C++ Returning reference to temporary [duplicate]

Answer 1

If your function looks like this:

AnotherObject& getAnotherObject()
{

    . . .

    Object::const_iterator it = find ("lang");
    if (it == this->end()) { return AnotherObject() };

    . . .

}

the problem is that the AnotherObject() you've returned will be destroyed as soon as the function exits, and so the caller to your function will have a reference to a bogus object.

If your function returned by value however:

AnotherObject getAnotherObject()

then a copy will be made before the original is destroyed and you'll be OK.

Answer 2

You shouldn't return a reference to a temporary which is destroyed at the end of the line, nor a reference to a local which is destroyed at the end of the function.

If you want to keep the current signature, you'd have to add a static constant instance that you can return as a default.

#include <iostream>

template <class T>
class X
{
    T value;
    static const T default_instance;
public:
    X(const T& t): value(t) {}
    const T& get(bool b) const
    {
        return b ? value : default_instance;
    }
};

template <class T>
const T X<T>::default_instance = T();

int main()
{
    X<int> x(10);
    std::cout << x.get(true) << ' ' << x.get(false) << '\n';
}

You may also return by value or return a pointer in which case you can return NULL.