[std-discussion] The access for a member class template in the deduction guide.

['Vlad from Moscow' via ISO C++ Standard - Discussion]

In the section 17.10 Deduction guides there is written (p.#3)

...A deduction-guide shall be declared in the same scope as the

corresponding class template and, for a member class template, *with the
same access*....

It is not clear whar access is meant. Does it mean that a deduction guide
can be declared in a class scope?

[Casey Carter]

Yes, a deduction guide for a nested class template can be declared at class
scope:

template<class T> struct X {
template<class U> struct Y {
template<class V> Y(V) {}
};

template<class W> Y(W) -> Y<W>;
};

int main() {
X<int>::Y{42}; // deduces X<int>::Y<int>
}


Per C++17 [class.mem], deduction-guide is one of the allowed replacements
for member-declaration in the grammar. (Although GCC apparently didn't get
the memo: https://godbolt.org/z/aJcWfr.)

Post by 'Vlad from Moscow' via ISO C++ Standard - Discussion
In the section 17.10 Deduction guides there is written (p.#3)
...A deduction-guide shall be declared in the same scope as the

corresponding class template and, for a member class template, *with the
same access*....

It is not clear whar access is meant. Does it mean that a deduction guide
can be declared in a class scope?

[Casey Carter]

Post by Casey Carter
Yes, a deduction guide for a nested class template can be declared at
template<class T> struct X {
template<class U> struct Y {
template<class V> Y(V) {}
};
template<class W> Y(W) -> Y<W>;
};
int main() {
X<int>::Y{42}; // deduces X<int>::Y<int>
}
Per C++17 [class.mem], deduction-guide is one of the allowed replacements
for member-declaration in the grammar. (Although GCC apparently didn't get
the memo: https://godbolt.org/z/aJcWfr.)

This is GCC PR 79501 (https://gcc.gnu.org/bugzilla/show_bug.cgi?id=79501).

['Vlad from Moscow' via ISO C++ Standard - Discussion]

Then sjould the bolded word *access* be appended with the word *control in
the quote*?

Post by Casey Carter

Post by Casey Carter
Yes, a deduction guide for a nested class template can be declared at
template<class T> struct X {
template<class U> struct Y {
template<class V> Y(V) {}
};
template<class W> Y(W) -> Y<W>;
};
int main() {
X<int>::Y{42}; // deduces X<int>::Y<int>
}
Per C++17 [class.mem], deduction-guide is one of the allowed replacements
for member-declaration in the grammar. (Although GCC apparently didn't get
the memo: https://godbolt.org/z/aJcWfr.)

This is GCC PR 79501 (https://gcc.gnu.org/bugzilla/show_bug.cgi?id=79501).