Is this possible with a template?

[Jamie Burns]

Hello,

I have an application which uses a class called "rSharedPointer ". I want to
replace it with the Boost shared pointer but would like to keep the old name
and not modify any existing code.

I have tried doing the following:

typedef boost::shared_p tr<T> rSharedPointer;

and:

template<typena me T> typedef boost::shared_p tr<T> rSharedPointer;

But it will not compile?

Any ideas?

Jamie.

[Victor Bazarov]

Re: Is this possible with a template?

"Jamie Burns" <reply_to_group @whatever.com> wrote...[color=blue]
> I have an application which uses a class called "rSharedPointer ". I want
> to replace it with the Boost shared pointer but would like to keep the old
> name and not modify any existing code.
>
> I have tried doing the following:
>
> typedef boost::shared_p tr<T> rSharedPointer;
>
> and:
>
> template<typena me T> typedef boost::shared_p tr<T> rSharedPointer;
>
> But it will not compile?
>
> Any ideas?[/color]

If 'rSharedPointer ' is a _class_, you cannot replace it with a template
in your code, the use for classes and templates is different. So, even
if there were (and there isn't) support for template typedefs, how would
you then use your 'rSharedPointer '?

If, contrary to your own words, 'rSharedPointer ' is, in fact, a template,
then you could try using a macro:

#define rSharedPointer boost::shared_p tr

V

[Jamie Burns]

Re: Is this possible with a template?

Thanks Victor.

Yes, it was a template, not a class!

The #define works a treat :o)

Jamie.

"Victor Bazarov" <v.Abazarov@com Acast.net> wrote in message
news:6BgBd.5947 03$wV.391026@at tbi_s54...[color=blue]
> "Jamie Burns" <reply_to_group @whatever.com> wrote...[color=green]
>> I have an application which uses a class called "rSharedPointer ". I want
>> to replace it with the Boost shared pointer but would like to keep the
>> old name and not modify any existing code.
>>
>> I have tried doing the following:
>>
>> typedef boost::shared_p tr<T> rSharedPointer;
>>
>> and:
>>
>> template<typena me T> typedef boost::shared_p tr<T> rSharedPointer;
>>
>> But it will not compile?
>>
>> Any ideas?[/color]
>
> If 'rSharedPointer ' is a _class_, you cannot replace it with a template
> in your code, the use for classes and templates is different. So, even
> if there were (and there isn't) support for template typedefs, how would
> you then use your 'rSharedPointer '?
>
> If, contrary to your own words, 'rSharedPointer ' is, in fact, a template,
> then you could try using a macro:
>
> #define rSharedPointer boost::shared_p tr
>
> V
>[/color]

[Andrey Tarasevich]

Re: Is this possible with a template?

Jamie Burns wrote:[color=blue]
> ...
> I have an application which uses a class called "rSharedPointer ". I want to
> replace it with the Boost shared pointer but would like to keep the old name
> and not modify any existing code.
>
> I have tried doing the following:
>
> typedef boost::shared_p tr<T> rSharedPointer;
>
> and:
>
> template<typena me T> typedef boost::shared_p tr<T> rSharedPointer;
>
> But it will not compile?
>
> Any ideas?
> ...[/color]

It looks like you need a template 'typedef'. C++ doesn't have template
'typedef's, but there is a well-known idiom that comes pretty close

template<typena me T> struct rSharedPointer {
typedef boost::shared_p tr<T> type;
};

However, in this case you'll have to declare your 'rSharedPointer 's as
follows

rSharedPointer< int>::type ptr;

This is probably different from what you currently have. If this is
unacceptable for you, you'll probably have to stick with
preprocessor-based solution suggested by Victor.

--
Best regards,
Andrey Tarasevich