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I declare a templated class with all parameters having default arguments, for example:
template<typename TYPE = int>
class Foo {};
Then the following two are equivalent:
Foo<int> one;
Foo<> two;
However, I'm not allowed to just do:
Foo three;
Is it possible to achieve that with a typedef to the same name but without the brackets, like this:
typedef Foo<> Foo;
228
You cannot give default arguments to the same template parameters in different declarations in the same scope. The compiler will not allow the following example: template<class T = char> class X; template<class T = char> class X { };
Just like in case of the function arguments, template parameters can have their default values. All template parameters with a default value have to be declared at the end of the template parameter list.
A template argument for a template template parameter is the name of a class template. When the compiler tries to find a template to match the template template argument, it only considers primary class templates. (A primary template is the template that is being specialized.)
typedef is defining a new type for use in your code, like a shorthand. typedef typename _MyBase::value_type value_type; value_type v; //use v. typename here is letting the compiler know that value_type is a type and not a static member of _MyBase . the :: is the scope of the type.
I do something like the following, I don't know if you will like it or not:
template<typename TYPE = int>
class basic_Foo {};
typedef basic_Foo<int> Foo;
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