Let's say I have two structs, Foo
and Bar
:
template<int...>
struct Foo{};
template<unsigned long...>
struct Bar{};
I want to create a type trait (call it match_class
) that returns true if I pass two Foo<...>
types or two Bar<...>
types, but false if I try to mix them:
int main()
{
using f1 = Foo<1, 2, 3>;
using f2 = Foo<1>;
using b1 = Bar<1, 2, 3>;
using b2 = Bar<1>;
static_assert(match_class<f1, f2>::value, "Fail");
static_assert(match_class<b1, b2>::value, "Fail");
static_assert(!match_class<f1, b1>::value, "Fail");
}
For C++1z (clang 5.0.0 and gcc 8.0.0) it's sufficient to do this (Demo):
template<class A, class B>
struct match_class : std::false_type{};
template<class T, template<T...> class S, T... U, T... V>
struct match_class<S<U...>, S<V...>> : std::true_type{};
But in C++14 I get the following error (same compilers*Demo):
error: class template partial specialization contains a template parameter that cannot be deduced; this partial specialization will never be used [-Wunusable-partial-specialization]
struct match_class<S<U...>, S<V...>> : std::true_type{};
^~~~~~~~~~~~~~~~~~~~~~~~~~~~~
note: non-deducible template parameter 'T'
template<class T, template<T...> class S, T... U, T... V>
Ideally the syntax for testing the type trait should remain the same.
Secondary question: Is the behavior for C++14 correct? (or alternatively is the behavior I see for C++17 unspecified?)
*Note that MSVC 19.00.23506 has the same kind of failure Demo
In C++14, you could not deduce T
in:
template<class T, template<T...> class S, T... U, T... V>
struct match_class<S<U...>, S<V...>> : std::true_type{};
but in C++17, you can. The behavior you see is correct.
In C++14, since you cannot deduce T
, you need a way of explicitly providing it. So you might require the class templates themselves to indicate their non-type template parameter type:
template <int...> struct Foo { using type = int; };
template <unsigned long...> struct Bar { using type = unsigned long; };
Or have an external trait for this. And then explicitly write out everything - two class templates match if they have the same non-type template parameter and then also have the same class template, in that order:
template <class... Ts> struct make_void { using type = void; };
template <class... Ts> using void_t = typename make_void<Ts...>::type;
template <class T1, class T2, class A, class B>
struct match_class_impl : std::false_type { };
template <class T, template <T...> class S, T... U, T... V>
struct match_class_impl<T, T, S<U...>, S<V...>> : std::true_type{};
template <class A, class B, class=void>
struct match_class : std::false_type { };
template <class A, class B>
struct match_class<A, B, void_t<typename A::type, typename B::type>>
: match_class_impl<typename A::type, typename B::type, A, B>
{ };
This is a consequence of adding support for template auto
. In C++14, [temp.deduct.type] contained:
A template type argument cannot be deduced from the type of a non-type template-argument. [Example:
template<class T, T i> void f(double a[10][i]); int v[10][20]; f(v); // error: argument for template-parameter T cannot be deduced
-end example]
But in C++17, it now reads:
When the value of the argument corresponding to a non-type template parameter
P
that is declared with a dependent type is deduced from an expression, the template parameters in the type ofP
are deduced from the type of the value. [ Example:template<long n> struct A { }; template<typename T> struct C; template<typename T, T n> struct C<A<n>> { using Q = T; }; using R = long; using R = C<A<2>>::Q; // OK; T was deduced to long from the // template argument value in the type A<2>
— end example ] The type of
N
in the typeT[N]
isstd::size_t
. [ Example:template<typename T> struct S; template<typename T, T n> struct S<int[n]> { using Q = T; }; using V = decltype(sizeof 0); using V = S<int[42]>::Q; // OK; T was deduced to std::size_t from the type int[42]
— end example ]
Question: What is a workaround for this in C++14?
A possible workaround in C++14 is based on traits.
As a minimal, working example (maybe even stupid, but it helps getting the idea):
#include <type_traits>
#include <utility>
template<int...>
struct Foo{};
template<unsigned long...>
struct Bar{};
template<typename>
struct traits;
template<int... V>
struct traits<Foo<V...>> { using type = Foo<0>; };
template<unsigned long... V>
struct traits<Bar<V...>> { using type = Bar<0>; };
template<typename T, typename U>
constexpr bool match = std::is_same<typename traits<T>::type, typename traits<U>::type>::value;
int main() {
using f1 = Foo<1, 2, 3>;
using f2 = Foo<1>;
using b1 = Bar<1, 2, 3>;
using b2 = Bar<1>;
static_assert(match<f1, f2>, "Fail");
static_assert(match<b1, b2>, "Fail");
static_assert(!match<f1, b1>, "Fail");
}
As a side note, in C++17 you can simplify things up as it follows:
template<template<auto ...> class S, auto... U, auto... V>
struct match_class<S<U...>, S<V...>> : std::true_type{};
About the reasons that are behind the error, @Barry's answer contains all what you need to understand it (as usual).
Here is a general C++14 solution that does not rely on manually specialized type traits or extending Foo
and Bar
.
A template metafunction that obtains a type representing the class template of its argument type:
namespace detail
{
// Type representing a class template taking any number of non-type template arguments.
template <typename T, template <T...> class U>
struct nontype_template {};
}
// If T is an instantiation of a class template U taking non-type template arguments,
// this has a nested typedef "type" that is a detail::nontype_template representing U.
template <typename T>
struct nontype_template_of {};
// Partial specializations for all of the builtin integral types.
template <template <bool...> class T, bool... Vs>
struct nontype_template_of<T<Vs...>> { using type = detail::nontype_template<bool, T>; };
template <template <char...> class T, char... Vs>
struct nontype_template_of<T<Vs...>> { using type = detail::nontype_template<char, T>; };
template <template <signed char...> class T, signed char... Vs>
struct nontype_template_of<T<Vs...>> { using type = detail::nontype_template<signed char, T>; };
template <template <unsigned char...> class T, unsigned char... Vs>
struct nontype_template_of<T<Vs...>> { using type = detail::nontype_template<unsigned char, T>; };
template <template <short...> class T, short... Vs>
struct nontype_template_of<T<Vs...>> { using type = detail::nontype_template<short, T>; };
template <template <unsigned short...> class T, unsigned short... Vs>
struct nontype_template_of<T<Vs...>> { using type = detail::nontype_template<unsigned short, T>; };
template <template <int...> class T, int... Vs>
struct nontype_template_of<T<Vs...>> { using type = detail::nontype_template<int, T>; };
template <template <unsigned int...> class T, unsigned int... Vs>
struct nontype_template_of<T<Vs...>> { using type = detail::nontype_template<unsigned int, T>; };
template <template <long...> class T, long... Vs>
struct nontype_template_of<T<Vs...>> { using type = detail::nontype_template<long, T>; };
template <template <unsigned long...> class T, unsigned long... Vs>
struct nontype_template_of<T<Vs...>> { using type = detail::nontype_template<unsigned long, T>; };
template <template <long long...> class T, long long... Vs>
struct nontype_template_of<T<Vs...>> { using type = detail::nontype_template<long long, T>; };
template <template <unsigned long long...> class T, unsigned long long... Vs>
struct nontype_template_of<T<Vs...>> { using type = detail::nontype_template<unsigned long long, T>; };
An alias template for ease of use:
// Alias template for nontype_template_of.
template <typename T>
using nontype_template_of_t = typename nontype_template_of<T>::type;
Then you can implement your match_class
trait just like this:
template <class A, class B>
struct match_class : std::is_same<nontype_template_of_t<A>, nontype_template_of_t<B>> {};
DEMO
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