the vector to be copied? That would be a costly performance penalty and therefore something to be avoided, if possible.

In this type of situation, a solid understanding of how return value optimization (or simply RVO, for short) works leads to inner peace. Any decent compiler will understand that numbers is a temporary variable whose value will be returned at the end of the random_numbers function, and since this returned value is an rvalue (a temporary object, see line 13), it can be used to initialize my_numbers directly, i.e., without going through any of the std::vector<int>'s constructors. In other words, no vectors should be copied or moved on the code above. How wonderful!

Let's take a closer look at how RVO removes the need for a copy operation. When random_numbers is called, memory on the call stack will be reserved for its return value (an std::vector<int>). Inside of random_numbers, numbers is clearly a temporary variable whose value will be returned at the end, so instead of copying this value to the memory location reserved for the function's return value at the very end, numbers is stored directly at that memory location. The elimination of such a copy operation when a function returns is what "return value optimization" stands for. In general, any type of optimization which causes copy operations to be eliminated is referred to as a copy elision optimization.

Despite all these facts, it is not uncommon for developers to be less confident than they should in this type of situation and prefer "being on the safe side" by writing this type of code:

std::vector<int> random_numbers(const size_t n)
{
std::vector<int> numbers;

/* generate n random numbers, store them on numbers */

return std::move(numbers);   /* don't do this! */
}


As innocent as this decision may be, it is severely flawed because of the way RVO works: only a local variable or a temporary object can be stored directly at the memory location for a function's return value (function parameters are not eligible for that), and this is only allowed if such an object is directly returned by the function and has the same type as the function's return type. By adding std::move on the return statement above, we converted the type of the returned object to std::vector<int>&& (an rvalue reference to an std::vector<int>), but random_numbers returns an std::vector<int>. This violates the conditions required for RVO, and the compiler will have no choice but to make numbers be constructed outside the memory area reserved for random_numbers's return value and then moved to that location when the function returns (a move operation is still possible here since std::move(numbers) is an rvalue and will therefore trigger the move constructor for the std::vector<int> which is constructed as the return value).

For many user-defined types, such an additionally incurred move operation will be cheap, but it will definitely not be cheaper than what RVO offers and therefore, by "playing safe", we ended up inevitably pessimizing our program. Also, notice that we were lucky to have a return type (std::vector<int>) with a move constructor; had this not been the case, the added std::move would have caused the return value to be initialized through a copy constructor, which is what we wanted to avoid in the first place. Ouch!

To finalize, here is an example which involves all concepts discussed so far:

#include <iostream>

class X
{
public:
/* default constructor */
X() { std::cout << "X::X()\n"; }

/* copy constructor */
X(const X&) { std::cout << "X::X(const X&)\n"; }

/* move constructor */
X(X&&) { std::cout << "X::X(X&&)\n"; }
};

X good()
{
std::cout << "good()\n";

X x;
return x;
}

{

X x;
return std::move(x);
}

int main()
{
X x1 = good();

return 0;
}


The program's output illustrates how the std::move on the bad function disables RVO and forces an unnecessary move operation (this will be the case even if you compile with lots of optimizations enabled, e.g. by using -O3 on gcc):

good()
X::X()
X::X()
X::X(X&&)


Try compiling and running this program, then remove the move constructor from X. The resulting output shows that std::move now causes X to be copied:

good()
X::X()
X::X()
X::X(const X&)