Returning a pair of objects

Question

The following is an anti-pattern:

auto f() {
  std::vector<int> v(100000);
  return std::move(v); // no need to use std::move thanks to RVO (return value optimization)
}

Using a std::move can even produce worst code (see here )

However, what should I do in the following situation:

auto f() {
  std::vector<int> v0(100000);
  std::vector<int> v1(100000);
  return std::make_pair(std::move(v0),std::move(v1)); // is the move needed?
}

Answer 1

For the second snippet,

auto f() {
  std::vector<int> v0(100000);
  std::vector<int> v1(100000);
  return std::make_pair(std::move(v0),std::move(v1)); // is the move needed?
}

return returns the result of the std::make_pair() function. That's an RValue.

However, the OP's question probably condenses to whether (or why not) Named Return Value Optimization still applies to v0 / v1 when returned as a std::pair .

Thereby, it's overlooked that v0 / v1 aren't subject of return anymore, but become arguments of std::make_pair() . As such, v0 / v1 are LValues – std::move(v0), std::move(v1) have to be applied to turn them into RValues if move-semantic is intended.

---

Demo on coliru :

#include <iostream>

template <typename T>
struct Vector {
  Vector(size_t n)
  {
    std::cout << "Vector::Vector(" << n << ")\n";
  }
  Vector(const Vector&)
  {
    std::cout << "Vector::Vector(const Vector&)\n";
  }
  Vector(const Vector&&)
  {
    std::cout << "Vector::Vector(const Vector&&)\n";
  }
  
};

auto f1() {
  Vector<int> v(100000);
  return std::move(v); // over-pessimistic
}

auto f2() {
  Vector<int> v(100000);
  return v; // allows NRVO
}

auto f3() {
  Vector<int> v0(100000);
  Vector<int> v1(100000);
  return std::make_pair(v0, v1); // copy constructor called for v0, v1
}    

auto f4() {
  Vector<int> v0(100000);
  Vector<int> v1(100000);
  return std::make_pair(std::move(v0),std::move(v1)); // move constructor called for v0, v1
}

#define DEBUG(...) std::cout << #__VA_ARGS__ << ";\n"; __VA_ARGS__ 

int main()
{
  DEBUG(f1());
  DEBUG(f2());
  DEBUG(f3());
  DEBUG(f4());
}

Output:

f1();
Vector::Vector(100000)
Vector::Vector(const Vector&&)
f2();
Vector::Vector(100000)
f3();
Vector::Vector(100000)
Vector::Vector(100000)
Vector::Vector(const Vector&)
Vector::Vector(const Vector&)
f4();
Vector::Vector(100000)
Vector::Vector(100000)
Vector::Vector(const Vector&&)
Vector::Vector(const Vector&&)

Answer 2

Yes, the move is needed to avoid copy in the latter case.

However, this would be even better:

return std::make_pair(
    std::vector<int>(100000),
    std::vector<int>(100000));