typedef a template with all default arguments

I declare a templated cl开发者_运维知识库ass 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;

---

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;

---

You can't redeclare a symbol with a different type, so whatever you will be able to do won't work as you expect. If you want to achieve this, use a different name as alias :

typedef Foo<> Foo_;

---

If the declaration typedef Foo<> Foo; is allowed, thereafter the name Foo as a template cannot be specified. That is, the following becomes invalid.

template< template< class > class > struct A {...};
A< Foo > a; // error

Though the above typedef isn't allowed in practice, if you still need to write Foo as Foo<>, a macro like the following will meet the purpose.

#define Foo Foo<>

---

typedef Foo<> Foo;

Gives:

prog.cpp:4: error: ‘typedef class Foo<int> Foo’ redeclared as different kind of symbol
prog.cpp:2: error: previous declaration of ‘template<class TYPE> class Foo’

The error pretty much tells what the problem is. Compiler sees Foo as being re-declared.

However, this shall compile and work:

template<typename TYPE = int> class Foo {};

typedef Foo<> FooClone;

int main()
{
   Foo<int> one;
   Foo<> two;
   FooClone three;

   return 0;
}

---

No. Although you can declare a typedef for a class with the same name as a class because you can use a typedef to redefine a name to refer to the type to which it already refers.

typedef class A A;

or if A was already declared as a class:

typedef A A;

You can't do that with the name of a template (the name of a template isn't a name of a class), you'd have to give it a different name.

typedef Foo<> Bar;

---

Unfortunately, no, because Foo is already the name for the class template itself, and thus can't be anything else in the same namespace.