1. מבוא מורחב למדעי המחשב בשפת Scheme תרגול 8

2. Environment Model 3.2, pages 238-251

3. Environments Binding: a pairing of a name and a value Frame: a table of bindings Environment: a sequence of frames A precise, completely mechanical, description of: name-rulelooking up the value of a variable define-rulecreating a new definition of a var set!-rulechanging the value of a variable lambda-rulecreating a procedure application rule applying a procedure The Environment Model

4. Name-rule: A name X evaluated in environment E gives the value of X in the first frame of E where X is bound Define-rule: A define special form evaluated in environment E creates or replaces a binding in the first frame of E Set!-rule: A set! of variable X evaluated in environment E changes the binding of X in the first frame of E where X is bound Lambda-rule: A lambda special form evaluated in environment E creates a procedure whose environment pointer points to E Application Rule: To apply a compound procedure P to arguments 1. Create a new frame A 2. Make A into an environment E: A's enclosing environment pointer goes to the same frame as the environment pointer of P 3. In A, bind the parameters of P to the argument values 4. Evaluate the body of P with E as the current environment

5. (define (square x) (* x x))

6. (square 5)

7. (define (square x) (* x x)) (define (sum-of-squares x y) (+ (square x) (square y))) (define (f a) (sum-of-squares (+ a 1) (* a 2)))

8. Application: (f 5)

9. Nested Procedures (define g (lambda () (lambda (x y) (* x y)))) (define f (g)) (f 3 4) => 12

10. GE p: b:(lambda (x y) (* x y)) g: (define g (lambda () (lambda (x y) (* x y))))

11. GE p: b:(lambda (x y) (* x y)) g: f: p: x y b: (* x y) E1 empty (define f (g))

12. GE p: b:(lambda (x y) (* x y)) g: f: p: x y b: (* x y) E1 empty (f 3 4) X=3 Y=4 E2

13. Nested Procedures (define g (lambda (z) (lambda (x y) (* x y z)))) (define f (g 2)) (f 3 4) => 24

14. GE p: z b:(lambda (x y) (* x y z)) g: f: (define g (lambda (z) (lambda (x y) (* x y z))))

15. GE p: z b:(lambda (x y) (* x y z)) g: f: p: x y b: (* x y z) E1 Z: 2 (define f (g 2))

16. GE p: z b:(lambda (x y) (* x y z)) g: f: p: x y b: (* x y z) E1 Z: 2 (f 3 4) X=3 Y=4 E2

17. Let expressions (let (( )) ) is syntactic sugar for ((lambda ( ) ) ) (define a 5) (define b 6) (let ((a 2) (c a)) (+ a b c)) = ((lambda (a c) (+ a b c)) 2 a)

18. Let – cont. GE a: 5b: 6 p: a c b: (+ a b c) E1 a: 2 c: 5 (define a 5) (define b 6) (let ((a 2) (c a)) (+ a b c)) = ((lambda (a c) (+ a b c)) 2 a)

19. The cash machine (define (make-withdraw balance) (lambda (amount) (if (>= balance amount) (begin (set! balance (- balance amount)) balance) "Insufficient funds"))) (define W1 (make-withdraw 100)) > W1 > >(W1 50) > >(W1 40) > >(W1 20) > 50 10 Insufficient funds #

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

21. (define W1 (make-withdraw 100))

22. (W1 50)

24. More than one cash machine >(define W1 (make-withdraw 100)) >(define W2 (make-withdraw 100)) > >(W1 50) > >(W2 40) > 50 60

25. (define W2 (make-withdraw 100))

26. question from past exams 26

27. Make-line (define (make-line a b) (lambda (x) (cond ((pair? x) (set! a (car x)) (set! b (cdr x))) (else (+ b (* x a)))) ) ) (define a 4) (define b 5) (define proc (make-line 1 2)) Q1

28. make-line: a: 4 b: 5 proc: GE p: a b b:(lambda (x)… a: 1 b: 2 E1 p: x b:(cond…

29. make-line: a: 3 b: 5 proc: GE p: a b b:(lambda (x)… a: 1 b: 2 E1 p: x b:(cond… x: 1 E2 (+ b (* x a)) (set! a (proc 1))

30. make-line: a: 3 b: 5 proc: GE p: a b b:(lambda (x)… a: 1 3 b: 2 4 E1 p: x b:(cond… x: 3.4 E3 (set! a (car x)) (set! b (cdr x)) (proc (cons 3 4))

31. make-line: a: 3 b: 5 proc:c: 7 GE p: a b b:(lambda (x)… a: 3 b: 4 E1 p: x b:(cond… x: 1 E4 (+ b (* x a)) (define c (proc 1))

32. ( (lambda (z) (define x (lambda (x) (lambda (y z) (y x)))) ( ( (x (lambda () z)) (lambda (z) z) 3 ) ) ) 2)

33. ExpressionEnvComment 1((lambda1(z) op1 op2) 2)GE(op1 op2) 2 E1 3Op1 (define x (lambda2 (x) (lambda3(y z) (y x)))) E1 4Op2 (((x (lambda5 () z)) (lambda4(z) z) 3)) ((op3 (lambda4 (z) z) 3)) E1 5op3 (x (lambda5 () z)) E1l3 6((l3 l4 3))E1 7(y x)E3(l4 l5) 8ZE4l5 9(l5)E1 10ZE52

34. L1:p: z b: (define x...) (...) E1 z: 2 x:L2 L2:p: x b: (lambda (y z) (y x))) L5:p: - b: z E2 x:L5 L3:p: y,z b: (y x) L4:p: z b: z E3 y: L4 z: 3 E4 z: L5 E5 GE