template class with a single method specialized in C++

Question

I only have an hpp file for a school assignment in C++ (I am not allowed to add a cpp file, declaration and implementation should be both written in the file).

I wrote this code inside it:

template<class T>
class Matrix
{
   void foo()
   {
       //do something for a T variable.
   }
};

I would like to add another foo method, but this foo() will be specialized for only an <int>. I have read in some places that I need to declare a new specialization class for this to work. But what I want is that the specialized foo will lie just beneath the original foo, so it will look like this:

template<class T>
class Matrix
{
   void foo(T x)
   {
       //do something for a T variable.
   }
   template<> void foo<int>(int x)
   {
       //do something for an int variable.
   }
};

• Why am I getting an error for this syntax ("expected unqualified-id before '<' token")?
• Why isn't this possible?
• How can I fix this without declaring a new specialized class?

Thanks

Answer 1

foo isn't a template. It's a member function of a template. Thus foo<int> is meaningless. (Also, explicit specializations must be declared at namespace scope.)

You can explicitly specialize a member function of a particular implicit instantiation of a class template:

template<class T>
class Matrix
{
   void foo(T x)
   {
       //do something for a T variable.
   }
};

// must mark this inline to avoid ODR violations
// when it's defined in a header
template<> inline void Matrix<int>::foo(int x)
{
     //do something for an int variable.
}