Lists of N length

Question

Is it to possible to create N elements of 0 array in another way than the following one:

makeInitialArray(N, A) :-
   N > 0,
   Nn is N-1,
   makeInitialArray(Nn, B),
   append([0], B, A).
makeInitialArray(0, []).

It is a poor way on my eye because of the fact of recurrence. I suppose that will be made to call very often operation like that.

Recurrence is not something wrong at all .. There may be another way to do it but it would also be based on a recurrence at some point. — meucaa, Jun 03 '16 at 22:42

score 3 · Accepted Answer · edited May 23 '17 at 12:07

Keep it simple! How? Delegate the "recursive part" to built-in / library predicates!

Based on length/2, meta-predicate maplist/2, and (=)/2 define n_zeroes/2:

:- use_module(library(lists), [maplist/2]).

n_zeroes(N, Zs) :-
   length(Zs, N),
   maplist(=(0), Zs).

Sample queries using SICStus Prolog 4.3.2:

| ?- n_zeroes(10, Zs).
Zs = [0,0,0,0,0,0,0,0,0,0] ? ;
no
| ?- n_zeroes(N, Zs).
N = 0, Zs = [] ? ;
N = 1, Zs = [0] ? ;
N = 2, Zs = [0,0] ? ;
N = 3, Zs = [0,0,0] ? ;
N = 4, Zs = [0,0,0,0] ? ;
N = 5, Zs = [0,0,0,0,0] ? 
...                             % ...goes on forever and ever...

Lists of N length

1 Answers1