CODEWITHSUNDEEP
prologswi-prologunification
Martin Kunze
Martin Kunze

Reputation: 1055

SWI-Prolog predicates member and nth1 without unification?

If I use the predicates member or the nt1 in SWI-Prolog as follows:

?- member(X, [A]).
X = A.

or

nth1(N, [A], X).
N = 1,
A = X.

The interpreter unifies the variable A as X.

Is their some alternative version of this functions which does not use the unification. Means, if I call something like this:

?- _member(X, [A]).

it would give

 false

as long as the call is not

?- member(X, [X]).

which would lead to

true

And in the same way

_nth1(N, [A], X).

would give false

but

_nth1(N, [X], X).

would give

N = 1

Upvotes: 1

Views: 114

Answers (1)

Nicholas Carey
Nicholas Carey

Reputation: 74277

Seems like you just need to roll your own using ==/2 instead of =/2:

See https://swish.swi-prolog.org/p/DqhYGuEf.pl

is_member_of( X , [Y|_]  ) :- X == Y .
is_member_of( X , [_|Ys] ) :- is_member_of(X,Ys) .

is_nth0( N , L , E ) :- is_nth(0,L,E,N) . 

is_nth1( N , L , E ) :- is_nth(1,L,E,N) .

is_nth( N , [Y|_]  , X , N ) :- X == Y .
is_nth( I , [_|Ys] , X , N ) :- J is I+1, is_nth(J,Ys,X,N) .

Upvotes: 2

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