Bsrgvty

I've been very surprised to see that it is not necessary to add typename when a dependent type appears as a base class:

struct B ;

struct wr
 typedef B type; ;

template<class T>
struct A : T::type
;

int main()

 A<wr> a;
 (void)a;

Why isn't typename required in front of T::type?

It's a special case, as others noted. To quote the standard on this:

[temp.res]

5 A qualified name used as the name in a class-or-decltype or an
elaborated-type-specifier is implicitly assumed to name a type,
without the use of the typename keyword. In a nested-name-specifier
that immediately contains a nested-name-specifier that depends on a
template parameter, the identifier or simple-template-id is implicitly
assumed to name a type, without the use of the typename keyword.
[ Note: The typename keyword is not permitted by the syntax of these
constructs.  — end note ]

And come C++20, there will be be even more exceptions to the need for typename.

Why isn't typename required in front of T::type?

Because you cannot inherit from a value. You use typename to tell the compiler that a given nested identifier is a type, but for inheritance, that must be the case anyhow so you can omit it - that's why the language provides an exception to the typename- rule for base-specifiers. From cppreference (emphasis mine):

The typename disambiguator for dependent names

In a declaration or a definition of a template, including alias template, a name that is not a member of the current instantiation and is dependent on a template parameter is not considered to be a type unless the keyword typename is used or unless it was already established as a type name, e.g. with a typedef declaration or by being used to name a base class.

Note that we will get more places where typename can be omitted, see P0634.

You only need to use typename if you need to tell the compiler to expect a type rather than something else.

Since only a type can be inherited from, there is no ambiguity, and so typename is superfluous.