1. namespace and routine_id()
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!
Confused,
Verne Tice
2. Re: namespace and routine_id()
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
