###Ex 3.9
Here is the diagram for the recursive version of factorial:
And here is the diagram for the iterative version:
###Ex 3.10
If we "desugar" the let expression, we get the following:
(define (make-withdraw initial-amount)
((lambda (balance)
(lambda (amount)
(if (>= balance amount)
(begin (set! balance (- balance amount))
balance)
"Insufficient funds"))) initial-amount))A let reduces to a lambda which we immediately apply.
Here is the resulting diagram from the expression (define W1 (make-withdraw 100)):
when we apply make-withdraw to 100, the first thing we do is create a new environment, E1. E1 contains a single binding, which binds initial-value to 100. Within E1, we then evaluate the body of make-withdraw. The body is a procedure application, but the procedure being applied must first be created by lambda, so we create a new procedure object, which takes a single parameter, balance, with the given body (which is another call to lambda) and whose environment pointer points to E1. Then we immediately apply the created procedure to initial-value in E1, which results in the creation of E2, which points to E1, and contains a binding from balance to 100. The body of the procedure created from the first lambda is itself a call to lambda, so we create a second procedure object, which takes as a parameter amount, and whose enviroment pointer points to E2. This procedure object is the result of the original application of make-withdraw, so back in the global environment, W1 is now bound to this procedure object.
After this, the application of (W1 50) is very similar to the example given in the book, so I won't bother making a diagram for it. (define W2 (make-withdraw 100)), will create seperate evironments E3 and E4 analogous to E1 and E2 respectively.
###Ex 3.11
Here is the diagram for (define acc (make-account 50)):
I dont show the separate procedure objects, so you'll have to use your imagination :).
Here is the diagram for ((acc 'deposit) 40):
Here, environment E2 is set up for the application of acc to 'deposit, which evaluates to the deposit procedure. Then E3 is set up in order to apply deposit to 40.
The local state for acc is stored in E2, where balance is bound. Likewise, for acc2, its state would be in the environment that was setup in order to apply make-account to 100. Other than the different instances of balance, acc and acc2 share th same withdraw, deposit and dispatch methods, both environments (acc's and acc2's) could point to the same procedure objects.