Question

I’m trying to specialize just one method of a class template.

The accepted answer there definitely does work, but what does this code mean to the compiler and why doesn’t it work as expected?

#include <stdio.h>

template <typename T>
struct Node
{
  void split() { puts( "Default method" ) ; }

  // this compiles, but it doesn't appear to do anything!
  template <int> void split() { puts( "Int method" ) ; }
} ;

// This definitely works
//template <> void Node<int>::split() { puts( "Int method" ) ; }

int main()
{
  Node<double> n ;
  n.split() ; // "Default method"

  Node<int> i;
  i.split();  // "Default method" again!
}

Answer 1

Call like this:

n.split<10>();
i.split<10>(); //or use any constant integral value

If you intend to call the function template. Note that the function template is not a specialization. It rather is a primary function template, and to call it, you’ve to provide the template argument explicitly.

You can do even this:

const int x = 100;
i.split<x>(); //ok - x is constant expression

But this is not allowed:

int y = 100;
i.split<y>(); //error - y is not constant expression