SICP Exercise 2.29 Confusion

Question

When I typed my original solution to subproblem b. of exercise 2.29 in SICP:

(define (total-weight m)
  (let ((left (left-branch m))
        (right (right-branch m)))
    (cond ((null? m) 0)
          ((not (pair? m)) m)
          (else
           (+ (total-weight (branch-structure left))
              (total-weight (branch-structure right)))))))

and tested it with the following data:

(define left1 (make-branch 5 8))
(define right1 (make-branch 7 10))
(define m1 (make-mobile left1 right1))
(define right2 (make-branch 1 3))
(define m2 (make-mobile left2 right2))
(define left0 (make-branch 12 m1))
(define right0 (make-branch 5 m2))
(define m0 (make-mobile left0 right0))
(total-weight m0)

The interpreter (MIT/GNU Scheme) reported error: "the object 3, passed as the first argument to cdr, is not the correct type". But when I eliminated the expression

(let ((left (left-branch m))
      (right (right-branch m)))
  ...)

with the following code:

(define (total-weight m)
  (cond ((null? m) 0)
        ((not (pair? m)) m)
        (else
         (+ (total-weight (branch-structure (left-branch m)))
            (total-weight (branch-structure (right-branch m)))))))

the program worked fine and printed the result

;Value: 27

I got confused. Can anybody take a trial on this problem and help me out?

Answer 1

The problem is that in the first version, (left-branch m) and (right-branch m) is called before you check whether or not m represents a mobile. i.e. m can be a number, or nil.