CODEWITHSUNDEEP
schemeracketletobject-state
user3450695
user3450695

Reputation: 2551

Local state of a variable

I am trying to fully understand Objects and local states of their variables

This code seems to produce different results for the same procedure called multiple times, meaning the local variable changes:

(define new-withdraw
  (let ((balance 100))
    (lambda (amount)
      (if (>= balance amount)
          (begin (set! balance (- balance amount))
                 balance)
          "Insufficient funds"))))

For this other code, it produces the same result, which means it creates a new local variable for every procedure call:

(define (make-account)
  (let ((balance 100))
    (define (withdraw amount)
      (if (>= balance amount)
          (begin (set! balance (- balance amount))
                 balance)
          "Insufficient funds"))
    (define (deposit amount)
      (set! balance (+ balance amount))
      balance)
    (define (dispatch m)
      (cond ((eq? m 'withdraw) withdraw)
            ((eq? m 'deposit) deposit)
            (else (error "Unknown request -- MAKE-ACCOUNT"
                         m))))
    dispatch))

My questions are:

Thank you

Upvotes: 1

Views: 462

Answers (1)

uselpa
uselpa

Reputation: 18917

Testing code 1:

> (new-withdraw 0)
100
> (new-withdraw 50)
50
> (new-withdraw 10)
40

Testing code 2:

> (define ac (make-account))
> ((ac 'withdraw) 0)
100
> ((ac 'withdraw) 50)
50
> ((ac 'withdraw) 10)
40

So both codes keep their local state. The difference between code 1 and code 2 is that code 1 works for one account only, whereas code 2 "creates a new account" on each call - the call to the procedure returns the dispatching procedure that you need to bind to a variable, and then use as shown above.

Therefore you get the impression that the local state is lost; it's not, you were probably creating a new account every time.

Upvotes: 2

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