In this question I unsuccessfully asked how to use different pimpl implementation depending on a template argument.
Maybe this example ilustrates better what I am trying to do :
#include <iostream>
template< int N, typename T >
struct B
{
B() : c( new C< N > )
{}
template< int M >
struct C;
C< N > *c;
};
template< int N, typename T >
template< int M >
struct B< N, T >::C
{
int a[M];
};
// version 1 that doesn't work
template< int N, typename T >
template< >
struct B< N, T >::C< 0 >
{
int a;
};
// version 2 that doesn't work
template< typename T >
template< int M >
struct B< 0, T >::C
{
int a;
};
int main()
{
B< 0, float > b0;
B< 1, int > b1;
std::cout << "b0 = " << sizeof(b0.c->a) << std::endl;
std::cout << "b1 = " << sizeof(b1.c->a) << std::endl;
}
It still fails if I try to specialize the struct C (the above doesn't compile)
So, is it possible to do?
I know a work around like this :
template< int M >
struct D
{
int a[M];
};
template< &g开发者_JS百科t;
struct D<0>
{
int a;
};
template< int N, typename T >
template< int M >
struct B< N, T >::C
{
D< M > helper;
};
but if possible, I would like to avoid it
What you're trying to do is not allowed by the language.
§ 14.7.3.16 (FCD 2010-03-26) states:
In an explicit specialization declaration for a member of a class template or a member template that appears in namespace scope, the member template and some of its enclosing class templates may remain unspecialized, except that the declaration shall not explicitly specialize a class member template if its enclosing class templates are not explicitly specialized as well. In such explicit specialization declaration, the keyword template followed by a template-parameter-list shall be provided instead of the template<> preceding the explicit specialization declaration of the member. The types of the template-parameters in the template-parameter-list shall be the same as those specified in the primary template definition.
[ Example:
template <class T1> class A {
template<class T2> class B {
template<class T3> void mf1(T3);
void mf2();
};
};
template <> template <class X>
class A<int>::B {
template <class T> void mf1(T);
};
template <> template <> template<class T>
void A<int>::B<double>::mf1(T t) { }
template <class Y> template <>
void A<Y>::B<double>::mf2() { } // ill-formed; B<double> is specialized but
// its enclosing class template A is not
—end example ]
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