0 
Some text vagueness
"provides determinism on the last element"
should be
"provides well-behaved determinism on the last element"
to make the text consistent with the definitions of SWI-Prolog: Deterministic/Semi-deterministic/Non-deterministic predicates, where
"deterministic"
means that the predicate succeeds exactly once (but may leave a choicepoint that, however, yields no further solution on backtracking) whereas
"well-behaved deterministic"
means that the predicate succeeds exactly once and leaves no choicepoint.
A nice illustration of nondeterministic behaviour
With call_cleanup/2
Fail, causing no effect on Det:
?- call_cleanup(member(0,[1,2,3]),Det=true). false.
Succeed three times and then leave no choicepoint, causing Det to become true:
?- call_cleanup(member(X,[1,2,3]),Det=true). X = 1 ; X = 2 ; X = 3, Det = true.
Succeed once leaving a choicepoint and then fail on redo, causing no effect on Det:
?- call_cleanup(member(1,[1,2,3]),Det=true). true ; false.
Succeed once leaving no choicepoint because the unification succeeds on the last list item ("member/2 provides determinism on the last element"), causing Det to become true:
?- call_cleanup(member(3,[1,2,3]),Det=true). Det = true.
See also nth0∕3
nth0/3 performs the same task, put also uses/outputs out the index:
Access by index:
?- nth0(3,[a,b,c,third],X). X = third.
Find the index of an entry:
?- nth0(Index,[a,b,c,d],b). Index = 1 ; false. ?- nth0(Index,[a,b,c,b,d],b). Index = 1 ; Index = 3 ; false. ?- nth0(Index,[a,b,c,b,d,b],b). Index = 1 ; Index = 3 ; Index = 5. % Deterministic on last element
Backtracking over solutions:
?- findall(I-V,nth0(I,[a,b,c,d],V),Bag). Bag = [0-a, 1-b, 2-c, 3-d].
Compare with memberchk∕2
Use memberchk/2 if you are hunting for efficiency and a single solution is sufficient (but watch out for unexpected failures with memberchk/2)
memberchk/2 should be the same as once/1 around member/2:
?- once(member(a,[b,b,a,b,a,b])). true. ?- member(a,[b,b,a,b,a,b]). true ; true ; false. ?- memberchk(a,[b,b,a,b,a,b]). true.
Some examples and differences with memberchk∕2
Obtain successive members; member/2 retrieves successive members on backtracking
?- member(A,[1,2,3]). A = 1 ; A = 2 ; A = 3.
memberchk/2 does not backtrack
?- memberchk(A,[1,2,3]). A = 1.
Check whether an element is present in List
member/2 backtracks; a retry can be attempted after the second success even though there is no further 1 down the list:
?- member(1,[1,2,1,3]). true ; true ; false.
More deterministic:
?- member(1,[1,2,1]). true ; true.
memberchk/2 does not backtrack
?- memberchk(1,[1,2,1,3]). true.
member∕2 behaves well for non-lists
Finding an element in a non-list:
?- member(1,stop). false. ?- member(1,[3,4,2,1,6|stop]). true ; false.
Note that memberchk/2 is more picky:
?- memberchk(1,stop). ERROR: Type error: `list' expected, found `stop' (an atom) ?- memberchk(1,[3,4,2,1,6|stop]). true. ?- memberchk(7,[3,4,2,1,6|stop]). ERROR: Type error: `list' expected, found `stop' (an atom)
Variables in the list? No problem
As the Elem is unified with list elements one after the other, at some point the Elem will "be the same" as any unbound variable in the list:
?- member(X,[A,foo,B,bar,C]). X = A ; X = foo ; X = B ; X = bar ; X = C.
Once can perform a "term equality" test using assertion/1 for this:
?- member(X,[A,foo,B,bar,C]), assertion((var(X) -> (X==A;X==B;X==C) ; true)). X = A ; X = foo ; X = B ; X = bar ; X = C.
member∕2 grows open lists and keeps them open
Finding an element in an open list (aka partial list) results in an infinite stream of longer and longer lists containing unbound variables where the element being looked for is the last entry of the suffix:
?- member(1,[3,4,2,1,6|More]). true ; More = [1|_140008] ; More = [_140006,1|_140744] ; More = [_140006,_140742,1|_141480] ; More = [_140006,_140742,_141478,1|_142216] ; ...
This applies in particular to the "maximally unspecified list", the unbound variable, which is an open list:
?- member(1,List). List = [1|_29654] ; List = [_29652,1|_30390] ; List = [_29652,_30388,1|_31126] ; List = [_29652,_30388,_31124,1|_31862] ; ...
Is that what one would expect? Difficult to say! It makes sense: List is now a list whereby the only thing that is certain is that there is an 1 at position 0, or 1, or 2, .... Not even the length's list is certain.
nth0/3 behaves likewise:
?- nth0(Index,List,1). Index = 0, List = [1|_131392] ; Index = 1, List = [_131390,1|_132520] ; Index = 2, List = [_131390,_132518,1|_133650] ; Index = 3, List = [_131390,_132518,_133648,1|_134780] ...
A (more general?) variation
This predicate is meant to work with open lists in the following way: It has a third argument which
- is set to
foundif the element was found in the prefix of the open list (or in a closed list) - is set to
appended(Fin)(Finis unified with the unbound variable at the second position of the last listbox) if the element was not found in the prefix or the prefix was exhausted and the element was added to the open end
Backtracking generates longer and longer open lists, same as for member/2.
Standard behaviour on proper lists:
?- member_openlist(x,[a,b,x],What). What = found. ?- member_openlist(x,[a,b,x,c],What). What = found ; false.
Only append to open list:
?- List=[a,b,x|_], member_openlist(x,List,appended(W)). List = [a,b,x,x|W] ; List = [a,b,x,_11742,x|W] ; List = [a,b,x,_11742,_12570,x|W] ; List = [a,b,x,_11742,_12570,_13398,x|W] ; List = [a,b,x,_11742,_12570,_13398,_14226,x|W] ; List = [a,b,x,_11742,_12570,_13398,_14226,_15054,x|W] ; ...
Only find in opn list:
?- List=[a,b,x|_], member_openlist(x,List,found). List = [a,b,x|_16558] ; false.
Find in or append to open list:
?- List=[a,b,x|_], member_openlist(x,List,What). List = [a,b,x|_18420], What = found ; List = [a,b,x,x|_20178], What = appended(_20178) ; List = [a,b,x,_21314,x|_21310], What = appended(_21310) ; List = [a,b,x,_21314,_22452,x|_21310], What = appended(_21310) ...
Append to and close open list (like last/2):
?- List=[a,b,x|_], member_openlist(x,List,appended([])). List = [a,b,x,x] ; List = [a,b,x,_26164,x] ; List = [a,b,x,_26164,_26910,x] ; List = [a,b,x,_26164,_26910,_27656,x] ; List = [a,b,x,_26164,_26910,_27656,_28402,x] ...
Here it is:
member_openlist(Element,List,appended(Fin)) :- % In this case, we are appending, so make sure "found" has not been requested. var(List), % We found the the "fin" of an open list (an unbound variable). List=[Element|Fin]. % Append by instantiating "fin" to a new listbox and succeed. member_openlist(Element,List,appended(FinalFin)) :- % In this case, we are appending, so make sure "found" has not been requested. var(List), % We found the the "fin" of an open list (an unbound variable). List=[_|NewFin], % Append by instantiating "fin" to a new listbox ... member_openlist(Element,NewFin,appended(FinalFin)). % ... and perform recursive call member_openlist(Element,List,What) :- nonvar(List), % List is instantiated to something, presumably a listbox [_|_] assertion((List == [] ; List = [_|_])), % List should be a listbox or an empty list! member_openlist_2(Element,List,What). % Handle various cases member_openlist_2(Element,[Element|B],found) :- % Unified at end of proper list. Cut to be deterministic. B==[], % We can't put [Element] in the head lest we unify an open end with [] !. member_openlist_2(Element,[Element|_],found). % Unified anywhere except at the end of a proper list. Do not cut! member_openlist_2(Element,[_|B],What) :- % Keep on looking member_openlist(Element,B,What).
With testcases
:- begin_tests(member_openlist).
test("nothing in an empty list",fail) :-
member_openlist(x,[],_).
test("not found in a closed list",fail) :-
member_openlist(x,[a,b,c],_).
test("found in the middle of a closed list") :-
once(member_openlist(x,[a,x,c],What)),
assertion(What==found).
test("found as last element of a closed list") :-
member_openlist(x,[a,b,x],What),
assertion(What==found).
test("instantiate them all over a closed list") :-
findall(
X-What,
member_openlist(X,[a,b,c],What),
Events),
assertion(Events == [a-found,b-found,c-found]).
test("found in the middle of an open list") :-
List=[a,x,c|_],
once(member_openlist(x,List,What)),
assertion(What==found),
assertion((List=[a,x,c|Fin],var(Fin))).
test("found as last element of an open list") :-
List=[a,b,x|_],
once(member_openlist(x,List,What)),
assertion(What==found),
assertion((List=[a,b,x|Fin],var(Fin))).
test("not found in an open list leads to append") :-
List=[a,b,c|_],
once(member_openlist(d,List,What)),
assertion((List = [a,b,c,d|Fin1],What=appended(Fin2),Fin1==Fin2,var(Fin1))).
test("found as element then appended using limit/2 inside findall/2") :-
List=[a,b,x|_],
findall(List-What, limit(2,member_openlist(x,List,What)), All),
% Note the list collected by findall is not physically the same list as "List"
assertion(
All = [[a,b,x|_]-found,
[a,b,x,x|Fin2]-appended(Fin2)]).
test("force appending and then close list") :-
List=[a,b,x|_],
once(member_openlist(x,List,appended(_))),
once(member_openlist(x,List,appended(_))),
once(member_openlist(a,List,appended([]))),
assertion(List==[a,b,x,x,x,a]).
:- end_tests(member_openlist).
With the above, you can assemble a list or disassemble a list of options with the same code. For example:
foo_pushpull(List,Value,What) :- once(member_openlist(foo(Cs),List,What)), atom_chars(Value,Cs). bar_pushpull(List,Value,What) :- once(member_openlist(bar(Cs),List,What)), atom_chars(Value,Cs).
Assemble:
?- foo_pushpull(List,hello,What), bar_pushpull(List,world,appended([])). List = [foo([h,e,l,l,o]),bar([w,o,r,l,d])], What = appended([bar([w,o,r,l,d])]).
Disassemble:
?- List=[foo([h,e,l,l,o]),bar([w,o,r,l,d])], foo_pushpull(List,Foo,found), bar_pushpull(List,Bar,found). List = [foo([h,e,l,l,o]),bar([w,o,r,l,d])], Foo = hello, Bar = world.