Simple nth1 predicate in Prolog

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陌清茗
陌清茗 2020-12-21 03:50

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

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  •  一整个雨季
    2020-12-21 04:32

    I didn't mean to be contradictory or get someone else to do my work actually; I just wanted some advice, sorry for not being clearer.

    I've implemented it myself now but could you guys possibly suggest improvements or better ways of doing it? What I often find myself doing in Prolog is writing a predicate with say a counter or set of counters and getting a predicate with fewer arguments to call the clauses with extra arguments. This often ends up producing quite a bit of code. Anyway, here's my implementation I just did:

    item_at( N, L, Item ) :-
        item_at( N, 0, L, Item ).   
    item_at( N, Count, [H|_], Item ) :-
        CountNew is Count + 1,
        CountNew = N,
        Item = H.
    item_at( N, Count, [_|T], Item ) :-
        CountNew is Count + 1,
        item_at( N, CountNew, T, Item ).
    

    Any comments? Thanks :). Usage:

    ?- item_at(3,[a,b,c,d,e],Item).
    Item = c ;
    

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