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Exercise 22.5.11 Develop a function multiply-all that takes in a list of numbers and returns the result of multiplying them all together.
For example: (check-expect (multiply-all (cons 3 (cons 5 (cons 4 empty)))) 60)
Hint: What is the “right answer” for the empty list? It may not be what you think at ?rst!

Solution: The data de?nition is similar to that for list-of-strings:

; A list-of-numbers is either
; empty or
; a nelon (non-empty list of numbers).
#|
(define (function-on-lon L)
; L a list of numbers
(cond [ (empty? L) ...]
[ (cons? L) (function-on-nelon L)]
))
|#
; A nelon looks like
; (cons number lon )
#|
(define (function-on-nelon L)
; L a cons
; (first L) a number
; (rest L) a lon
; (function-on-lon (rest L)) whatever this returns
...)
|#

Any suggestions?

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1 Answer

For the simplest solution, use apply for this:

(define (multiply-all lst)
  (apply * lst))

If you need to build the procedure from scratch, just remember that the base case (an empty list) should return 1, and the recursive step should multiply the current value using the standard solution template, like this:

(define (multiply-all lst)
  (if (empty? lst)
      1
      (* (first lst)
         (multiply-all (rest lst)))))

For a nicer answer, you can try using tail recursion:

(define (multiply-all lst)
  (let loop ([lst lst] [acc 1])
    (if (empty? lst)
        acc
        (loop (rest lst) (* (first lst) acc)))))

Anyway the procedures work as expected:

(multiply-all '())
=> 1
(multiply-all '(3 5 4))
=> 60

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