Question

I have used an example from

http://en.wikipedia.org/wiki/Template_metaprogramming

const unsigned long long y = Factorial<0>::value; // == 1  

I understand the compiler can do a type check but I did not think you could put 0 where the type should be and it would be taken as the value.

Could someone explain how this is working ?

Thanks

Answer 1

I did not think you could put 0 where the type should be and it would be taken as the value

The problem is the assumption: it is not “where the type should be”. Rather, it is “where the template argument may be specified”.

In this case, there is a non-type template argument.

It works, simply because it is a different feature than type template arguments.

It is just that way. I can recommend C++ Templates: the Complete Guide or many other books on that list to read up on the basics of C++ templates.

• The Definitive C++ Book Guide and List