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copied!<p>A variable match a <strong>term</strong>, and the anonimus variable is not exception. A list is just syntax sugar for a binary relation, between head and tail. So a variable can match the list, the head, or the tail, but not an unspecified sequence.</p> <p>Some note I hope will help you:</p> <blockquote> <p>listcount(L, N) :- listcountA(LS, [], N).</p> </blockquote> <p>In Prolog, predicates are identified by name <strong>and</strong> num.of.arguments, so called <strong>functor</strong> and <strong>arity</strong>. So usually 'service' predicates with added arguments keep the same name.</p> <blockquote> <p>listcountA([X|Tail], [? [X, B], ?], N) :- B is B+1, listcountA(Tail, [? [X,B] ?], N).</p> </blockquote> <p>B is B+1 will never succeed, you <strong>must</strong> use a new variable. And there is no way to match inside a list, using a 'wildcard', as you seem to do. Instead write a predicate to find and update the counter.</p> <p>A final note: usually <strong>pairs</strong> of elements are denoted using a binary relation, conveniently some (arbitrary) operator. For instance, most used is the dash.</p> <p>So I would write </p> <pre><code>listcount(L, Counters) :- listcount(L, [], Counters). listcount([X | Tail], Counted, Counters) :- update(X, Counted, Updated), !, listcount(Tail, Updated, Counters). listcount([], Counters, Counters). update(X, [X - C | R], [X - S | R]) :- S is C + 1. update(X, [H | T], [H | R]) :- update(X, T, R). update(X, [], [X - 1]). % X just inserted </code></pre> <p>update/3 can be simplified using some library predicate, 'moving inside' the recursion. For instance, using select/3:</p> <pre><code>listcount([X | Tail], Counted, Counters) :- ( select(X - C, Counted, Without) -> S is C + 1 ; S = 1, Without = Counted ), listcount(Tail, [X - S | Without], Counters). listcount([], Counters, Counters). </code></pre>

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