Difference between copy_term/2 and functor/3

Help!
1

While working on a directive to allow iterating through succeeding depths of recursion for an arbitrary predicate (even if the predicate does not terminate in normal prolog), it was pointed out that I should use functor/3 instead of copy_term/2 to obtain a most generic term.

What is the difference between these two if the argument of the compound terms is not instantiated?

34 ?- copy_term(myterm(T),NewTerm1).
NewTerm1 = myterm(_1194).

35 ?- functor(NewTerm2,myterm,1).
NewTerm2 = myterm(_1104).

In my code it turns out the term generated by functor/3 is more generic, than the one produced by copy_term/2 as Jan pointed out.

I can’t grasp why the difference, if T is not instantiated.

EDIT: I understand copy_term/2 shares ground terms and functor/3 does not (of course), but here we’re dealing with an argument which is not instantiated.

2

I see. In your code the term that you copy is the most general term, so your copy will be most general too.

Probably using functor/3 is faster though. copy_term/2 has to do a lot of work to figure out variable sharing, subtem sharing, cycles and ground terms. functor/3 simply allocates a term and fills it with variables.

3

In my original code I was doing something like this:

   functor(Head,Pred,Arity),
   % ...some code here...
   wrap_predicate(M:Head, iterate_depth, OrigPred,
      (
	 % We need to run iterate_depth_/4 only
	 % one time, on entry. Recursive calls
	 % need to be done normally (as if they
	 % were not wrapped)
	 prolog_current_frame(F),
	 prolog_frame_attribute(F,parent,P),
	 copy_term(Head,Head1),
	 (  prolog_frame_attribute(P,parent_goal,Head1)
	 -> call(OrigPred)
	 ;  iterate_depth_(OrigPred, Start, Step, End, _Result)
	 )
      )
   ).

But this worked erratically until I changed it to functor/3 like you suggested.

4

Thanks for this, I was using copy_term/2 because I wrongly thought it would be faster.
I’ll keep using functor/3.