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Why isn't this Racket code ending?

I am reading about lazy evaluation and do not understand how they took the main example.

#lang racket
(define (bad-if x y z)
  (if x y z))
(define (factorial-wrong x)
  (bad-if (= x 0)
          1
          (* x (factorial-wrong (- x 1)))))

(factorial-wrong 4)

I am a little confused as to why this program never ends. I know the following code works really well:

(define (factorial x)
  (if (= x 0)
      1
      (* x (factorial (- x 1)))))

(factorial 4)

So I guess it has something to do with scope. I've tried step by step debugging and factorial-wrong performs the recursion function even when x is mapped to 0.

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2 answers


Standard if 

(if test-expr then-expr else-expr)


will evaluate only then-expr

or else-expr

, depending on test-expr

, because this one if

is either a special form or a syntactic extension on a special form, which means it does not conform to the normal evaluation rules.

bad-if

, on the other hand, is a standard procedure. In this case, the schema evaluates both expressions first , since they are procedure parameters bad-if

before actually executing bad-if

. So even at x = 0 there will be an estimate (* x (factorial -1))

, which in turn will estimate (* x (factorial -2))

, etc. In an endless loop.

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Use a pedometer!

More specific:



  • Remove the #lang racket from the top of your program.
  • Change the language level to "Intermediate learner"
  • Click the Step button. Watch closely to see what happens off the tracks.
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