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From: gao_bolin on 6 Mar 2006 13:32
The following code doesn't compile with g++ 4.0.2:

--

#include <iostream>

template < typename T >
class A {};

template < typename T >
class B
{
public:
struct C;
};

template < typename T >
struct B<T>::C
{
void foo() { std::cout << "C" << std::endl; }
};

template < typename T >
struct B<A<T> >::C
{
void foo() { std::cout << "C spec" << std::endl; }
};

int main(int argc, char * argv[])
{
B<A<double> >::C c;
c.foo();
return 0;
}

--

it fails with the following message:
main.cpp:22: error: qualified name does not name a class before '{'
token
(line 22 is the line following "struct B<A<T> >::C").

Am I doing something wrong when partially specializing member class C?
Or is it something that is forbidden by the standard? If so, what could
be the reason?

Thx

G.


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From: amparikh on 7 Mar 2006 19:47

gao_bolin(a)voila.fr wrote:
> The following code doesn't compile with g++ 4.0.2:
>
> --
>
> #include <iostream>
>
> template < typename T >
> class A {};
>
> template < typename T >
> class B
> {
> public:
> struct C;
> };
>
> template < typename T >
> struct B<T>::C
> {
> void foo() { std::cout << "C" << std::endl; }
> };
>
> template < typename T >
> struct B<A<T> >::C
> {
> void foo() { std::cout << "C spec" << std::endl; }
> };
>
> int main(int argc, char * argv[])
> {
> B<A<double> >::C c;
> c.foo();
> return 0;
> }
>
> --
<snip>

You need to first specialize B for A<T> and then use specialization for
foo for that B.
Here.

template < typename T >
class A {};

template < typename T >
class B
{
public:
struct C;
};

template < typename T >
struct B<T>::C
{
void foo() { std::cout << "C" << std::endl; }
};

template<typename T>
class B < A<T> >
{
public:
struct C;
};

template < typename T >
struct B< A<T> >::C
{
void foo() { std::cout << "C spec" << std::endl; }
};

int main(int argc, char * argv[])
{
B<A<double> >::C c;
c.foo();
return 0;
}


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