为什么函数模板不允许使用私有类型的显式特化？

Question

我不明白这背后的原因：为什么 class 模板允许私有类型的显式特化而不是函数模板？

假设我们有一个 class:

class pepe
{
    struct lolo
    {
        std::string name = "lolo";
    };
public:
    static lolo get()
    {
        return {};
    }
};

Class 模板可以明确特化。
并且函数模板在隐式实例化时没有问题。
尽管您无法创建 spec_class<pepe::lolo>{}，因为 pepe::lolo 无法访问。

template <typename>
struct spec_class
{};

// this is ok
template <>
struct spec_class<pepe::lolo>
{};

// this will be ok also upon implicit instantiation
template <typename T>
void template_func(const T &t)
{
    std::cout << "implicit: " << t.name << std::endl;
}

但是：

// this is not ok!
// template <>
// void template_func(const pepe::lolo &p)
// {
//     std::cout << "explicit: " << p.name << std::endl;
// }

// this is ok, but more or less understandable why
template <>
void template_func(const decltype(pepe::get()) &p)
{
    std::cout << "explicit: " << p.name << std::endl;
}

// not ok, but expected
// void func(const pepe::lolo&)
// {}

所以： 为什么函数模板禁止显式特化？

Answer 1

从 C++20 开始，在函数模板特化的参数中使用私有成员是完全有效的，因为 PR0692. In particular, the following wording was added to temp.spec.general#6:

The usual access checking rules do not apply to names in a declaration of an explicit instantiation or explicit specialization, with the exception of names appearing in a function body, default argument, base-clause, member-specification, enumerator-list, or static data member or variable template initializer.

[ Note 1: In particular, the template arguments and names used in the function declarator (including parameter types, return types and exception specifications) can be private types or objects that would normally not be accessible. - end note]

（强调我的）

代码未编译是由于 GCC bug 97942, and it compiles just fine in Clang. demo。

为什么函数模板不允许使用私有类型的显式特化？

Why isn't explicit specialization with private type allowed for function templates?

c++

language-lawyer