Why does std::declval add a reference?

Question

std::declval is a compile-time utility used to construct an expression for the purpose of determining its type. It is defined like this:

template< class T > typename std::add_rvalue_reference<T>::type declval() noexcept; 

Would this not be simpler instead?

template< class T > T declval() noexcept; 

What is the advantage of a reference return type? And shouldn't it be called declref?

The earliest historical example I find is n2958, which calls the function value() but already always returns a reference.

Note, the operand of decltype does not need to have an accessible destructor, i.e. it is not semantically checked as a full-expression.

template< typename t > t declprval() noexcept;  class c { ~ c (); }; decltype ( declprval< c >() ) * p = nullptr; // OK 

Answer 1

The "no temporary is introduced for function returning prvalue of object type in decltype" rule applies only if the function call itself is either the operand of decltype or the right operand of a comma operator that's the operand of decltype (§5.2.2 [expr.call]/p11), which means that given declprval in the OP,

template< typename t > t declprval() noexcept;  class c { ~ c (); };  int f(c &&);  decltype(f(declprval<c>())) i;  // error: inaccessible destructor 

doesn't compile. More generally, returning T would prevent most non-trivial uses of declval with incomplete types, type with private destructors, and the like:

class D;  int f(D &&);  decltype(f(declprval<D>())) i2;  // doesn't compile. D must be a complete type 

and doing so has little benefit since xvalues are pretty much indistinguishable from prvalues except when you use decltype on them, and you don't usually use decltype directly on the return value of declval - you know the type already.

Answer 2

Arrays cannot be returned by value thus even just the declaration of a function returning an array by value is invalid code.

You can however return an array by reference.