Typedef of a template?

[Richard van Wegen]

Dear All

I'm hoping someone can show me the correct way to typedef a template
class -
i.e. something along the lines of

typedef boost::shared_p tr<T> shared_ptr_mult ithread<T>;

(syntax is obviously wrong but you get the idea)

i.e. I want to be able to write

int main()
{
shared_ptr_mult ithread<MyClass > my_class_multit hreaded_ptr;
shared_ptr<MyCl ass> my_class_single threaded_ptr;
}

so that for the time being it is equivalent to
int main()
{
shared_ptr<MyCl ass> my_class_multit hreaded_ptr;
shared_ptr<MyCl ass> my_class_single threaded_ptr;
}

but later on I can change the way that shared_ptr_mult ithread<MyClass >
is
implemented without affecting all my other shared_ptr<MyCl ass>
declarations.

I'm sure it must be simple, but I can't find the syntax for it
anywhere and I've tried all the combinations of "template" and
"typename" etc I can think of with no success. (GCC 2.96). At the
moment I'm forced to rely on #define (eugh).

Thanks for any pointers!
Richard

[John Harrison]

"Richard van Wegen" <sp***********@ volcanomail.com > wrote in message
news:c7******** *************** ***@posting.goo gle.com...

Dear All

I'm hoping someone can show me the correct way to typedef a template
class -
i.e. something along the lines of

typedef boost::shared_p tr<T> shared_ptr_mult ithread<T>;

(syntax is obviously wrong but you get the idea)

i.e. I want to be able to write

int main()
{
shared_ptr_mult ithread<MyClass > my_class_multit hreaded_ptr;
shared_ptr<MyCl ass> my_class_single threaded_ptr;
}

so that for the time being it is equivalent to
int main()
{
shared_ptr<MyCl ass> my_class_multit hreaded_ptr;
shared_ptr<MyCl ass> my_class_single threaded_ptr;
}

but later on I can change the way that shared_ptr_mult ithread<MyClass >
is
implemented without affecting all my other shared_ptr<MyCl ass>
declarations.

I'm sure it must be simple, but I can't find the syntax for it
anywhere and I've tried all the combinations of "template" and
"typename" etc I can think of with no success. (GCC 2.96). At the
moment I'm forced to rely on #define (eugh).

Thanks for any pointers!
Richard

You cannot write template typedefs, its a limitiation of the C++ syntax
(soon to be corrected?).

The best you can do is wrap your typedefs in a struct, which can be
templated.

template <class T>
struct Type
{
typedef boost::shared_p tr<T> shared_ptr_mult ithread;
typedef boost::shared_p tr<T> shared_ptr; // or something
};

int main()
{
Type::shared_pt r_multithread<M yClass> my_class_multit hreaded_ptr;
Type::shared_pt r<MyClass> my_class_single threaded_ptr;
}

john

[John Harrison]

"Sarah Thompson" <sa***@telergy. remove.this.bit .com> wrote in message
news:be******** ***********@new s.demon.co.uk.. .

John Harrison wrote:

You cannot write template typedefs, its a limitiation of the C++ syntax
(soon to be corrected?).

The best you can do is wrap your typedefs in a struct, which can be
templated.

template <class T>
struct Type
{
typedef boost::shared_p tr<T> shared_ptr_mult ithread;
typedef boost::shared_p tr<T> shared_ptr; // or something
};

int main()
{
Type::shared_pt r_multithread<M yClass> my_class_multit hreaded_ptr;
Type::shared_pt r<MyClass> my_class_single threaded_ptr;
}

That should be written as:

int main()
{
Type<MyClass>:: shared_ptr_mult ithread my_class_multit hreaded_ptr;
Type<MyClass>:: shared_ptr my_class_single threaded_ptr;
}

shouldn't it?

Yes indeedy, thanks.

john

[Richard van Wegen]

> > > You cannot write template typedefs, its a limitiation of the C++ syntax

(soon to be corrected?).

The best you can do is wrap your typedefs in a struct, which can be
templated.

template <class T>
struct Type
{
typedef boost::shared_p tr<T> shared_ptr_mult ithread;
typedef boost::shared_p tr<T> shared_ptr; // or something
};

int main()
{
Type<MyClass>:: shared_ptr_mult ithread my_class_multit hreaded_ptr;
Type<MyClass>:: shared_ptr my_class_single threaded_ptr;
}

Ah ok, that explains why I was having so much trouble.

Thanks very much to you both for the help!

Cheers
Richard