问题描述
在模板特化中，我有一个带有 enable_if 参数的模板参数，该参数导致 enable_if 没有类型"成员，因此模板特化应该失败，但不会产生错误:
In a template specialization I have a template argument with an enable_if with parameter that results in the enable_if not having a 'type' member, and so that template specialization should fail, but not create an error:
#include <type_traits>
template <typename value_t_arg, typename T = void>
struct underlyingtype
{
using underlyingtype_t = value_t_arg;
};
template <typename value_t_arg>
struct underlyingtype < value_t_arg, typename std::enable_if<false>::type>
// std::enable_if<false> has no 'type' member, and so substitution should fail,
// but no create an error, right?
{
//using underlyingtype_t = value_t_arg::integral_t;
};
为什么会在此处创建错误?
Why is there an error created here?
推荐答案
您的代码格式错误(无需诊断)，因为无论模板参数如何，条件始终为假，这意味着特化对于每个可能的模板参数.
Your code is ill-formed (no diagnostic required) because the condition is always false regardless of the template argument, meaning the specialization would be ill-formed for every possible template argument.
程序格式错误，无需诊断，如果:
The program is ill-formed, no diagnostic required, if:
——不能为模板生成有效的特化......并且模板没有被实例化，......
— no valid specialization can be generated for a template ... and the template is not instantiated, ...
部分专业化似乎算作模板"；就本节而言.
Partial specializations appear to count as "templates" for the purposes of this section.
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