无法推断概念中的占位符类型 [英] Unable to deduce placeholder type in concept

问题描述

我正在尝试使用GCC 8中的Concepts TS复制标准C ++ 20概念，以便在标准库中可用它们之前就可以使用它们。我主要复制了最新草案中的所有内容，然后遇到了一个问题：

I'm trying to replicate standard C++20 concepts using Concepts TS in GCC 8 so I can use them before they are available in standard library. I mostly copy pasted everything from the latest draft and I came across a problem:

#include <type_traits>
#include <utility>

// [concept.same]
template <typename T, typename U>
concept bool Same = std::is_same_v<T, U>;

// [concept.assignable]

// TODO: Proper implementation requires std::common_reference that is not in
// libstdc++ yet and implementing it myself is too hard.
template <typename LHS, typename RHS>
concept bool Assignable = std::is_lvalue_reference_v<LHS> &&
    requires(LHS lhs, RHS&& rhs)
    {
        {lhs = std::forward<RHS>(rhs)} -> Same<LHS>;
    };

template <typename T>
    requires Assignable<T&, T>
void Test(T a) {}

int main()
{
    Test(42);
}

很多其他概念需要可分配的类型，当尝试使用此概念时，我得到：

A lot of other concepts require assignable types and when trying to use this concept I get:

Concepts.h:54:14: note: within 'template<class LHS, class RHS> concept const bool ftz::General::Assignable<LHS, RHS> [with LHS = int&; RHS = int]'
 concept bool Assignable = std::is_lvalue_reference_v<LHS> &&
              ^~~~~~~~~~
Concepts.h:54:14: note:     with 'int& lhs'
Concepts.h:54:14: note:     with 'int&& rhs'
Concepts.h:54:14: note: unable to deduce placeholder type 'ftz::General::Same<int&>' from 'lhs =(forward<int>)(rhs)'

这里出了什么问题？

推荐答案

这是由于 P1084 。问题是它曾经是：

This was a recent change made to Concepts that was adopted in the San Diego (November 2018) as a result of P1084. The problem is that it used to be that:

{ E } -> Same<T>;

实际上是指 f（E）对于以下形式的发明函数模板有效：

Actually meant that the expression f(E) is valid for an invented function template of the form:

template <class U> requires Same<U, T> void f(U );

显然，对于引用类型 T （与OP中一样）。

Which is obviously never going to hold for reference types T (as in the OP).

换句话说，旧规则是： {E}->相同的 表示相同的，T> 。新规则是指 Same< decltype（（E）），T> 。似乎gcc的 -fconcepts 和clang的concept分支都尚未实现这些新规则。

In other words, the old rule was that: { E } -> Same<T> meant Same<remove_cvref_t<decltype((E))>, T>. The new rule is that it means Same<decltype((E)), T>. It appears that neither gcc's -fconcepts nor clang's concepts branch implements these new rules yet.

当前的解决方法是更改​​：

A current workaround is to change:

{ E } -> Same<LHS> // i.e. Same<T&>

至：

{ E } -> Same<std::remove_reference_t<LHS>>& // i.e. Same<T>&

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