Is a lambda an atom in the book "the little schemer"?

Question

I'm reading the 10th chapter of the book "The little schemer - 4th edition", which implements a simple schemer interpreter. All the other content is fine to me, except the function :atom? in Page 188:

(define :atom?
  (lambda (x)
    (cond
      ((atom? x) #t)
      ((null? x) #f)
      ((eq? (car x) (quote primitive)) #t)
      ((eq? (car x) (quote non-primitive) #t)
      (else #f)))))

I'm not clear about this line:

((eq? (car x) (quote non-primitive) #t)

From the previous of the book, the non-primitive is corresponding to a lambda definition.

(lambda (x) (+ x 1))

has the value of (with the passing environment table):

(non-primitive (table (x) (+ x 1)))

Does this mean a lambda definition is an atom, in the book?

I guess it is but not quite sure about it since I can't find their relationship mentioned in the book.

score 1 · Accepted Answer · answered Apr 06 '16 at 12:14

Source code (lambda (x) (+ x 1)) would not be an atom, but the evaluated value which results in a closure should be identified by the evaluator as an atom.

In lisps everything that is not a a list is an atom. You could define it like this:

(define atom?
  (lambda (x)
    (not (or (null? x)
             (pair? x)))))

Now if you modeled your closures as pairs with first element as a tag identifying it as such you need to make sure those evaluates as atomic too and that's what your :atom? does by looking for primitive and non-primitive

Is a lambda an atom in the book "the little schemer"?

1 Answers1