# Local State Variables

## A Preview

Let's take a quick glance at what we will be going over in this section: (define (make-account balance)
(define (withdraw amount)
(set! balance (- balance amount)) balance)
(define (deposit amount)
(set! balance (+ balance amount)) balance)
(define (dispatch msg)
(cond
((eq? msg 'withdraw) withdraw)
((eq? msg 'deposit) deposit) ) )
dispatch)


What do you think? Do you have any idea about what this function does?

## Withdraw

Let's withdraw money from the bank account. We will do this using a procedure withdraw, which takes as argument an amount to be withdrawn. If there is enough money in the account to accommodate the withdrawal, then withdraw should return the balance remaining after the withdrawal. Otherwise, withdraw should return the message "Insufficient funds". For example, if we begin with \$100 in the account, we should obtain the following sequence of responses using withdraw:

(withdraw 25)
75
(withdraw 25)
50
(withdraw 60)
"Insufficient funds"
(withdraw 15)
35


Observe that the expression (withdraw 25), evaluated twice, yields different values.

Wait, but I thought that a particular function call with the same argument returns the same value!

Up to now it has, but this is a new kind of behavior for a procedure. All our procedures could be viewed as functions that pass the vertical line test. A call to a procedure computes the value of the function applied to the given arguments, and two calls to the same procedure with the same arguments always produces the same result. But in this situation, the balance needs to be changed after each transaction. Otherwise, we all are going to be rich!

To implement withdraw, we can use a variable balance to indicate the balance of money in the account and define withdraw as a procedure that accesses balance. The withdraw procedure checks to see if balance is at least as large as the requested amount. If so, withdraw decrements balance by amount and returns the new value of balance. Otherwise, withdraw returns the Insufficient funds message. Here are the definitions of balance and withdraw:

(define balance 100)

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


Decrementing balance is accomplished by the expression

(set! balance (- balance amount))


This uses the set! special form, whose syntax is

(set! [name] [new-value])


Here [name] is a symbol and [new-value] is any expression. Set! changes [name] so that its value is the result obtained by evaluating [new-value]. In the case at hand, we are changing balance so that its new value will be the result of subtracting amount from the previous value of balance.

Withdraw also uses the begin special form to cause two expressions to be evaluated in the case where the if test is true: first decrementing balance and then returning the value of balance. In general, evaluating the expression

(begin [exp1] [exp2] ... [expk])


causes the expressions [exp1] through [expk] to be evaluated in sequence and the value of the final expression [expk] to be returned as the value of the entire begin form.

Play with withdraw, set! and begin on your STk interpreter!

## Something's Fishy...

Before we move on, examine again how withdraw and balance are defined:

(define balance 100)

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


Do you see anything that could cause a trouble?

## Trouble Detected

The problem is with the variable balance. As specified above, balance is a name defined in the global environment and is freely accessible to be examined or modified by any procedure. It would be much better if we could somehow make balance internal to withdraw, so that withdraw would be the only procedure that could access balance directly and any other procedure could access balance only indirectly (through calls to withdraw). This would more accurately model the notion that balance is a local state variable used by withdraw to keep track of the state of the account.

We can make balance internal to withdraw by rewriting the definition as follows:

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


What we have done here is use let to establish an environment with a local variable balance, bound to the initial value 100. Within this local environment, we use lambda to create a procedure that takes amount as an argument and behaves like our previous withdraw procedure. This procedure -- returned as the result of evaluating the let expression -- is new- withdraw, which behaves in precisely the same way as withdraw but whose variable balance is not accessible by any other procedure.

> (new-withdraw 10)
90
> (new-withdraw 30)
60


Play with new-withdraw on the STk interpreter and make sure you understand how it works.

## make-account

Here is a simplified version of the make-account procedure in SICP:

(define (make-account balance)
(define (withdraw amount)
(set! balance (- balance amount)) balance)
(define (deposit amount)
(set! balance (+ balance amount)) balance)
(define (dispatch msg)
(cond ((eq? msg 'withdraw) withdraw)
((eq? msg 'deposit) deposit) ) )
dispatch)


Now, let's try to rewrite this using local state variables. Fill in the blank in the following code so that the result works exactly the same as the make-account procedure above. That is, it responds to the same messages and produces the same return values. The differences between the two procedures are that the inside of make-account above is enclosed in the let statement below, and the names of the parameters to make-account are different.

(define (make-account init-amount)
(let (______________________)
(define (withdraw amount)
(set! balance (- balance amount)) balance)
(define (deposit amount)
(set! balance (+ balance amount)) balance)
(define (dispatch msg)
(cond ((eq? msg 'withdraw) withdraw)
((eq? msg 'deposit) deposit) ) )
dispatch) )


Now, modify either version of make-account so that, given the message balance, it returns the current account balance, and given the message init-balance, it returns the amount with which the account was initially created. For example,

> (define acc (make-account 100))
acc
> (acc 'balance)
100


Make another modification such that, given the message transactions (any deposit or withdrawal), it returns a list of all transactions made since the account was opened. For example:

> (define acc (make-account 100))
acc
> ((acc 'withdraw) 50)
50
> ((acc 'deposit) 10)
60
> (acc 'balance)
60
> (acc 'transactions)
((deposit 10) (withdraw 50))


Before viewing the entire solution below, try out your definition in the STk interpreter and make sure you understand the entire code for make-account.

Here is our solution:

(define (make-account init-amount)
(let ((balance init-amount)
(transactions '()))
(define (withdraw amount)
(set! balance (- balance amount))
(set! transactions (cons (list 'withdraw amount) transactions))
balance)
(define (deposit amount)
(set! balance (+ balance amount))
(set! transactions (cons (list 'deposit amount) transactions))
balance)
(define (dispatch msg)
(cond ((eq? msg 'withdraw) withdraw)
((eq? msg 'deposit) deposit)
((eq? msg 'balance) balance)
((eq? msg 'transactions) transactions) ) )
dispatch) )


## The Substitution Model of Evaluation

Given this definition:

(define (plus1 var)
(set! var (+ var 1))
var)


Follow the substitution model to find the result of computing

(plus1 5)


That is, show the expression that results from substituting 5 for var in the body of plus1, and then compute the value of the resulting expression.

Now, try it in the STk interpreter. Did you get the same answer? Why or why not?

Introducing assignments accompanies a pretty big cost. At this point, you may realize that we cannot use the substitution model of evaluation anymore because it yields the wrong value. The trouble here is that substitution is based ultimately on the notion that the symbols in our language are essentially names for values. But as soon as we introduce set! and the idea that the value of a variable can change, a variable can no longer be simply a name. Now a variable somehow refers to a place where a value can be stored, and the value stored at this place can change.

Then how can I evaluate the procedures?

The new model of evaluation is waiting for you in the next subsection.

## Takeaways

In this section, you learned:

1. How to implement local state variables
2. Costs of assignments
3. How to use set! and begin

## What's Next?

Let's go to the next subsection and learn about the new model of evaluation!

8 - Assignment, State, and Environment