TITLE: templated typedefs

(Newsgroups:   comp.lang.c++.moderated, 12 Jun 97)


GOOD: "Sebastian Good" <Sebastian.Good@trilogy.com>

Simple question : why aren't templated typedefs allowed?

I would like to be able to do the following

	template <class T, class V> 
	class Stuff {...};

	template <class T>
	typedef Stuff<T,int> IntStuff;

Instead, I must resort to a macro:

	#define IntStuff(T) Stuff<T,int>

which makes the syntax non-uniform.  The only reason I could think of was
that there is no existing concept of overloading typedefs...


VANDEVOORDE: David Vandevoorde <daveed@vandevoorde.com>

It's a good question.

The answer might be that people realized too late how
useful such a feature would be. I'm not sure.
The usual workaround is to use a template-class local
typedef or inheritance. E.g., for your example:

        template<typename T>
        struct IntStuff: public Stuff<T, int> {};


FRESL: Kresimir Fresl <fresl@grad.hr>

But constructors and operators = are not inherited.


TIANXING: gztxwang@public1.guangzhou.gd.cn (Wang TianXing)

So you want to try the other option given by Daveed, "template-class
local typedef":

	template <typename T>
	struct IntStuff { typedef Stuff<T,int> t; };

Use it like this:

	IntStuff<double>::t x;

It needs a little more typing, though.