为什么函数调用是xvalue（如果返回类型是rvalue）？

Question

Let's say we have a function:

struct A {
    int m;
};

A&& f();

As far as I know the expressions:

f();
f().m; 

are both xvalue. But why? Why aren't they prvalue? I'm a little bit confused.

Answer 1

Because you are returning by reference, not by value, from f . This implies that the A has a lifetime longer than f() , eg

A&& f()
{
     static A res;
     return std::move(res);
}

or

A global;

A&& f()
{
     return std::move(global);
}

But not

A&& f()
{
     return {}; // dangling reference
}

In f().m; , the use of m inherits the value category of the earlier sub-expression, as normal for member access.