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Name, Places, Frames, and EnvironmentsStateful Evaluation RulesExam 1Tandem Repeats
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Class 22: Stateful Evaluation Rules
cs1120 Fall 2011David Evans14 October 2011
2
Plan
Names and PlacesEnvironmentsRevised Evaluation RulesExam 1
3
Dennis Ritchie, 1941-2011
1972 (with Ken Thompson and PDP-11)
4
Brie
f His
tory
of
Prog
ram
min
g La
ngua
ges
Fortran (1955, Backus)LISP
(1959, McCarthy)
Algol(1960)
C(1969, Ritchie)
Scheme(1975, S&S)
Python(1991, von Rossum)
Java(1995, Sun)
C++(1983, Stroustrup)
Simula(1967)
JavaScript(1995, Eich)
PS1-5
PS6-7
PS8
5
Why design a new programming language?
C: wanted both portability andlow-level machine access
6Image: cc http://en.wikipedia.org/wiki/File:Unix_history-simple.svg
Multics
Ken Thompson and Dennis Ritchie
Android iOS
Windows and Symbian:not direct descendants of Unix, but they are mostly programmed in C
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“Here’s how to succeed: by being lucky. Grab on to something that’s moving pretty fast. Let yourself be carried on when you’re in the right place at the right time.”
Dennis Ritchie
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Recap: Names and Places
A name refers to a place for storing a value.define creates a new place(set! name expr) changes the value in the place name to the value of expr
mcons creates a mutable pair of two new places(set-mcar! pair expr) changes the value in the mcar
place of pair to the value of expr(set-mcdr! pair expr) changes the value in the mcdr
place of pair to the value of expr
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Application and Places
(lambda (x) …) also creates a new place named xThe passed argument is put in that place
> (define x 3)> ((lambda (x) x) 4)4> x3 How are these
places different?
x : 3
x : 4
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Location, Location, Location
Places live in framesAn environment is a frame and a pointer to a
parent environmentAll environments except the global environment
have exactly one parent environment, global environment has no parent
Application creates a new environment
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Environments
globalenvironment
> (define x 3)
+ : #<primitive:+>
null? : #<primitive:null?>
The global environment points to the outermost frame. It starts with all Scheme built-ins defined.
x : 3
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Stateful Definition Evaluation Rule
A definition creates a new place with the definition’s name in the frame associated with the evaluation environment. The value in the place is value of the definition’s expression.
If there is already a place with the name in the current frame, the definition replaces the old place with a new place and value.
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Stateful Name Evaluation Rule To evaluate a name expression, search the evaluation environment’s frame for a place with a name that matches the name in the expression. If such a place exists, the value of the name expression is the value in that place. Otherwise, the value of the name expression is the result of evaluating the name expression in the parent environment. If the evaluation environment has no parent, the name is not defined and the name expression evaluates to an error.
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Evaluating NamesTo evaluate a name expression, search the evaluation environment’s frame for a place with a name that matches the name in the expression. If such a place exists, the value of the name expression is the value in that place. Otherwise, the value of the name expression is the result of evaluating the name expression in the parent environment. If the evaluation environment has no parent, the name is not defined and the name expression evaluates to an error.
globalenv x : 3
x : 17
y : 3
How are environments like this created?
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Procedures
globalenvironment
> (define double (lambda (x) (+ x x)))
+ : #<primitive:+>null? : #<primitive:null?>
double: ??
x : 3
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How to Draw a Procedure
• A procedure needs both code and an environment– We’ll see why soon
• We draw procedures like this:Environmentpointer
environment: parameters: xbody: (+ x x)
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How to Draw a Procedure (for artists only)
Environmentpointer
x (+ x x)Input parameters(in mouth) Procedure Body
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Procedures
globalenvironment
> (define double (lambda (x) (+ x x)))
+ : #<primitive:+>null? : #<primitive:null?>
double:
x : 3
environment:parameters: xbody: (+ x x)
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Application
• Old rule: (Substitution model)
Apply Rule 2: Constructed Procedures. To apply a constructed procedure, evaluate the body of the procedure with each formal parameter replaced by the corresponding actual argument expression value.
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Stateful Application Rule(Constructed Procedures)
To apply a constructed procedure:1. Construct a new environment, whose parent is the
environment of the applied procedure.2. For each procedure parameter, create a place in the
frame of the new environment with the name of the parameter. Evaluate each operand expression in the environment or the application and initialize the value in each place to the value of the corresponding operand expression.
3. Evaluate the body of the procedure in the newly created environment. The resulting value is the value of the application.
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1. Construct a new environment, parent is procedure’s environment pointer
2. Make places in that frame with the names of each parameter, and operand values
3. Evaluate the body in the new environment
globalenvironment
> (double 4)8
+ : #<primitive:+>
x : 3
x : 4
(+ x x)
double:
environment:parameters: xbody: (+ x x)
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x : 3
x : 17
y : 3
What would create this environment?
Think about this, we’lldiscuss it next week…
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Exam 1
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Overall Results
5 below 70 5 above 100
Come to my office hours or arrange a meeting to go over things you didn’t understand
Solutions/comments will be posted on course site later by tomorrow.
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Honor15. Do you trust your classmates to follow the honor expectations in this class? — Yes, I trust them completely.— I worry that there may be a few transgressions, but I believe the vast majority
of the class is honorable and it is fair and beneficial to rely on this.— I think this class places too high a burden on students’ honor, and there are
enough dishonorable students that it is unfair on the honorable students.— I have reason to suspect that other students violated the honor policy on
problem sets.— I have direct knowledge of other students violating the honor policy on
problem sets.— I have reason to suspect that other students violated the honor policy on this
exam.— I have direct knowledge of other students violating the honor policy on this
exam.0000
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Honor Responses
Too High
Vast Majority
Between 1 and 2
Trust Completely
0 5 10 15 20 25 30 35
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Problem 11
Define a procedure, count-tandem-repeats, that takes as input a list, p, and a number, n. The output should be the total number of times the first n elements of p are consecutively repeated at the beginning of p. Repetitions may not overlap.
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(define (count-tandem-repeats p n) (count-prefix-repeats p (list-prefix p n)))
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(define (list-prefix p n) (if (= n 0) null (cons (car p) (list-prefix (cdr p) (- n 1)))))
What is the asymptotic running time of list-prefix?
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(define (contains-matching-prefix p q) ; question 10 (or (null? q) (and (eq? (car p) (car q)) (contains-matching-prefix (cdr p) (cdr q)))))
(define (list-prefix p n) (if (= n 0) null (cons (car p) (list-prefix (cdr p) (- n 1)))))
(define (count-prefix-repeats p q) (if (contains-matching-prefix p q) (+ 1 (count-prefix-repeats ((n-times cdr (length q)) p) q)) 0)) (define (count-tandem-repeats p n) (count-prefix-repeats p (list-prefix p n))) Not defensive: errors if
(< (length p) n)
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Returning Exam 1
(define (list-quicksort cf p) (if (null? p) null (list-append (list-quicksort cf (list-filter (lambda (el) (cf el (car p))) (cdr p))) (cons (car p) (list-quicksort cf (list-filter (lambda (el) (not (cf el (car p)))) (cdr p)))))))
(list-quicksort (lambda (s1 s2) (< (get-email-id s1) (get-email-id s2))) cs1120-students)
Code from Chapter 8
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