Question

I guess there is something about temporary objects that I don’t understand.
Given the relationships:

class C {};

class F {
public:
    C getC() { return C(); };
};

class N {
public:
    N( C & base ){};
};

This works:

N n(C());

This doesn’t work:

F f;
N n(f.getC()); //compile error

Why?

Answer 1

A non-const reference (like B& base) can only bind to an lvalue.

F::getC() returns a C object by value, so the call expression f.getC() is an rvalue, not an lvalue.

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The reason that N n(C()); works, however, is due to an unrelated problem.

This does not declare an object. It declares a function named n that returns N and takes a parameter of type “pointer to a function that has no parameters and returns C.”

This is one manifestation of a language peculiarity known as the most vexing parse. To change this to declare an object, you’d need one of the following:

N n = C();  // Use copy initialization
N n((C())); // Use more parentheses

Both of these would fail to compile, though, because both would attempt to bind the result of the rvalue expression C() to the non-const reference B& base.

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A const reference (like B const& base) can bind to an rvalue, as can an rvalue reference (like B&& base) in C++11.