In an attempt to rewrite a predicate combinator like this
auto constexpr all = [](auto const&... predicates){
return [predicates...](auto const&... x){
return (predicates(x...) && ...);
};
};
(little generalization of this) in a way that it would give meaningful errors when fed with non-predicates/predicates with different arities/arguments, I started writing something like this:
template<typename T, typename = void>
struct IsPredicate : public std::false_type {};
template<typename T>
struct IsPredicate<T, std::enable_if_t<std::is_same_v<bool, return_type_of_callable_T>, void>>
: public std::true_type {};
and then I stared at it for a while... How do I even check what is the return type of a function, if I don't even know how to call it?
I see this:
decltype(overloaded_predicate_function)
to IsPredicate
, because template type deduction can't occur with an overloaded name,operator()
, in case it is overloaded.So my question is: is it even possible to determine the return type of an arbitrary callable?
I'm mostly interested in a C++17 answer, but, why not?, I'd also like to know what C++20's concept offer in this respect.
So my question is: is it even possible to determine the return type of an arbitrary callable?
No. You can only do this in very narrow circumstances:
That's it. If you have a function object whose call operator is either overloaded or a template, you can't really figure out what its return type is. Its return type could depend on its parameter type, and you may not have a way of knowing what the parameter types could be. Maybe it's a call operator template that only accepts a few specific types that you have no way of knowing about, but it is a predicate for those types?
The best you can do is defer checking until you know what what arguments are. And then C++20 already has the concept for you (predicate):
inline constexpr auto all = []<typename... Ps>(Ps const&... predicates){
return [=]<typename... Xs>(Xs const&... x)
requires (std::predicate<Ps const&, Xs const&...> && ...)
{
return (std::invoke(predicates, x...) && ...);
};
};
Note that you should use std::invoke
to allow for pointers to members as predicates as well (and this is what std::predicate
checks for).