Question

如何编写一个概念来描述基于范围的for循环的类型？

一次尝试是：

template < typename Range > concept bool RRange
    = requires(Range range) {{std::begin(range),std::end(range)};};

但我真正想要的是这样的事情：

template < typename Range > concept bool RRange
    = requires(Range range) {{for(auto&& item : range);};}; // compile error

即，RRange是表达式for(auto&& item : range);有效的所有类型的概念。实现这一目标的最佳方法是什么？

我正在使用带有g++ -std=c++1z -fconcepts的GCC7快照。

Answer 1

以下是我在审核[stmt.ranged]时提出的建议。

#include <utility>
#include <experimental/type_traits>

template <class T> using begin_non_mf_t = decltype(begin(std::declval<T>()));
template <class T> using begin_mf_t     = decltype(std::declval<T>().begin());
template <class T> using begin_t        = decltype(T::begin);
template <class T> using end_non_mf_t   = decltype(end(std::declval<T>()));
template <class T> using end_mf_t       = decltype(std::declval<T>().end());
template <class T> using end_t          = decltype(T::end);

template <class T>
constexpr bool has_member_begin_or_end {
    std::experimental::is_detected_v<begin_mf_t,T> ||
    std::experimental::is_detected_v<begin_t,T> ||
    std::experimental::is_detected_v<end_mf_t,T> ||
    std::experimental::is_detected_v<end_t,T>};

template <class T>
std::add_lvalue_reference_t<T>  declref() noexcept;
template <class T> using declref_t = decltype(declref<T>());

template <class T>
concept bool Range =
    requires /*Arrays*/ {
        requires std::is_array_v<T>;
        requires std::extent_v<T>!=0; // Extent is known.
    } ||
    /*Classes with member begin/end*/
    requires {
        requires std::is_class_v<T> && has_member_begin_or_end<T>;
    } &&
    requires (begin_mf_t<declref_t<T>> _begin,
                end_mf_t<declref_t<T>> _end) {
        { _begin!=_end } -> bool;
        { *_begin } -> auto&&;
        { ++_begin };
    } ||
    /*Types with non-member begin/end*/
    requires {
        requires !std::is_class_v<T> || !has_member_begin_or_end<T>;
    } &&
    requires (begin_non_mf_t<declref_t<T>> _begin,
                end_non_mf_t<declref_t<T>> _end) {
        { _begin!=_end } -> bool;
        { *_begin } -> auto&&;
        { ++_begin };
    };

和测试用例。

#include <vector>

// Evaluates to true or diagnoses which constraints failed.
template <Range> constexpr bool is_range {true};

static_assert(!Range<void>);
static_assert(!Range<int>);
static_assert(!Range<int*>);
static_assert(!Range<int[]>);
static_assert(is_range<int[1]>);
static_assert(is_range<std::vector<int>>);

struct A { };
struct B {
    int begin;
};
struct C {
    int* begin();
    int* end();
};
struct D { };
struct E {
    int end;
};
enum F { };
struct G {
    int* begin() &&;
    int* end();
};
struct H {
    int* begin() &&;
    int* end() &&;
};
int* begin(D);
int* end(D);
int* begin(E);
int* end(E);
int* begin(F);
int* end(F);
int* begin(H);
int* end(H);

static_assert(!Range<A>);
static_assert(!Range<B>);
static_assert(is_range<C>);
static_assert(is_range<D>);
static_assert(!Range<E>);
static_assert(is_range<F>);
static_assert(!Range<G>);
static_assert(!Range<H>);

int main() { }

Answer 2

根据P0587，这应该足够了：

#include <vector>

template<typename T>
concept bool RangeForAble = requires (T t) {
   requires requires (decltype(begin(t)) b, decltype(end(t)) e) {
     b != e;
     ++b;
     *b;
   };
 };

int main()
{
static_assert(RangeForAble<std::vector<int>>);
static_assert(RangeForAble<double>);
}

Answer 3

在 C++20 中，它看起来像这样：

template< class T >
concept RealContainer = requires(T&& t) {
  std::begin(std::forward<T>(t));
  std::end  (std::forward<T>(t));
};

可能并不完美，但适用于 std::vector 和 C 数组，演示：https://gcc.godbolt.org/z/M4xhnqG46

如何写一个简单的范围概念？

3 个答案: