In Scheme, does `or` and `and` short circuit?

Question

Do and and or short circuit in Scheme?

The following are two implementation of lat? (list of atoms). One uses cond … else and the other uses or and and. I was wondering if they are equivalent and the answer to that hinges on whether or and and have short circuit evaluation in Scheme.

(define lat?
  (lambda (l)
    (cond
         ((null? l) #t)
         ((atom? (car l)) (lat? (cdr l)))
         (else #f))))

• uses cond and else

---

(define lat?
  (lambda (l)
    (or (null? l)
        (and (atom? (car l))
             (lat? (cdr l))))))

• uses or and and

I think or short-circuits. Why? I know (car ()) and (cdr ()) each produce Error: Attempt to apply…. If or didn’t short-circuit, then (lat? ()) would eventually evaluate (car ()) and produce the error. However, (lat? ()) does not produce the error, therefore (via Modus Tollens) or short-circuits. Is this correct? And does and short-circuit?

Answer 1

Yes, they both short-circuit per r6rs specification (didn't find html version of r7rs online but here a link to the pdf version of r7rs specification, see section 4.2):

If there are no <test>s, #t is returned. Otherwise, the expressions are evaluated from left to right until a <test> returns #f or the last <test> is reached. In the former case, the and expression returns #f without evaluating the remaining expressions. In the latter case, the last expression is evaluated and its values are returned.

and & or is subsequently defined in terms of test.