Template instantiation does not "do inheritance"

Question

The title is quoted from this SO answer. It is discussing using SFINAE to detect the existence of a member function with the given signature and points out a failing of the method in the accepted answer when dealing with inherited member functions. In particular, the explanation given is as follows

If you are not already wise to this gotcha, then a look at of the definition of std::shared_ptr<T> in the header will shed light. In that implementation, std::shared_ptr<T> is derived from a base class from which it inherits operator*() const. So the template instantiation SFINAE<U, &U::operator*> that constitutes "finding" the operator for U = std::shared_ptr<T> will not happen, because std::shared_ptr<T> has no operator*() in its own right and template instantiation does not "do inheritance".

This snag does not affect the well-known SFINAE approach, using "The sizeof() Trick", for detecting merely whether T has some member function mf (see e.g. this answer and comments).

Using the terminology from the answer, what is the difference between using T::mf as a template argument to instantiate a type vs having the compiler determine it through a template function argument deduction? What does "template instantiation does not do inheritance" mean? And lastly, why doesn't this affect simply checking for existence of a member, like here?

score 3 · Accepted Answer · answered Aug 27 '14 at 17:20

Minimized example:

struct A {
    void a() const;
};

struct B : A {};

template<typename U, void (U::*)() const> struct SFINAE {};
template<typename U> void Test(SFINAE<U, &U::a>*) { }

int main(void)
{
    Test<B>(0); // doesn't compile
    return 0;
}

Demo.

The problem is that when B::a is inherited from A, the type of &B::a is actually "pointer to member of A" - and, while normally a pointer-to-member-of-base can be implicitly converted to pointer-to-member-of-derived, this conversion doesn't apply for non-type template arguments, per §14.3.2 [temp.arg.nontype]/p5:

The following conversions are performed on each expression used as a non-type template-argument. If a non-type template-argument cannot be converted to the type of the corresponding template-parameter then the program is ill-formed.

[...]

For a non-type template-parameter of type pointer to member function, if the template-argument is of type std::nullptr_t, the null member pointer conversion (4.11) is applied; otherwise, no conversions apply. If the template-argument represents a set of overloaded member functions, the matching member function is selected from the set (13.4).

Template instantiation does not "do inheritance"

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