Can NRVO be turned off in debug builds?

Question

In the program as follows

#include <iostream>

struct A 
{ 
    A() { std::cout << "0"; }
    A( const A & ) { std::cout << "1"; }
    A( A && ) noexcept { std::cout << "2"; }
};

A foo()
{
    A res;
    return res;
}
 
int main()
{
    const A a = foo();
    (void)a;
}

I expected named returned value optimization take place in foo(), so neither copy nor move constructor will be called. And indeed gcc and clang print “0”.

But Visual Studio 2019 in Debug configuration prints “02”. Is it allowed for the compiler to avoid NRVO in debug builds?

score 2 · Accepted Answer · answered May 14 '21 at 17:17

2

Since mandatory copy elision doesn't apply here, yes, compiler is not obligated to optimize away a move in any builds, be it debug or optimized ones.

It is allowed, but not required.

answered May 14 '21 at 17:17

SergeyA

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2

Improve your answer by adding why it is not mandatory here. – Surt May 14 '21 at 19:24

score 1 · Answer 2 · answered Aug 16 '23 at 14:19

It's unnamed returned value optimization(URVO) that is mandatory since C++17, not NRVO. Nevertheless in practice, most mainstream compilers employ NRVO because of the fact that copy elision is even better than a move operation, since it just constructs the object in place. Now even if I try to force MSVS 2022 in debug mode by calling a function like foo() in your code and adding more complex statements inside, it doesn't call copy/move ctor.

Can NRVO be turned off in debug builds?

2 Answers2