Re: namespace and routine_id()

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Verne Tice wrote:
> 
> If I do something like
>   include sort.e as eus1
>   include sort.e as s2
> 
> I can sort by
>   a = eus1:sort(a)
>   a = s2:sort(a)
> with no problems.
> 
> But, if I want to set up a sequence of sort routines 
> (for performance comparisons), how can I get the routine_id("eus1:sort")??
> 
> When I try to use the namespace qualifier in routine_id() I get an id of -1!

That's a bug in 2.5 (both alpha and beta). It works in 2.4.
Thanks for reporting it. I'll fix it for the official release.

Regards,
   Rob Craig
   Rapid Deployment Software
   http://www.RapidEuphoria.com

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