I have carefully read many answers concerning this topic, but nevertheless I cannot figure out EXACTLY when these two keywords ARE or AREN'T needed in the scope of a non-template function which is member of a nested template class.
My reference compilers are GNU g++ 4.9.2 and clang 3.5.0.
They behave hardly different on the following code where I put embedded comments trying to explain what happens.
#include <iostream>
// a simple template class with a public member template struct
template <class Z>
class Pa
{
// anything
public:
template <class U>
struct Pe // a nested template
{
// anything
void f(const char *); // a non-template member function
};
template <class U> friend struct Pe;
};
// definition of the function f
template <class AAA>
template <class BBB>
void Pa<AAA> :: Pe<BBB> :: f(const char* c)
{
Pa<AAA> p; // NO typename for both clang and GNU...
// the following line is ACCEPTED by both clang and GNU
// without both template and typename keywords
// However removing comments from typename only
// makes clang still accepting the code while GNU doesn't
// accept it anymore. The same happens if the comments of template
// ONLY are removed.
//
// Finally both compilers accept the line when both typename AND
// template are present...
/*typename*/ Pa<AAA>::/*template*/ Pe<BBB> q;
// in the following clang ACCEPTS typename, GNU doesn't:
/*typename*/ Pa<AAA>::Pe<int> qq;
// the following are accepted by both compilers
// no matter whether both typename AND template
// keywords are present OR commented out:
typename Pa<int>::template Pe<double> qqq;
typename Pa<double>::template Pe<BBB> qqqq;
std::cout << c << std::endl; // just to do something...
}
int main()
{
Pa<char>::Pe<int> pp;
pp.f("bye");
}
So, in the scope of f
is Pa<double>::Pe<BBB>
a dependent name or not?
And what about Pa<AAA>::Pe<int>
?
And, after all, why this different behaviour of the two quoted compilers?
Can anyone clarify solving the puzzle?
The important rule in [temp.res] is:
When a qualified-id is intended to refer to a type that is not a member of the current instantiation (14.6.2.1) and its nested-name-specifier refers to a dependent type, it shall be prefixed by the keyword
typename
, forming a typename-specifier . If the qualified-id in a typename-specifier does not denote a type, the program is ill-formed.
The question revoles around two qualified-id s:
Pa<double>::Pe<BBB>
Pa<AAA>::Pe<int>
First, what is a dependent type? According to [temp.dep.type]:
A type is dependent if it is
— a template parameter,
— a member of an unknown specialization,
— a nested class or enumeration that is a dependent member of the current instantiation,
— a cv-qualified type where the cv-unqualified type is dependent,
— a compound type constructed from any dependent type,
— an array type whose element type is dependent or whose bound (if any) is value-dependent,
— a simple-template-id in which either the template name is a template parameter or any of the template arguments is a dependent type or an expression that is type-dependent or value-dependent, or
— denoted bydecltype
( expression ), where expression is type-dependent (14.6.2.2).
Pa<double>
(the nested-name-specifier of the first example) is not a dependent type, as it fits none of the bullet points. Since we don't meet that criteria, we don't need to prefix the typename
keyword.
Pa<AAA>
, however, is a dependent type since it is a simple-template-id in which one of the template arguments is a dependent type ( AAA
is trivially a dependent type since it is a template parameter).
What, then, is "a member of the current instantiation"?
A name refers to the current instantiation if it is
— [...]
— in the definition of a primary class template or a member of a primary class template, the name of the class template followed by the template argument list of the primary template (as described below) enclosed in <> (or an equivalent template alias specialization)" — in the definition of a nested class of a class template, the name of the nested class referenced as a member of the current instantiation, or
The current instantiation, in this case, is Pa<AAA>
(or, also, Pa
). And:
A name is a member of the current instantiation if it is [...] A qualified-id in which the nested-name-specifier refers to the current instantiation and that, when looked up, refers to at least one member of a class that is the current instantiation or a non-dependent base class thereof.
So Pe
is a member of the current instantiation. Thus, while the nested-name-specifier of Pa<AAA>::Pe<int>
is a dependent type, it is a type that is a member of the current instantiation, so you do not need the keyword typename
. Note that Pa<AAA>::Pe<int>
is a dependent type itself (it's a nested class that is a dependent member of the current instantiation), but that in itself does mean that the typename
keyword is required.
The fact that gcc doesn't accept typename here:
/*typename*/ Pa<AAA>::Pe<int> qq;
because it wants
typename Pa<AAA>::template Pe<int> qq;
is a bug.
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