namespace and routine_id()
- Posted by Verne Tice <fredfarkle at highstream.net> Jan 31, 2005
- 408 views
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
