Reputation: 1334
#include <iostream>
template <typename T1, typename T2>
class B{
public:
void update(){ std::cerr<<__PRETTY_FUNCTION__<<std::endl; }
void func1(){ std::cerr<<__PRETTY_FUNCTION__<<std::endl; }
void func2(){ std::cerr<<__PRETTY_FUNCTION__<<std::endl; }
};
template <typename T1>
class B<T1, int>{
public:
void update(){ std::cerr<<__PRETTY_FUNCTION__<<"(specialization)"<<std::endl;}
};
int main(){
B<int, double> b1;
b1.update();
b1.func1();
B<int, int> b2;
b2.update();
//b2.func1();//there's no function 'func1' in B<int,int>
}
I want to specialize update
function for specific template parameter(data type).
So I tried to specialize template class B
but it seems that I have to implement whole member functions again.
Because other members are exactly same between specializations, reimplementing whole members look cumbersome.
Is there any workaround for this case?
Upvotes: 2
Views: 847
Reputation: 48447
Tag-dispatch the call to update
:
template <typename> struct tag {};
template <typename T1, typename T2>
class B
{
public:
void update()
{
return update(tag<B>());
}
private:
template <typename U1>
void update(tag<B<U1, int> >)
{
// specialization
}
template <typename U1, typename U2>
void update(tag<B<U1, U2> >)
{
// normal
}
};
Upvotes: 5