Prolog - how to do setof that returns empty list instead of failing

Question

I need an ordered list of Objects that satisfy Goal. setof takes care of the ordering, but fails when no Objects satisfy Goal. I want to return an empty list instead like findall does.

This works, but is there a way of accomplishing this without a cut? I'm using SWI-Prolog.

setof(Object, Goal, List), !; List = [].

Answer 1

First,

..., ( setof(Object, Goal, List), ! ; List = [] ), ...

does not work, as you suggest. It always succeeds for List = [], and it only shows the first answer of setof/3. But setof/3 may produce several answers. The general method that works in any Prolog is:

..., ( \+ Goal -> List = [] ; setof(Object, Goal, List) ), ...

Many implementations offer an implementation specific control construct for this which avoids that Goal is called twice. E.g. if/3 (SICStus, YAP), or (*->)/2 (SWI, GNU):

..., if( setof(Object, Goal, ListX), ListX = List, List = [] ), ...

..., ( setof(Object, Goal, ListX) *-> ListX = List ; List = [] ), ...

The new variable ListX is necessary for the (admittedly rare) case that List is already instantiated.

_{Note that both other answers do not produce exactly what you asked for.}

Answer 2

(setof(Object, Goal, List) ; List = [])

will work as well (setof itself is deterministic).

To be sure to get rid of the choice point, we need a bit more verbose

(setof(Object, Goal, List) -> true ; List = [])

edit as it stands, my answer is plainly wrong, or at least very incomplete. After false comments, and answer, I would suggest

setof(Object, Goal, List) *-> true ; List = [].

Answer 3

If you do not need the potential nondeterminism or the variable-quantification features of setof, you can stick with findall/3. This is deterministic and doesn't fail:

?- findall(X, fail, Xs).
Xs = []
yes

You can then sort the results yourself using sort/2:

findall(Object, Goal, UnsortedWithDuplicates),
sort(UnsortedWithDuplicates, List)