For what T does `std::declval()` not have a matching function?

Question

I was surprised to find that for some T, decltype(std::declval<T>()) is not legal:

#include <utility>

template<typename T>
using Alias = decltype(std::declval<T>());

// as expected
using A1 = Alias<int>;
using A2 = Alias<int(int)>;

// error: no matching function for call to 'declval<...>()'
using A3 = Alias<int(int) const>;
using A4 = Alias<int(int) volatile>;
using A5 = Alias<int(int) &>;
using A6 = Alias<int(int) &&>;
// and all combinations of the above

cppreference doesn't seem to indicate that this error is expected.

Are there any other types for which declval<T> cannot be used? Where does the spec define these?

Putting attributes after function prototype type seems to throw an error. You may have to use `std::function` over the types: `Alias const>` — Nathanael Demacon, Oct 06 '19 at 11:46
@D.Nathanael That alters semantics. `std::function` is about type erasure. — L. F., Oct 06 '19 at 13:36
`std::function const` is the analog to `int(const*)(int)`, not to `int(*)(int) const` — Eric, Oct 06 '19 at 21:19

L. F. · Accepted Answer · 2019-10-06T12:04:21.033

Per [declval], the signature of declval is:

template <class T>
add_rvalue_reference_t<T> declval() noexcept;

Therefore, the call is ill-formed if add_rvalue_reference_t<T> cannot occur as a return type specifier.

Qualified function types have a special rule:

A function type with a cv-qualifier-seq or a ref-qualifier (including a type named by typedef-name ([dcl.typedef], [temp.param])) shall appear only as:

(6.1) the function type for a non-static member function,

(6.2) the function type to which a pointer to member refers,

(6.3) the top-level function type of a function typedef declaration or alias-declaration,

(6.4) the type-id in the default argument of a type-parameter, or

(6.5) the type-id of a template-argument for a type-parameter ([temp.arg.type]).

They cannot be a return type specifier.

Looking through Types, I'm pretty sure qualified function types are the only case.

For what T does `std::declval()` not have a matching function?

1 Answers1