Re: Replace occurrence of a list with another?

From: brianbasquille@xxxxxxxxx
Date: 28 Apr 2006 16:24:23 -0700

Realised i needed a cut... it works now like so:

replaceall(Item,Repl,[H|T],[Repl|List]) :-
H = Item, !,
replaceall(Item,Repl,T,List).

But what would be a suitable stopping condition for it (after the 'H =
Item' is satisfied) and not go into the other replaceall predicate?

.

References:
- Replace occurrence of a list with another?
  - From: brianbasquille
- Re: Replace occurrence of a list with another?
  - From: Bill Spight
- Re: Replace occurrence of a list with another?
  - From: brianbasquille
- Re: Replace occurrence of a list with another?
  - From: Bill Spight

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