Simple nth1 predicate in Prolog

Question

With SWI Prolog, there's a predicate that finds the nth item in a list called nth1. I want to implement my own version of the predicate but SWI's is so complicated if you look at the listing(nth1) code. Is there a simpler way of doing it?

Thank you :).

Answer 1

The SWI code is a bit complex because the predicate can be used to generate from a variable index:

?- nth1(Idx,[a,b,c],X).
Idx = 1,
X = a ;
Idx = 2,
X = b ;
Idx = 3,
X = c ;
false.

If you don't want that behavior, nth1/3 can be implemented easily in terms of nth0:

nth1(Idx,List,X) :-
    Idx0 is Idx-1,
    nth0(Idx0,List,X).

Edit: it's also possible to do without nth0 in just a few lines of code:

nth1(1,[X|_],X) :- !.
nth1(Idx,[_|List],X) :-
    Idx > 1,
    Idx1 is Idx-1,
    nth1(Idx1,List,X).