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How do I specialize a static member of a template class on a templated type?

开发者 https://www.devze.com 2023-01-20 14:48 出处:网络
Say I have the f开发者_开发百科ollowing class: template<class T> struct A { static int value;

Say I have the f开发者_开发百科ollowing class:

template<class T>
struct A
{
    static int value;
};

template<class T>
int A<T>::value = 0;

I can specialize A::value on a concrete type without problems:

struct B
{
};

template<>
int A<B>::value = 1;

I'd like to specialize A::value on a template type, I tried the following:

template<class T>
struct C
{
};

// error: template definition of non-template 'int A<C<T> >::value'
template<>
template<class T>
int A<C<T> >::value = 2;

Is there any way to do this or is it only possible to specialize A::value on non-template types?


Instead of introducing a whole explicit specialization, you could just specialize the initialization

template<class T>
struct Value {
  static int const value = 0;
};

template<class T>
struct Value< C<T> > {
  static int const value = 2;
};

template<class T>
int A<T>::value = Value<T>::value;


You can use partial specialization of A via C:

#include <iostream>

using namespace std;

template<class T>
struct A
{
    static int value;
};

template<class T>
int A<T>::value = 0;

//(1) define the C template class first:
template<class T>
struct C
{
};

//(2) then define the partial specialization of A, in terms of C:
template<typename T>
struct A<C<T> > 
{
    static int value;
};

template<typename T>
int A<C<T> >::value = 2;

int main(void)
{
    cout<<A<C<int> >::value<<endl;

    cout<<"ok!"<<endl;
    return 0;
}


Partial specialization via a template-template argument (see my comment above):

#include <iostream>

using namespace std;

template<class T>
struct A
{
    static int value;
};

template<class T>
int A<T>::value = 0;



//solution 2:
//define a template-template argument partial specialization
//for any generic class U depending on a template argument,
//(which is a typename, in this case, but there's no reason why
//you wouldn't define specializations for non-types, etc.)
//this specialization has the advantage of not depending on
//the template definition of U (or C, etc.); in this case
//both T and U are free to vary, with the only constraint
//that U is parameterized by T:
template<typename T, template<typename> class U>
struct A<U<T> >
{
    static int value;
};

template<typename T, template<typename> class U>
int A<U<T> >::value = 3;

//define the C template class, somewhere, where/when/if needed
//possibly in a different namespace, "module" (etc.)
template<class T>
struct C
{
};

int main(void)
{
    cout<<A<C<int> >::value<<endl;//this now should print out: 3

    cout<<"ok!"<<endl;
    return 0;
}
0

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