Exporting predicate as an operator

General
1

Which one of 1, 2, 3 below is correct to export a predicate in module 0 with operator declaration ? I think 3 is correct. Once 1 caused compile error. 2 seems OK, but also redundant. 3 seem to work as far as my trials. I am not familiar with exporting operators. Real case is complicated for others to reproduce here.

(0)

:- module(m, []).
:- op(500, fy, a).
a(X).

(1)

:- module(m, [a/1, op(500, fy, a)]).
:- op(500, fy, a).
a(X).

(2)

:- module(m, [(a)/1, op(500, fy, a)]).
a(X).

(3)

:- module(m, [op(500, fy, a)]).
a(X).
2

I remember this being really hard the first time I did it getting it to work and still do not consider exporting/importing operators a friend.

One thing I do remember was that operators need to be thought of as an entry into the operator table using op/3 and also as a predicate thus the need of the predicate indicator. So 2 would be my answer. The way I eventually came to that reasoning when first learning to use operators was by using listing/N to look at generated code of working code in the SWI-Prolog repository on GitHub. If however none of the possible answers is correct then again when Jan W. replies I will have learned something.

3

Operators and predicates are not related, so you need to export both. (2) is the correct answer.

4

Can you add more detail to the meaning of that, thanks.

5

An operator defines a human friendly syntax for a term with arity 1 or 2 as op Arg, Arg1 op Arg2 or Arg Op, This has no relation to a predicate. For example, + is an operator, but there is no predicate +/2. Some people seem to confuse (compound) terms and predicates. They are unrelated. A term is simply a data structure. A predicate links to executable code. For example, p(1) is just data. Only if you use it in the body position of a clause or give it to call/1 it gets related to the predicate p/1.

4 Likes
6

Thanks Jan and Eric

Now, due to the answer, queries on zdd gets more easy to write by exporting (zdd)/1 and op(1150, fy, zdd) than before like this.

% ?- zdd((append([], [], X))).    % before
%@ X = [].
% ?- zdd append([], [], X).   % after
%@ X = [].

% ?- zdd((true, true)).    % before
% ?- zdd true, true.      % after
7

You can think of the op/3 directive as something that modifies the current_op/3 data (which the parser uses). If an op/3 term appears in a module declaration, that does a implicit op/3 directive. You can look at all the operators by:

?- forall(current_op(Precedence, Type, Name), 
  writeln(current_op(Precedence, Type, Name))).
8

Your comment explains well, I think, why paren like (a)/1 is necessary in the exporting list. Thanks.

9

I have read what help(op/3) explains for the first time, and now understand clearly the meaning the symbol x y in operator type.

BTW, I took a ridiculous reading at first the word “identical” that, for example, op(10, xfy, [a, b]) makes a==b is true. Of course, I noticed right after “identical” means that the atom a and b have the same precedece 10 and the type xfy.