[Moderator's note: this defect report has been
 forwarded to the C++ committee. -moderator(fjh).]

Section: 14.8.2.4/4 [temp.deduct.type]:
"If a type is specified as void f(A<T>::B, A<T>), the T in A<T>::B is
nondeduced but the T in A<T> is deduced."

As far as T is supposed to be a template parameter then void f(A<T>::B,
A<T>) declaration is ill-formed because  14.6/2 that says that A<T>::B is
not a type name.

Suggested resolution: Change f's declaration to the following: void
f(typename A<T>::B, A<T>).

With regards,
Michael Kochetkov.
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