Why is a list with an empty list true?

Boris · March 13, 2020, 11:33am

I think the reason is in here:

SWI-Prolog/swipl-devel/blob/95b540cb962b65f5f993c4a969b1494e9b71a253/boot/init.pl#L2033


      
          load_files(Files) :-
              load_files(Files, []).
          load_files(Module:Files, Options) :-
              '$must_be'(list, Options),
              '$load_files'(Files, Module, Options).
          
          '$load_files'(X, _, _) :-
              var(X),
              !,
              '$instantiation_error'(X).
          '$load_files'([], _, _) :- !.
          '$load_files'(Id, Module, Options) :-   % load_files(foo, [stream(In)])
              '$option'(stream(_), Options),
              !,
              (   atom(Id)
              ->  '$load_file'(Id, Module, Options)
              ;   throw(error(type_error(atom, Id), _))
              ).
          '$load_files'(List, Module, Options) :-
              List = [_|_],
              !,

See consult/1. I can’t be certain, but the sentence:

Abbreviation using ?- [file1,file2]. does not work for the empty list ( [] ).

probably means that this doesn’t work:

?- [].

If I read the code in init.pl correctly:

?- [[]].

becomes

?- consult(user:[]).

which calls load_files/2 which calls ‘$load_files’/3 which has the clause:

'$load_files'([], _, _) :- !.

which is why ?- [[]]. succeeds without choice points.

This is the “what happens”. The answer to “why” is beyond me.

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Why is a list with an empty list true?

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