如何静态断言元组的所有类型是否满足某些条件？

Question

我有一些类型特征SomeTraits，可以T通过 从中提取类型是否满足某些条件SomeTraits<T>::value。如何检查给定的所有类型std::tuple<>并检查（通过静态断言）它们是否都满足上述条件？例如

using MyTypes = std::tuple<T1, T2, T3>;
// Need some way to do something like
static_assert(SomeTupleTraits<MyTypes>::value, "MyTypes must be a tuple that blabla...");

在哪里SomeTupleTraits检查SomeTraits<T>::value == true里面的每种类型是否MyTypes？

我仅限于 C++14。

Answer 1

作为单行（换行符可选），您可以执行以下操作：

// (c++20)
static_assert([]<typename... T>(std::type_identity<std::tuple<T...>>) {
    return (SomeTrait<T>::value && ...);
}(std::type_identity<MyTypes>{}));

或者您可以创建一个辅助特征来执行此操作：

// (c++17)
template<template<typename, typename...> class Trait, typename Tuple>
struct all_of;

template<template<typename, typename...> class Trait, typename... Types>
struct all_of<Trait, std::tuple<Types...>> : std::conjunction<Trait<Types>...> {};

static_assert(all_of<SomeTrait, MyTypes>::value);

或者在 C++11 中，您可以std::conjunction在 helper 特征内重新实现：

template<template<typename, typename...> class Trait, typename Tuple>
struct all_of;

template<template<typename, typename...> class Trait>
struct all_of<Trait, std::tuple<>> : std::true_type {};

template<template<typename, typename...> class Trait, typename First, typename... Rest>
struct all_of<Trait, std::tuple<First, Rest...>> :
    std::conditional<bool(Trait<First>::value),
                     all_of<Trait, std::tuple<Rest...>>,
                     std::false_type>::type::type {};

static_assert(all_of<SomeTrait, MyTypes>::value, "");