Problem with 'let' syntax in scheme

Question

I'm going through "Structure and Interpretation of Computer Programs" and I'm having a bit of trouble doing one of the exercises ( 2.1 ) . I'm coding in DrRacket in R5RS mode.

here's my code :

(define (make-rat n d) 
  (let (((c (gcd n d))
         (neg (< (* n d) 0))
         (n (/ (abs n) c))
         (d (/ (abs d) c)))
        (cons (if neg (- n) n) d))))

and here's the error message DrRacket is giving me:

let: bad syntax (not an identifier and expression for a binding) in: ((c (gcd n d)) (neg (< (* n d) 0)) (pn (/ (abs n) c)) (pd (/ (abs d) c)))

I think I've messed up let's syntax. but I'm not sure how to fix it.

Answer 1

As your edit indicates, you're using the c identifier prematurely. (Which is why it isn't working after fixing the syntax issue of the extra parenthesis.) Identifiers in "let" don't see each other. You'd need to nest your second three lets under the first.

    (let ((c (gcd ...)))
      (let ((...))
        exps ...))

I don't recall when/if SICP introduces other let forms, but if you are stuck using a lot of nested lets, you can use let* in which each subsequent identifier is in the scope of all the previous. That is, the following two definitions are equivalent:

(define foo
  (let* ((a 1)
         (b (+ 1 a))
         (c (+ 1 b)))
    (+ 1 c)))

(define foo
  (let ((a 1))
    (let ((b (+ 1 a)))
      (let ((c (+ 1 b)))
        (+ 1 c)))))

The scoping rules of the different let forms can be a bit much for a beginner, unfortunately.

Answer 2

I added an extra set of parentheses around the variable declarations, whoops.

Also, since I used c to define n and d, I had to change let into let* to make it work properly

my fixed code:

(define (make-rat n d) 
  (let* ((c (gcd n d))
         (neg (< (* n d) 0))
         (n (/ (abs n) c))
         (d (/ (abs d) c)))
        (cons (if neg (- n) n) d)))

Answer 3

Try this:

(define (make-rat n d)
  (let ([c (gcd n d)]
        [neg (< (* n d) 0)]
        [n (/ (abs n) c)]
        [d (/ (abs d) c)])
    (cons (if neg
              (- n)
              n) 
          d)))