So the univ operator. I don't exactly understand it.
For example this:
foo(PredList,[H|_]) :- bar(PredList,H).
foo(PredList,[_|T]) :- foo(PredList,T),!.
bar([H|_],Item) :- G =.. [H,Item],G.
bar([_|T],Item) :- bar(T,Item).
What is this doing? This looks to see if another predicate is true. I don't understand what the ".." does.
How would you rewrite this without the univ operator?
Univ (=..
) breaks up a term into a list of constituents, or constructs a term from such a list. Try:
?- f(x,y) =.. L.
L = [f, x, y].
?- f(x,y,z) =.. [f|Args].
Args = [x, y, z].
?- Term =.. [g,x,y].
Term = g(x, y).
bar
seems to call each predicate in PredList
on Item
, with foo
backtracking over the Item
s. (Using a variable as a predicate is not portable; the call
predicate should be preferred.)
Edit: Kaarel is right, univ can be replaced by functor/3
and arg/3
, as follows:
bar([H|_],Item) :-
functor(Goal,H,1), % unifies Goal with H(_)
arg(1,Goal,Item), % unifies first argument of Goal with Item
call(Goal). % use this for portability
The most appropriate rewrite in my opinion would be:
bar( [H|_], Item ) :- call(H, Item).
call/n
are not yet part of the ISO core standard, but they might become in the near future (*). A lot of Prolog systems already support them.
There is one reason why call/n
should be preferred over simple (=..)/2
and functor/3
+ arg/3
solutions. The call/n
solution is capable to handle closures (**).
With the simple (=..)/2
and functor/3
+ arg/3
solution one can invoke bar/2
only with atoms in the first list argument. For example:
p1(1).
p2(2).
?- bar( [p1, p2], 1 ).
Yes
?- bar( [p1, p2], 2 ).
Yes
?- bar( [p1, p2], 3 ).
No
With closures we are not restricted to atoms, and we might save some coding effort. For example we can do the following directly:
?- bar( [=(1), =(2)], 1 ).
Yes
?- bar( [=(1), =(2)], 2 ).
Yes
?- bar( [=(1), =(2)], 3 ).
No
Best Regards
(*)
Draft Technical Corrigendum 2
http://www.complang.tuwien.ac.at/ulrich/iso-prolog/dtc2#call
(**)
Who Invented It?: call/n
Predicates
http://www.complang.tuwien.ac.at/ulrich/Prolog-inedit/naish.html
来源:https://stackoverflow.com/questions/4129484/prolog-gnu-univ-operator-explanation-of-it