Is there a reason declval returns add_rvalue_reference instead of add_lvalue_reference

Question

changing a type into a reference to a type, allows one to access the members of the type without creating an instance of the type. This seems to be true for bot

Answer 1

With add_rvalue_reference:

• declval() is of type Foo&&.
• declval() is of type Foo& (reference collapsing: “Foo& &&” collapses to Foo&).
• declval() is of type Foo&& (reference collapsing: “Foo&& &&” collapses to Foo&&).

With add_lvalue_reference:

• declval() would be of type Foo&.
• declval() would be of type Foo& (reference collapsing: “Foo& &” collapses to Foo&).
• declval() would be of type Foo& (!) (reference collapsing: “Foo&& &” collapses to Foo&).

that is, you would never get a Foo&&.

Also, the fact that declval() is of type Foo&& is fine (you can write Foo&& rr = Foo(); but not Foo& lr = Foo();). And that declval&&>() would be of type Foo& just feels “wrong”!

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Edit: Since you asked for an example:

#include 
using namespace std;

struct A {};
struct B {};
struct C {};

class Foo {
public:
    Foo(int) { } // (not default-constructible)

    A onLvalue()   &  { return A{}; }
    B onRvalue()   && { return B{}; }
    C onWhatever()    { return C{}; }
};

decltype( declval().onLvalue()   ) a;
decltype( declval().onRvalue()   ) b;
decltype( declval().onWhatever() ) c;

If declval used add_lvalue_reference you couldn't use onRvalue() with it (second decltype).