std::invoke_result on std::tie

Question

I am using GCC 7.3 with C++17 and I do not understand why this line is failing:

template <typename... Args>
using X = std::invoke_result<std::tie, Args...>::type;

Error is:

error: type/value mismatch at argument 1 in template 
parameter list for ‘template<class _Functor, class ... _ArgTypes> 
struct std::invoke_result’
using X = std::invoke_result<std::tie, Args...>::type;
note:   expected a type, got ‘std::tie’

Answer 1

It's all in the error message:

note: expected a type, got 'std::tie'

invoke_result is a metafunction that takes a bunch of types . std::tie() is a function template - it's not a type. And it's not even an object, so you cannot do invoke_result<decltype(std::tie), Args...> either.

What invoke_result gives you is one syntax that works for all kinds of callables. But you don't need that with std::tie - it's a function template, so you can just directly invoke it in an unevaluated context:

template <typename... Args>
using X = decltype(std::tie(std::declval<Args>()...));

---

Note: Unless you really, specifically need the metafunction itself, just always use the _t alias. That is, std::invoke_result_t<...> rather than std::invoke_result<...>::type . The latter is wrong anyway, since you're missing the typename keyword - and the alias obviates that need.