有状态的元编程是病态的（还）吗？

Question

我有幸遇到的最 beloved/evil 的发明之一是 constexpr counter，又名状态元编程。正如 post 中提到的，它在 C++14 下似乎是合法的，我想知道 C++17 有什么变化吗？

以下是主要基于post

的实现

template <int N>
struct flag
{
    friend constexpr int adl_flag(flag<N>);
    constexpr operator int() { return N; }
};

template <int N>
struct write
{
    friend constexpr int adl_flag(flag<N>) { return N; }
    static constexpr int value = N;
};

template <int N, int = adl_flag(flag<N>{})>
constexpr int read(int, flag<N>, int R = read(0, flag<N + 1>{}))
{
    return R;
}

template <int N>
constexpr int read(float, flag<N>)
{
    return N;
}

template <int N = 0>
constexpr int counter(int R = write<read(0, flag<0>{}) + N>::value)
{
    return R;
}

我们use it作为

static_assert(counter() != counter(), "Your compiler is mad at you"); 

template<int = counter()>
struct S {};

static_assert(!std::is_same_v<S<>, S<>>, "This is ridiculous");

顺便说一下，这与 Storing States in C++ Metaprogramming?

直接矛盾

Answer 1

这是CWG active issue 2118:

Defining a friend function in a template, then referencing that function later provides a means of capturing and retrieving metaprogramming state. This technique is arcane and should be made ill-formed.

Notes from the May, 2015 meeting:

CWG agreed that such techniques should be ill-formed, although the mechanism for prohibiting them is as yet undetermined.

这仍然是一个活跃的问题，C++17 至少目前不会有任何改变。虽然当确定了这样的禁止机制时，这可能会被追溯为DR。

有状态的元编程是病态的（还）吗？

Is stateful metaprogramming ill-formed (yet)?

c++

metaprogramming

language-lawyer

c++17