To be more specific why
std::is_assignable_v<int, int> << '\n';
returns false? Is it because an int has no overloaded assignment operator (being a primitive type and all)?
(by the way std::is_trivially_assignable_v<int, int> gives false too.)
Note that this:
struct Structure {};
std::is_assignable<class Structure, class Structure>::value;
would return true, because an overloaded assignment operator is implicitly defined for Structure.
Am i correct so far? If so then I suppose it wouldn't be trivial to enhance is_assignable to accept primitive types as well? Otherwise, any hints for such a possible work-around?
If the expression
std::declval<T>() = std::declval<U>()is well-formed in unevaluated context
std::is_assignable<int, int>::value << '\n' // 1 = 1; wouldn't compile