对？ 功能应用于球拍中的报价

Question

In order to implement derivative of given polynomial, I need to factorize the polynomial, ie, (* 3 xy) should be the product of 3 and (* xy) .

So I implemented a function multiplicand to get the second factor of a product:

(define (multiplicand p) 
  (let ((second-factor (cdr (cdr p))))
  (if (pair? second-factor) (cons '* second-factor)
   second-factor)))

and the test code is

(multiplicand '(* x y))

But the output is '(* y) . It seems that the condition (pair? second-factor) equals #true with second-factor values 'y .

Can anybody help me with this, thanks a lot.

Answer 1

Bear in mind that (cdr (cdr ...)) is returning a list (not an element!), so pair? will return true if the list has enough elements (three or more). Perhaps you were aiming for something like this?

(define (multiplicand p) 
  (if (null? (cdddr p)) ; assuming list has at least 3 elements
      p
      `(* ,(second p) (* ,(third p) ,(fourth p)))))

(multiplicand '(* x y))
 => (* x y)

(multiplicand '(* 3 x y))
 => (* 3 (* x y))

Answer 2

Manipulating symbolic expressions is what match is made for:

(define (multiplicand p)
  (match p
     [(list '* m n) n]
     [_ (error 'multiplicand (~a "expected product, got: " p))]))