what's the purpose of typename assignment inside templates

Question

I have come across this piece of code (I\'m trying to include all details in case I\'m missing something):

template< typename TYPE = TYPE_with_an_arbitrar         


        

Answer 1

A function-template may not be the best construct to demonstrate default template arguments. Here's something similar with template-structs:

#include 
#include 

template
struct foo {
   static void f() {
      std::cout << typeid(T).name() << "\t";
   }
};

template
struct bar {
   static void f() {
      std::cout << typeid(T).name() << "\t";
   }
};

int main() {
  foo<>::f(); foo::f();  foo::f();  std::cout << std::endl;
  bar<>::f(); bar::f();  bar::f();  std::cout << std::endl;
}

Running this, I get:

% ./a.out 
i   i   d   
d   i   d