CODEWITHSUNDEEP
schemeracket
Chloe_1993
Chloe_1993

Reputation: 15

if and cond in scheme

I figured out how to use if statements in scheme, but not quite sure of how to use cond here. I need to change both if to cond. Any suggestions?

(define (pi err)
  (let loop ([n 0]
             [p +nan.0]
             [c 0.0])
    (if (< (abs (- c p)) (/ err 4))
        (* 4 p)
        (loop (add1 n)
              c
              ((if (even? n) + -) c (/ 1 (+ 1 n n)))))))

Upvotes: 0

Views: 259

Answers (1)

Sylwester
Sylwester

Reputation: 48775

With only one predicate and a then and else expression you won't get any benefits of converting it to a cond, but it is very easy:

(if predicate-expression
    then-expression
    else-expression)

is the same as:

(cond
  (predicate-expression then-expression)
  (else else-expression))

Using your if example it becomes:

(cond
  ((< (abs (- c p)) (/ err 4)) (* 4 p))
  (else (loop ...)))

It's much more interesting if you have nested if like this:

(if predicate-expression
    then-expression
    (if predicate2-expression
        then2-expression
        else-expression))

which turns flatter and usually easier:

(cond
  (predicate-expression then-expression)
  (predicat2e-expression then2-expression)
  (else else-expression))

Upvotes: 1

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