CODEWITHSUNDEEP
c++
SEONGWOO CHAE
SEONGWOO CHAE

Reputation: 65

understand about what decltype returns in c++

I'm very confused with decltype.

In the wiki(https://en.wikipedia.org/wiki/Decltype#Semantics), there is an example.

const int&& foo();
const int bar();
int i;
struct A { double x; };
const A* a = new A();
decltype(foo()) x1; // type is const int&&
decltype(bar()) x2; // type is int
decltype(i) x3; // type is int
decltype(a->x) x4; // type is double
decltype((a->x)) x5; // type is const double&

But, I don't understand at all about it.

  1. decltype(bar()) x2; // type is int

I think type should be const int because bar() returns const int and decltype(foo()) returns const int&&. But, why is it just an int?

  1. decltype((a->x)) x5; // type is const double&

As wiki said,

The reason for the difference between the latter two invocations of decltype is that the parenthesized expression (a->x) is neither an id-expression nor a member access expression, and therefore does not denote a named object.[13] Because the expression is an lvalue, its deduced type is "reference to the type of the expression", or const double&

wiki said that the expression (a->x) is an lvalue. But, AFAIK, lvalue is something that points to a specific memory location. Namely, it has its name or identifier. I think a->x does got its name, but (a->x) dosen't. Why it is a lvalue?

Upvotes: 1

Views: 140

Answers (1)

theWiseBro
theWiseBro

Reputation: 1529

According to cppreference, we can classify that:

  1. return type of const int&& foo(); is an xvalue.
  2. return type of const int bar(); is a prvalue.

Searching for this brought me to this which states:

non-class prvalues always have cv-unqualified types

So its understood that return type of bar() is an int since its not a class type.

As to you second query:

Why it is a lvalue?

Because it has been mentioned in the standard as such. cppreference note:

Note that if the name of an object is parenthesized, it is treated as an ordinary lvalue expression

You can think of (a->x) as being the lhs of an assignment, and thus decltype((a->x)) returns a reference const int&.

Upvotes: 1

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