是否可以在C ++标准库中实现always_false? [英] Is it possible to implement always_false in the C++ standard library?
问题描述
在某些情况下，您可能会使用 always_false 助手，例如如果尝试实例化某些模板，则会导致无条件的 static_assert 失败:
There are cases where one uses an always_false helper to e.g. cause unconditional static_assert failure if instantiation of some template is attempted:
template <class... T> struct always_false : std::false_type {};
template<class T>
struct UsingThisShouldBeAnError {
static_assert(always_false<T>::value, "You should not use this!");
};
此帮助程序是必需的，因为模板定义必须(至少在理论上)具有至少一组模板参数，可以为其生成有效的专业名称，以使程序格式正确:
This helper is necessary because a template definition must (at least theoretically) have at least one set of template parameters for which a valid specialization can be produced in order for the program to be well-formed:
[temp.res]/8 :程序不正确-如果满足以下条件，则无需进行诊断:
[temp.res]/8: The program is ill-formed, no diagnostic required, if:
- 无法为模板生成有效的专业化，并且模板没有实例化，或者
[...]
(编写 static_assert(false，``您不应该使用此！''实例化，这不是故意的.)
(Writing static_assert(false, "You should not use this!"); above would thus be ill-formed and a compiler could always fire the static assert, even without the template being instantiated, which is not the intention.)
以下是涉及此模式的问题的快速样本(包括进一步的解释):
Here is a quick sampling of questions involving this pattern (including further explanation):
将 always_false 作为标准库中的工具可能会很有用，因此我们不必不断地编写它.但是，以下问题的答案让我怀疑这是否可能:
It might be useful to have always_false as a tool in the standard library so we don't have to constantly write it again. However, the answer to the following question makes me wonder whether this is even possible:
有一个论点(也针对[temp.res]/8)，即 std :: enable_if_t< T> 始终是 void 或不是类型，并且任何人进一步对其进行专门化都是非法的.因此，依赖于理论上的专业性"的模板被称为模板". std :: enable_if 的代码，以避免[temp.res]/8子句实际上导致程序格式错误，无需诊断.
There the argument is made (also with respect to [temp.res]/8) that std::enable_if_t<T> is always either void or not a type and that it is illegal for anyone to specialize it further. Therefore, a template that relies on the theoretical "specializability" of std::enable_if to avoid the [temp.res]/8 clause actually causes the program to be ill-formed, no diagnostic required.
回到我的问题:如果该标准提供了 always_false ，它将不得不禁止图书馆用户照常对其进行专业化(出于明显的原因).但是根据上述推理，这会破坏 always_false 的全部要点(即，理论上它可以专门用于 std :: false_type 之外的其他事物)-关于[temp.res]/8，与直接使用 std :: false_type 的效果相同.
Coming back to my question: If the standard provided always_false, it would have to forbid library users from specializing it as usual (for obvious reasons). But by the above reasoning, that would defeat the whole point of always_false (namely that it could theoretically be specialized to something other than std::false_type) - with respect to [temp.res]/8 it would be the same as using std::false_type directly.
我在这个推理上错了吗?还是标准库实际上不可能以有意义/有用的方式(无需更改核心语言)提供 always_false 吗?
Am I wrong in this reasoning? Or is it actually impossible for the standard library to provide always_false in a meaningful/useful way (without core language changes)?
推荐答案
释义Jarod的想法，可能是
To paraphrase Jarod's idea, It could be something like
template <class... T> struct always_false : std::false_type {};
template <> struct always_false</* implementation defined */> : std::true_type{};
/*实现定义的*/可以由 std :: __ ReservedIdentifer 填充.用户代码无法访问它，因为标识符是保留给库的，但是存在一种特殊化，即 true .那应该避免对专业化中的ODR和lambda产生疑问.
Where /* implementation defined */ can be filled by std::_ReservedIdentifer. User code can't access it, since the identifier is reserved to the library, but there exists a specialization that is true. That should avoid questions about the ODR and lambdas in specializations.
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