CODEWITHSUNDEEP
prolog
H.G.W
H.G.W

Reputation: 33

Defining operators in Prolog

Having defined two operators in Prolog: 

op(100, xfy, #).
op(100, fy, ϴ).

what is the expression

a # ϴ b # c

equivalent to?

a # ϴ (b # c)

or maybe

a # ((ϴ b) # c)

And why?

Upvotes: 3

Views: 557

Answers (1)

lurker
lurker

Reputation: 58234

You can see what Prolog will do with it using write_canonical/1:

?- op(100,xfy,#).
true.

?- op(100,fy,@).
true.

?- write_canonical(a # @ b # c).
#(a,@(#(b,c)))
true.

So, it appears that your first speculation is correct. The interpretation of a # @ b # c is a # (@(b # c)). The key comment in the documentation for op/3 bearing on this is:

The f indicates the position of the functor, while x and y indicate the position of the arguments. y should be interpreted as "on this position a term with precedence lower or equal to the precedence of the functor should occur". For x the precedence of the argument must be strictly lower.

The use of fy results in the precedence grouping of @(b # c) rather than (@b) # c.

Upvotes: 4

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