CODEWITHSUNDEEP
prolog
Enforcerke
Enforcerke

Reputation: 267

Append initialized list

If I have a pre-initialized list, for example:

List = [_1663, _1665, _1667]

How can I put a variable (like a simple integer) in this list, taking up the first free spot?

I tried:

append(5, List).

But that doesn't work. The list must be pre-initialized like above. Any ideas? Thank you.

Upvotes: 0

Views: 130

Answers (1)

lurker
lurker

Reputation: 58244

Prolog will not automatically instantiate the next free variable in a list. You can easily, though, do this with a predicate:

bind_first_free_element(Element, List) :-
    once(bind_free_element(Element, List)).

bind_free_element(X, [Y|_]) :-
    var(Y), X = Y.
bind_free_element(X, [_|T]) :-
    bind_free_element(X, T).

When you query, you get:

2 ?- L = [a,X,Y,b,Z], bind_free_element(5, L).
L = [a, 5, Y, b, Z],
X = 5 ;
L = [a, X, 5, b, Z],
Y = 5 ;
L = [a, X, Y, b, 5],
Z = 5 ;
false.

So you can see that bind_free_element shows all of the possible solutions for binding a free element. Thus, we use bind_first_free_element/2 that uses once/1 to only seek the first solution:

3 ?- L = [a,X,Y,b,Z], bind_first_free_element(5, L).
L = [a, 5, Y, b, Z],
X = 5.

Upvotes: 2

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