Defining Predicates

In the discussion of custom data types we discussed the variety of types in Scheme is extended using predicates. Instead of defining “classes” or “prototypes” and instantiate objects as having a certain type predicates are used to determine if a given expression/value matches certain criteria. Generically speaking a predicate is a procedure expecting one argument that evaluates to #t or #f depending on criteria specific to the requested type. In this chapter we get back to that topic as an exercise that will help us get a better understanding of procedure definitions.

In that earlier chapter we introduced the color? predicate. Now we're in the position to investigate its actual definition which can be found in the file scm/output-lib.scm within LilyPond's installation directory:

(define-public (color? x)
  (and (list? x)
       (= 3 (length x))
       (every number? x)
       (every (lambda (y) (<= 0 y 1)) x)))

We define a procedure with the name of color? and one argument x. Appending a question mark to the name is the Scheme convention for predicates. (And it is created with define-public because output-lib.scm is a Scheme module, and earlier I told you that definitions have to be explicitly made public within modules.)

The body of the expression is one single and expression, so the procedure will evaluate to a true value if all of the sub-conditions are met. If on the other hand any single subexpression evaluates to #f the predicate will also evaluate to #f. The (four) conditions that a value has to meet in order to be a “color” are:

  • (list? x) The value has to be a list
  • (= 3 (length x)) This list must have exactly three elements (representing the red, green and blue components)
  • (every number? x) All three elements must be (real) numbers
  • (every (lambda (y) (<= 0 y 1)) x)

The last condition should be inspected more closely. every is like and but with lists. TODO: Reference to subchapter of "list operations": it applies a procedure to a list of arguments, one after another, and returns either #f or the value of the application to the last list element. But what is the procedure that is applied here? it's a lambda expression, in other words: an unnamed local procedure:

(lambda (y) (<= 0 y 1))

which every will apply to each element of the x list. The local procedure expects a single argument y and will check if it is a number between (including) 0 and 1. This <= expression expects numbers and would trigger errors otherwise, but from the previous subexpression in the and we know that it is a number. <= will evaluate to #t or #f and not to an arbitrary value, so the every expression will do so as well - it is not possible that any other “true value” than #t will be the outcome. Therefore finally the value of the whole predicate will be either #t or #f. This is the specific requirement when writing predicates: they have to return real #t and not just true values.

Practising With Predicates

To get a better understanding of predicates and types let's write a few predicates as an exercise and investigate some characteristics.

Specifying Type More Narrowly

start with something really simple: checking for a positive integer number:

#(define (positive-integer? x)
   (and (integer? x)
        (> x 0)))

Again we have an and expression in the body, this time we first check if x is an integer number and then if it's greater than zero. Now let's see a somewhat more involved predicate, checking if a color is “reddish” (which we define as the “red” component being stronger than the sum of the “green” and “blue” parts):

#(define (reddish? col)
   (and (color? col)
        (>= (first col)
            (+ (second col) (third col)))))

#(display (reddish? red))
% => #t
#(display (reddish? blue))
% => #f
#(display (reddish? (list 0.7 0.35 0.4)))
% => #f
#(display (reddish? magenta))
% => #t

First we check if the tested object is a color in the first place, and of course we don't reimplement that check but use the existing color? predicate. As the second subexpression of the and we build the sum of the second and third list elements and compare that to the first list element.

Choice

A common situation is that values with one out of several types can be accepted in a certain place. For these cases there already are a number of X-or-Y? predicates available, and one can easily write custom predicates as well. Imagine for some obscure reason you expect a variable to be either a list, a color or a symbol, then you can create the following predicate:

#(define (list-or-color-or-symbol? x)
   (or (list? x)
       (color? x)
       (symbol? x)))

While the previous - “narrowing” - predicates used the and conditional this type of predicates tends to use or instead.

Another typical “choice” type of predicate would check if a value is part of a predefined list:

#(define (mode? x)
   (and (symbol? x)
        (or (eq? x 'major)
            (eq? x 'minor))))

This would return #t when (and only when) applied to 'major or 'minor.

Caveat: “True Values”

As a last example I'm going to show you a somewhat more involved example with a caveat: checking if an object is an association list that contains a specific key:

#(define (alist-with-color? x)
   (and (list? x)
        (every pair? x)
        (assq 'color x)))

#(display
  (alist-with-color?
   '((amount . 5)
     (color . red))))

Surprisingly, when we compile this code the output on the console isn't #t or #f but (color . red). This is because the predicate procedure has the value its last expression has, and this is the assq in this case. From the discussion of association lists we recall that the return value is either #f or the retrieved pair - but what we need is a simple #t in this case.

This means whenever a test used in a predicate returns “a true value” it has to be wrapped in order to really return #t or #f. Which is fortunately very easy to do:

#(define (alist-with-color? x)
   (and (list? x)
        (every pair? x)
        (if (assq 'color x)
            #t
            #f)))

The assq is wrapped in an if expression, so if assq returns “a true value” the expression manually returns #t instead.

A good exercise to do on your own now would be to write a predicate alist-with-key? where you can additionally specify the key whose presence you'd like to check.


Last update: January 31, 2020