表达式作为在 C++ 中调用 function 的结果

Question

How does result of function call can be an expression? I am reading this article - http://eel.is/c++draft/expr and it is written there that

An expression is an xvalue if it is the result of calling a function, whether implicitly or explicitly, whose return type is an rvalue reference to object type

I don't understand how does an expression can be a result of function call.

Answer 1

Consider these two examples.

const auto someValue = 43 * otherValue;
const auto someOther = func();

In both cases, the right hand side of the assignent is an expression. In the first case, it's an integer multiplication. In the second call, it's a function call. Let's be as clear as possible: the expression is not the result of calling a function -- instead, func() is an expression.

Let's imagine the signature of this function call is

struct Obj { /* ... */ };

Obj&& func();

Here, the return type is an rvalue reference of object type. This is meant by the wording in the standard. And in the above case, the expression func() is an xvalue (an "eXpiring value").