DRY arithmetic expression evaluation in Prolog

Question

I wanted to write evaluating predicate in Prolog for arithmetics and I found this:

eval(A+B,CV):-eval(A,AV),eval(B,BV),CV is AV+BV.
eval(A-B,CV):-eval(A,AV),         


        

Answer 1

Your almost DRY solution does not work for several reasons:

• Formula =.. [Op, L, R] refers to binary operators only. You certainly want to refer to numbers too.

• The arguments L and R are not considered at all.

• Op(L,R) is not valid Prolog syntax.

on the plus side, your attempt produces a clean instantiation error for a variable, whereas positive/1 would fail and eval/2 loops which is at least better than failing.

Since your operators are practically identical to those used by (is)/2 you might want to check first and only then reuse (is)/2.

eval2(E, R) :-
   isexpr(E),
   R is E.

isexpr(BinOp) :-
   BinOp =.. [F,L,R],
   admissibleop(F),
   isexpr(L),
   isexpr(R).
isexpr(N) :-
   number(N).

admissibleop(*).
admissibleop(+).
% admissibleop(/).
admissibleop(-).

Note that number/1 fails for a variable - which leads to many erroneous programs. A safe alternative would be

t_number(N) :-
   functor(N,_,0),
   number(N).