Access control in template parameters

问题

Inspired from this answer, which claims to subvert the access control system, I wrote the following minimal version of the hack

template<typename T>
inline T memptr{};

template<auto Val>
struct setptr
{
    struct setter { setter() { memptr<decltype(Val)> = Val; } };
    static setter s;
};

template<auto Val>
typename setptr<Val>::setter setptr<Val>::s{};

which is then used as

class S
{
    int i;
};

template struct setptr<&S::i>;

auto no_privacy(S& s)
{
    return s.*memptr<int S::*>;
}

Why doesn't template struct setptr<&S::i>; violate access control?

Is it because [class.access]

Access control is applied uniformly to all names, whether the names are referred to from declarations or expressions.

specifically doesn't include instantiations? In which case, why doesn't it include instantiations?

_{Errata: Explicit instantiations are also classified as declarations.}

回答1:

From [temp.spec]/6 (emphasis mine):

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

So this technique you see abuses this rule, which is primarely there to allow implementers of a class specialize templates (such as traits) with private types or other private entities

来源：https://stackoverflow.com/questions/54909496/access-control-in-template-parameters