What is value type of *T?

Question

Let's consider very short piece of code in C++:

class T{
};
T *t = new T();

What is type of *t? rvalue/lvalue/xvalue/glvalue? why? Thanks in advance.

I don't follow. Are you asking what the type of the *expression* `*t` is? Or are you asking what the type of the *variable* `t` is? — Brian Rodriguez, May 23 '17 at 18:45
already answered here https://stackoverflow.com/a/40169699/3521116 — warl0ck, May 23 '17 at 18:46

R Sahu · Accepted Answer · 2017-05-23T18:53:30.067

1

The check for lvalue is very simple. Can it appear on the left hand side of an assignment operator?

In your case, the answer is "yes".

*t = T();

is valid. Hence, the value of type of *t is lvalue.

By virtue of being an lvalue, it is also a glvalue.

A glvalue (“generalized” lvalue) is an lvalue or an xvalue.

edited May 23 '17 at 18:53

answered May 23 '17 at 18:48

R Sahu

1

It's also a glvalue. – Mooing Duck May 23 '17 at 18:51
1

Well, overloading `operator =` for `T` would make it able to be called with an rvalue, too, so this check is not really accurate. – Quentin May 23 '17 at 19:21
@Quentin, true. I am sure the OP is not looking at anything that deep. – R Sahu May 23 '17 at 19:31
1

@RSahu I wrote too fast. It actually does that even without overloading (which, in restrospect, is obvious)... So `struct Foo { }; Foo{} = Foo{};` compiles just fine. – Quentin May 23 '17 at 19:37
@Quentin, by strict definition of the standards, an lvalue *designates a function or an object*. It doesn't say anything about the life time of the object. If we go by that imprecise definition, even a temporary object, which does designate an object, should qualify as an lvalue. – R Sahu May 23 '17 at 19:45

1 Answers1