Btukfyl

I have many EnableIf traits that basically check whether the input type satisfies an interface. I was trying to create a generic Resolve trait that can be used to transform those into a boolean trait.

Something like this - https://wandbox.org/permlink/ydEMyErOoaOa60Jx

template <

  template <typename...> class Predicate,

  typename T,

  typename = std::void_t<>>

struct Resolve : std::false_type {};

template <template <typename...> class Predicate, typename T>

struct Resolve<Predicate, T, Predicate<T>> : std::true_type {};

Now if you have an EnableIf trait like so

template <typename T>

using EnableIfHasFoo = std::void_t<decltype(std::declval<T>().foo())>;

You can create a boolean version of that very quickly

template <typename T>

struct HasFoo : Resolve<EnableIfHasFoo, T> {};

Or the analogous variable template.

But for some reason the partial specialization is not working as expected. Resolve does not work as intended. See the output here - https://wandbox.org/permlink/ydEMyErOoaOa60Jx. The same thing implemented "manually" works - https://wandbox.org/permlink/fmcFT3kLSqyiBprm

I am resorting to manually defining the types myself. Is there a detail with partial specializations and template template arguments that I am missing?

I cannot find the exact reason why your example don't work. If you want to dig more into the details of std::void_t, here's an interesting explanation

Even if I cannot explain it in depth, I would like to add another reliable syntax that is used in the detection idiom.

template<

    template <typename...> class Predicate,

    typename T,

    typename = void>

struct Resolve : std::false_type {};



template <template <typename...> class Predicate, typename T>

struct Resolve<Predicate, T, std::void_t<Predicate<T>>> : std::true_type {};



template <typename T>

using EnableIfHasFoo = decltype(std::declval<T>().foo());

live on compiler explorer

The reason why your approach will fail is that Predicate<T>> in the third template parameter is not a non-deduced context. This causes the deduction to directly fail (see [temp.alias]/2), instead of using the deduced template arguments from elsewhere as in a non-deduced context.

You can wrap your Predicate<T>> to a non-deduced context to make it work:

template<class T>

struct identity {

    using type = T;

};



template <template <typename...> class Predicate, typename T>

struct Resolve<Predicate, T, typename identity<Predicate<T>>::type> : std::true_type {};

Live Demo

Because inside a non-deduced context, the deduction won't happen for Predicate<T> part, instead, it will use the Predicate and T obtained from elsewhere.

As for why the usual detection-idiom (see Guillaume Racicot's answer) will work, it is because std::void_t as a template alias, will be replaced by void in deduction phase (see [temp.alias]/2), thus no deduction will happen.

Here are some examples to illustrate it more clearly:

template<class T>

using always_int = int;



template<template<class> class TT>

struct deductor {};



template<template<class> class TT, class T>

void foo(T, deductor<TT>) {}



template<template<class> class TT, class T>

void bar(T, deductor<TT>, TT<T>) {}



template<class T>

void baz(T, always_int<T>) {}



int main() {

    // ok, both T and TT are deduced

    foo(0, deductor<always_int>{});



    // ERROR, TT<T> is NOT a non-deduced context, deduction failure

    bar(0, deductor<always_int>{}, 0);



    // ok, T is deduced, always_int<T> is replaced by int so no deduction

    baz(0, 0);

}