How do I transfer ownership of a unique_ptr when returning a tuple?

Question

I'm trying to return a tuple, one of the elements of which is a std::unique_ptr. I want to transfer the ownership of the unique_ptr to the caller. How do I do this?

#include <tuple>
#include <memory>
#include <iostream>

using namespace std;

class B
{
public:
    B(int i) : i_(i) {}

    int getI() const { return i_; }
private:
   int i_;
};

tuple<unique_ptr<B>, int>
getThem()
{
    unique_ptr<B> ptr(new B(10));
    return make_tuple(ptr, 50);
}

int
main(int argc, char *argv[])
{

    unique_ptr<B> b;
    int got = 0;

    tie(b, got) = getThem();

    cout << "b: " << b->getI() << endl;
    cout << "got: " << got << endl;

    return 0;
}

This fails to compile because the copy constructor of unique_ptr is deleted, for obvious reasons. But how to I indicate I want to move the unique_ptr into the tie?

Answer 1

Essentially you just need to explicitly move the non-copyable types into the tuple, thus using std::move. std::tuple has the appropriate constructors to copy and move the types internally (the move being appropriate here).

As follows;

#include <tuple>
#include <memory>
#include <iostream>

using namespace std;

class B
{
public:
    B(int i) : i_(i) {}

    int getI() const { return i_; }
private:
    int i_;
};

tuple<unique_ptr<B>, int>
getThem()
{
    unique_ptr<B> ptr(new B(10));
    return make_tuple(std::move(ptr), 50); // move the unique_ptr into the tuple
}

int
main(int argc, char *argv[])
{
    unique_ptr<B> b;
    int got = 0;

    tie(b, got) = getThem();

    cout << "b: " << b->getI() << endl;
    cout << "got: " << got << endl;

    return 0;
}

Answer 2

Use std::move to invoke the move operator instead.

return make_tuple(std::move(ptr), 50);