David Evans cs.virginia/evans

1. Lecture 17: Environments David Evans http://www.cs.virginia.edu/evans CS200: Computer Science University of Virginia Computer Science

2. Menu • Environments • Evaluation Rules • Exam 1 CS 200 Spring 2003

3. Environments CS 200 Spring 2003

4. (define nest (lambda (x) (lambda (x) (+ x x)))) > ((nest 3) 4) 8 Does the substitution model of evaluation tell us how to evaluate this? CS 200 Spring 2003

5. Review: Names, Places, Mutation • A name is a place for storing a value. • define creates a new place • cons creates two new places, the car and the cdr • (set! nameexpr) changes the value in the place name to the value of expr • (set-car! pairexpr) changes the value in the car place of pair to the value of expr • (set-cdr! pairexpr) changes the value in the cdr place of pair to the value of expr CS 200 Spring 2003

6. Lambda and Places • (lambda (x) …) also creates a new place named x • The passed argument is put in that place > (define x 3) > ((lambda (x) x) 4) 4 > x 3 x : 3 x : 4 How are these places different? CS 200 Spring 2003

7. Location, Location, Location • Places live in frames • An environment is a pointer to a frame • We start in the global environment • Application creates a new frame • All frames except the global frame have one parent frame CS 200 Spring 2003

8. Environments + : #<primitive:+> global environment null? : #<primitive:null?> x : 3 The global environment points to the outermost frame. It starts with all Scheme primitives. > (define x 3) CS 200 Spring 2003

9. Procedures + : #<primitive:+> global environment null? : #<primitive:null?> x : 3 double: ?? > (define double (lambda (x) (+ x x))) CS 200 Spring 2003

10. Environmentpointer Code pointer parameters: xbody: (+ x x) How to Draw a Procedure • A procedure needs both code and an environment • We’ll see why soon • We draw procedures like this: CS 200 Spring 2003

11. Procedures + : #<primitive:+> global environment null? : #<primitive:null?> x : 3 double: > (define double (lambda (x) (+ x x))) parameters: xbody: (+ x x) CS 200 Spring 2003

12. Application • Old rule: (Substitution model) Apply Rule 2: Compounds. If the procedure is a compound procedure, evaluate the body of the procedure with each formal parameter replaced by the corresponding actual argument expression value. CS 200 Spring 2003

13. Application • Construct a new frame, enclosed in the environment of this procedure • Make places in that frame with the names of each parameter • Put the values of the parameters in those places • Evaluate the body in the new environment CS 200 Spring 2003

14. global environment + : #<primitive:+> • Construct a new frame, enclosed in the environment of this procedure • Make places in that frame with the names of each parameter • Put the values of the parameters in those places • Evaluate the body in the new environment double: x : 3 parameters: xbody: (+ x x) 4 x : > (double 4) 8 (+ x x) (+ 4 4) 8 CS 200 Spring 2003

15. global environment (define nest (lambda (x) (lambda (x) (+ x x)))) > ((nest 3) 4) + : #<primitive:+> nest: x : 3 parameters: xbody: (lambda (x) (+ x x)) 3 x : (nest 3) ((lambda (x) (+ x x)) 4) 4 x : CS 200 Spring 2003 (+ x x)

16. Evaluation Rule 2 (Names) If the expression is a name, it evaluates to the value associated with that name. To find the value associated with a name, look for the name in the frame pointed to by the evaluation environment. If it contains a place with that name, use the value in that place. If it doesn’t, evaluate the name using the frame’s parent environment as the new evaluation environment. If the frame has no parent, error (name is not a place). CS 200 Spring 2003

17. evaluate-name (define (evaluate-name name env) (if (null? env) (error “Undefined name: …”) (if (frame-contains name (get-frame env)) (lookup name (get-frame env)) (evaluate-name name (parent-environment (get-frame env)))))) Hmm…maybe we can define a Scheme interpreter in Scheme! CS 200 Spring 2003

18. How does CS200 compare to other CS classes? CS 200 Spring 2003

19. CS 101 (Spring 2001) 7. Why do programmers comment? (a) Give compiler extra instructions (b) Aid readability (c) To hide code for debugging purposes (d) (b) and (c) (e) none of the above “Correct” answer CS 200 Spring 2003

20. Why Comment? • To assign credit, blame or (non) liability /* I was drunk when I wrote this. */ • To document external assumptions (e.g., this parameter must be a sorted list) • To remind you what you are doing (; This is a bug!) • To keep code readers entertained with jokes, poems, and stories • To make sure your employer has to rehire you to fix the code (comments in foreign languages are especially effective) • Because you are taking a course where programs are judged based on the number of comments in it, not the clarity and economy of your code CS 200 Spring 2003

21. CS 201: Software Development Methods(Fall 2001) • 5. [13 points total] Recursion. • The “SuperCool” function is defined for positive integers greater than or equal to zero as: • supercool(i) = supercool(i-1) * supercool(i-2); • supercool(0) = 1; • supercool(1) = 2; • For example: • supercool(2) = 2 * 1 = 2 • supercool(3) = 2 * 2 = 4 • supercool(4) = 4 * 2 = 8 • supercool(5) = 8 * 4 = 32 • Write a recursivefunction that calculates supercool. Your function must be recursive to receive any credit. CS 200 Spring 2003

22. CS 415: Programming Languages (Spring 2002) 3. Define a function (zip op x y) that takes lists (x1 x2 x3 ....) and (y1 y2 y3 ...) and evaluated to the list ((op x1 y1) (op x2 y2) (op x3 y3) ...). Examples: > (zip + '(1 2 3 4) '(5 10 15 20)) (6 12 18 24) > (zip list '(a b c d) '(6 7 8 9)) ((a 6) (b 7) (c 8) (d 9)) CS 200 Spring 2003

23. zip (define (zip f list1 list2) (if (null? list1) null (cons (f (car list1) (car list2)) (zip f (cdr list1) (cdr list2))))) (define zip map) (Easy way) CS 200 Spring 2003

24. CS588: Cryptology Problem Set 1, question 1: QPIV AUH DKV UA PKEXHE QA ATHU QPKU QPH AQPHE AUH - DIVH CZO THIS ONE WAS NO HARDER TO OPEN THAN THE OTHER ONE - WISE GUY (From Safecracker Meets Safecracker in “Surely You're Joking, Mr. Feynman!: Adventures of a Curious Character”, Richard P. Feynman.) CS 200 Spring 2003

25. CS588: Cryptology Midterm, last question: The following statements were taken from Microsoft's recently released "Safe Internet: Microsoft Privacy & Security Fundamentals" web site and their "The Ten Immutable Laws of Security" page). For each of these statements, indicate whether the statement is true, misleading or an outright falsehood. Support your answer with 1-3 sentences. An incorrect, but well-supported answer is worth more credit that a correct, poorly-supported answer. Some parts of the statements are bold to focus your attention. a. (7) Unless your e-mail application provides encryption features such as the use of digital signatures, your messages are about as private as a letter sent in an unsealed envelope. CS 200 Spring 2003

26. CS588: Cryptology Midterm, last question: Outright Falsehood. Digital signatures (like physicial signatures) provide authenticity, not privacy. Anyone can read a signed message by decrypting using the signer's public key. CS 200 Spring 2003

27. CS 655 (Spring 2001) 1. Mergering a. Write a Scheme procedure listadder that combines two lists by adding their elements. For example, (listadder '(3 1 3) '(3 4 2)) should produce (6 5 5). CS 200 Spring 2003

28. listadder (define (listadder list1 list2) (if (null? list1) '() (cons (+ (car list1) (car list2)) (listadder (cdr list1) (cdr list2))))) CS 200 Spring 2003

29. PhD Qualifying Exam Write a procedure posfilter that takes a list of integers as input and evaluates to the list with all the non-positive elements removed. For example, (posfilter (list 34 -2 50 0 23)) evaluates to (34 50 23) (define posfilter enronize) CS 200 Spring 2003

30. If you don’t already have enronize… (define (posfilter lst) (if (null? lst) null (if (> (car lst) 0) (cons (car lst) (posfilter (cdr lst))) (posfilter (cdr lst))))) If you got a 7 or better on question 5, you did better than more than half of our PhD students! (18 of you got 10/10) CS 200 Spring 2003

31. Part d (of that question) Write a procedure filter that takes a list of integers and a filter function as input and evaluates to the list with only the elements for which the filter function evaluates to true. (define (filter lst f) (if (null? lst) null (if (f (car lst)) (filter (cdr lst)) (cons (car lst) (filter (cdr lst)))))) CS 200 Spring 2003

32. make-enronize Why couldn’t I ask the PhD students to implement make-enronize? CS 200 Spring 2003

33. PhD Qualifying Exam • Define a procedure that takes a binary tree as input, and produces as output the mirror image of that tree, swapping the left and right sides at every node. • How much work is your procedure? 12 of you did better on this than our best PhD student this year did! CS 200 Spring 2003

34. Charge • No class Friday – enjoy your “break”! • But that doesn’t mean you have no work to do… • Mutation makes evaluation rules more complicated • Environment diagrams quickly get complicated – but like substitution evaluations, just follow rules mechanically • Don’t wait until after Spring Break to start PS5 • Finish GEB part I before next class. CS 200 Spring 2003