Jerry Feldman wrote:

On Sunday 21 August 2005 3:51 pm, John D Lamb wrote:

...

Here's a puzzle: I'd like to define something like

typedef std::pair P;

I can obviously do

class P : public std::pair{};

or

typedef std::pair P; In this case, you have not yet defined P. You could use a forward declaration, such as: class P; Then, typedef std::pair P; should work. I have not tried your example specifically.

Sadly it doesn't work though I actually had to try this to see. It does confirm that I'm not being totally stupid in thinking that C++ might reasonably allow this sort of thing. The C++ standard doesn't allow you to use typedef for a name already declared in the same scope. I think C++ won't allow you to do something like: class X x; i.e. create a template class whose template parameter is a pointer to an object of this class. It will allow class Y : public class X {} x; which is nearly the same. For my own problem, I've worked out I can use static_cast rather than reinterpret_cast, which is a bit of an improvement. -- JDL

vamsi krishna wrote:

I tried this and It worked for me on g++ (3.4.4).

#include<list> using namespace std;

std::list P; /*I tried for list template, it should work for pair also*/ typedef class P* Y; typdef std::list<Y> P_new;

/*P_new is the class what u are expecting.*/

Unfortunately it doesn't quite work though it took me a while to work out why: I hadn't thought of declaring a class inside the brackets <...>. There are two classes P here I think: one declared but not defined and the other defined as std::list. The problem is that neither has type std::list*> that I need. The code compiles correctly because the declarations are correct. You can construct a list of type std::list*> but you can't do anything with the objects it contains because their class has never been defined. Still it proves there are things I hadn't thought of trying.. -- JDL