I know that the below code is a partial specialization of a class:
template <typename T1, typename T2>
class MyClass {
…
};
// partial specialization: both template parameters have same type
template <typename T>
class MyClass<T,T> {
…
};
Also I know that C++ does not allow function template partial specialization (only full is allowed). But does my code mean that I have partially specialized my function template for one/same type arguments? Because it works for Microsoft Visual Studio 2010 Express! If no, then could you please explain the partial specialization concept?
#include <iostream>
using std::cin;
using std::cout;
using std::endl;
template <typename T1, typename T2>
inline T1 max (T1 const& a, T2 const& b)
{
return a < b ? b : a;
}
template <typename T>
inline T const& max (T const& a, T const& b)
{
return 10;
}
int main ()
{
cout << max(4,4.2) << endl;
cout << max(5,5) << endl;
int z;
cin>>z;
}
Function partial specialization is not yet allowed as per the standard. In the example, you are actually overloading & not specializing the max<T1,T2>
function.
Its syntax should have looked somewhat like below, had it been allowed:
// Partial specialization is not allowed by the spec, though!
template <typename T>
inline T const& max<T,T> (T const& a, T const& b)
{ ^^^^^ <--- [supposed] specializing here
return 10;
}
In the case of a function templates, only full specialization is allowed by the C++ standard, -- excluding the compiler extensions!