Explicit specialization of function templates

In a function template specialization, a template argument is optional if the compiler can deduce it from the type of the function arguments. The following example demonstrates this:

template<class T> class X { };
template<class T> void f(X<T>);
template<> void f(X<int>);

The explicit specialization template<> void f(X<int>) is equivalent to template<> void f<int>(X<int>).

You cannot specify default function arguments in a declaration or a definition for any of the following:

• Explicit specialization of a function template
• Explicit specialization of a member function template

For example, the compiler will not allow the following code:

template<class T> void f(T a) { };
template<> void f<int>(int a = 5) { };

template<class T> class X {
  void f(T a) { }
};
template<> void X<int>::f(int a = 10) { };

Related information

• Function templates (C++ only)