Does RVO optimisation happens when a copy constructor is not defined for a class?

Question

Lets say I have this piece of code:

class Base {
public:
    void f() {}
private:
    int n;
};

Base foo()
{
    Base b;

    // processing

    return b;
}

Base doesn't have a copy constructor defined and looking at its members the compiler won't synthesise one.
Is Named Return Value (NRV) optimisation applied for this code ? What is the exact rule when a compiler would apply NRV ?

Why do you think the compiler won't synthesize a copy constructor for that class? — Benjamin Lindley, Feb 17 '14 at 09:04
Because it performs only member wise copy. Please correct if I'm wrong. — Adrian, Feb 17 '14 at 09:08
Performing member-wise copy is what a compiler-synthesized copy constructor *does*. — Sneftel, Feb 17 '14 at 09:10

score 1 · Accepted Answer · answered Feb 17 '14 at 09:14

First of all, there's no exact rule, because NRVO is not explicitly part of the standard, but rather is allowed by other parts of the standard. The compiler is never required to employ NRVO, and different compilers will be inhibited from using NRVO by different constructs.

In your situation, there's nothing obvious to inhibit the use of NRVO. There's no explicit destructor or copy constructor, and the returned variable is declared outside any conditional or loop block. It depends on what's going on in that // processing area, though.

As my experiments on g++/clang++ in a Linux env, NRVO will not be implemented if there is no copy ctor in the class. — Wizmann, Feb 26 '18 at 06:53

Does RVO optimisation happens when a copy constructor is not defined for a class?

1 Answers1