如何定义递归概念?
cppreference.com 指出:
概念不能递归地引用自己
Concepts cannot recursively refer to themselves
但是我们如何定义一个概念来表示整数或整数向量，或整数向量的向量等.
But how can we define a concept that will represent an integer or a vector of integers, or a vector of vector of integers, etc.
我可以拥有这样的东西:
I can have something this:
template < typename Type > concept bool IInt0 = std::is_integral_v<Type>;
template < typename Type > concept bool IInt1 = IInt0<Type> || requires(Type tt) { {*std::begin(tt)} -> IInt0; };
template < typename Type > concept bool IInt2 = IInt1<Type> || requires(Type tt) { {*std::begin(tt)} -> IInt1; };
static_assert(IInt2<int>);
static_assert(IInt2<std::vector<int>>);
static_assert(IInt2<std::vector<std::vector<int>>>);
但我想要像 IIntX 这样的东西，这意味着任何 N 的 IIntN.
But I want to have something like IIntX that will mean IIntN for any N.
有可能吗?
推荐答案
概念总是可以遵从类型特征:
Concepts can always defer to a type trait:
template <typename T> concept C = some_trait<T>::value;
而且这个特征可以递归:
And that trait can be recursive:
template <typename T>
struct some_trait : std::false_type { };
template <std::Integral T>
struct some_trait<T> : std::true_type { };
template <typename T, typename A>
struct some_trait<std::vector<T, A>> : some_trait<T> { };
如果你的意思不只是vector，那么最后的部分特化可以推广为:
If you don't mean just vector, then the last partial specialization can be generalized to:
template <std::Range R>
struct some_trait<R> : some_trait<std::range_value_t<R>> { };
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