optional typename without name

[Frank Winter]

Hi!

Codeguru about allocators: They say the std::allocator that
compiler vendors normally ship looks something like:

template<typename T>
class Allocator {
   typedef T value_type;
   typedef value_type* pointer;
   typedef const value_type* const_pointer;
   typedef value_type& reference;
   typedef const value_type& const_reference;
   typedef std::size_t size_type;
   typedef std::ptrdiff_t difference_type;
//...
public :
   inline pointer allocate(
     size_type cnt,
     typename std::allocator<void>::const_pointer = 0) {
       return reinterpret_cast<pointer>(::operator new(cnt * sizeof (T)));
     }
//...
}

Would it be possible for someone to explain why the optional argument
typename std::allocator<void>::const_pointer = 0 is there?

It's a typename that can be suppressed. Why to write it in the first 
place then? Is this written to disambiguate something?


Frank