מבוא מורחב למדעי המחשב בשפת Scheme תרגול 1. Outline Administration Dr. Scheme Functional vs. Imperative Programming Compiler vs. Interpreter Evaluation.

מבוא מורחב למדעי המחשב Schemeבשפת

1תרגול

Outline

• Administration

• Dr. Scheme

• Functional vs. Imperative Programming

• Compiler vs. Interpreter

• Evaluation Rule

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תרגולים ועזרים

ספר הקורס )ניתן לקריאה ברשת, קישור •)מאתר הקורס

www.cs.tau.ac.il/~schemeאתר הקורס: •

מתרגלים•

מעבדה•

פורום )קישור מאתר הקורס(•

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http://www.cs.tau.ac.il/~scheme

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תרגיליםתרגילים שבועיים שהגשתם חובה.•הכנת התרגילים היא אישית.• שלהן )רלוונטי דוגמאות הרצהתוכניות יוגשו יחד עם •

החל מהתרגיל השני(. מהתרגילים בהצלחה.80%נדרש: לפחות •[email protected]הגשה בדוא"ל: •לכתוב שם, ת.ז., מספר קבוצה, לצרף הצהרה חתומה •

.ולשמור את אישור ההגשהערעור לבודק בכתב – לתיבת הדואר האישית.•פרטים באתר הקורס.•

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Dr. Scheme התקנת

(:209התקנת המשערך )גירסא •התקנה בבית, דרך אתר הקורס, או

http://www.drscheme.org

:Linuxמותקן במעבדה בבניין שרייבר במערכת הפעלה drscheme-209&

בהפעלה הראשונה )בלבד(Language->Choose Language>-

Graphical )under PLT tab((Run)לאחר השינוי יש ללחוץ על כפתור ה-

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חלון ההגדרות

חלון האינטרקציות

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בדיקת תחביר

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שמירת קבצים והדפסתם

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עבודה בחלון האינטרקציות

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עבודה בחלון ההגדרות

Click on “Run”

Machine vs. High-Level Language

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Assembly Code

BinaryCode

Meaning High-Level Language Code

ADD 5 7 15 35 5 7 15

Op. code of ADD command is 35

Adds contents of the memory cell number)address( 5 to the content of cell #7 and put the results into the cell #15

X:=Y+Z

)+ Y Z(

JEQ 5 8 9 48 5 8 9

Op. code of JEQ command is 48

Examines the content of the memory cell #5 . If it’s equal to zero it executes command located in the memory cell #8 , otherwise executes command located in the memory cell #9

If)a==0( then …..else ……end

)IF a exp1 exp2(

A Compiler is a software program that translates)compiles( programs written in high-level language into machine)binary( code

Compiler vs. InterpreterCompiler Interpreter

Translate the whole program into a binary code understandable by hardware

Evaluate commands )or expressions( one by one.

Analyze code only once Analyze every expression each time it is executed

Recompile whole program for any change

Only changed expression/command is reevaluated

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Dr. Scheme is an interpreter

Functional vs. Imperative Programming

Functional Imperative

Basic paradigm

Evaluation of expressions without side effects )changes in the memory state(

Result of one expression can be used as an argument to another

Execution of commands that change an environment )memory state(

One command change )prepare( environment for another

Example Compute: f)n(=“n* f)n-1( n>1 or 1 if n=1”on n=5;

s:=1; k:=1Repeat 5 times: s:=s*k; k:=k+1;

Effect Evaluates function n! on n=5. No changes in the environment.

Change state of the variable s so it contains 5!. Also change state of the variable k.

15Scheme is a functional programming language )not purely functional(

Why functional programming?

• Program Optimization– Memorize value of expression instead of repeated

evaluation– Do not evaluate an expression if its value is not

used– No dependency => expressions can be evaluated

in different order or in parallel

• Languages– Lisp, Scheme, Haskell, Erlang

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Syntax vs. Semantics

Syntax Semantics

Which expression is a legal expression?

How to evaluate/execute a legal expression?

A list is a legal expression in Scheme language

Scheme uses the “evaluation rule” to compute value of an expression

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)# 5 7( -is syntactically correct but has no meaning

)f 5 7( –meaning)value( depends on the meaning of the symbol f

Postfix, Infix , Prefix notation

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Name Description Example

Infix Operator is in-between operands

)5+7(*8

Postfix Operator is after the operands

5 7 + 8 *

Prefix Operator is before the operands

* + 5 7 8

)* )+ 5 7( 8( : Scheme uses the prefix notation for arithmetic expressions

Reminder: The Scheme Language

Element Example Semantics

Simple Expression

<,+,586 Element’s value

Means of Combination

)+ 1 3 )* 5 7(( Value of an expression

Means of abstraction

)define x 5( Associate name with a value

Means of abstraction

)lambda)x( )* x x(( Create compound procedure )function(

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Primitive vs. Compound Procedures

• Primitive procedures are part of the Scheme language )not defined using lambda expressions(– Arithmetic)+,*(, IF, DEFINE

• Compound procedures are created by programmer using the lambda expressions– )lambda)x( )+ x 2((

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Some Scheme ExpressionsExample Description

)+ )- 9 7( 3( Arithmetic expression

)> 5 2()AND #t )> 10 5(()OR #f )>10 20((

Boolean expressions

)define x )+ 5 2(( Name definition

)define g )lambda)a( )+a 1((( Function )compound procedure( definition

)define )g a( )+ a 1((

)define )g b c( )+ b c((

Function definition using syntactic sugar

)g x())lambda)d( )+ d 1(( 2(

Function call )application of compound procedure(

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Reminder: Syntactic Sugar

• Shortcuts in the syntax, do not add power to the language

– Can write )define )f x( )* x x( (

– Instead of )define f )lambda)x( )* x x(( (

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Example in DrScheme

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מספרים רציונליים ומרוכבים

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מספרים גדולים

•Scheme תומך ב"מספרים גדולים", כאלה שגדולים מהמספרים -264, > 232-1המקסימליים שארכיטקטורת מחשב תומכת בהם ) >

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Reminder: Environment table

• )define x 5( associates name x to value of 5 in an environment table

• )define f )lambda)x()* x x((( associates name f to the function lambda)x()* x x( in an environment table

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Name Value

x 5

f lambda)x()* x x(

Environment Table>)+ x 5(10>)f 5(25

Reminder: How to evaluate an expression?

• An expression is usually built out of sub-expressions– )expr_0 expr_1 … exp_n(

• The first sub-expression evaluates to a primitive or compound procedure– )f )+ 2 3( )g 2((

• Need well-defined rule to evaluate any expression

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Remainder: Evaluation rule1. Reduce step:

Evaluate all sub-expressions )in any order(

If the procedure to apply is a primitive, just do it.

)+ )* 5 2( )* 5 4((

2. Expand step)substitution model( If the procedure to apply is a compound procedure:

Substitute by the body of the procedure while replacing each formal parameter with the corresponding actual arguments

)define )f x( )* x x((

)f 5(-> )* 5 5(

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What is the order of the application of step 1 and step 2?

)f )* 5 2((

Normal vs. Applicative order

Applicative Normal

Step 0 )f )+ 2 3( ( )f )+ 2 3( (

Step 1 Reduce: )f 5( Expand: )+ )+ 2 3( 1(

Step 2 Expand: )+ 5 1( Reduce: )+ 5 1(

Step 3 6 6

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)define )f x( )+ x 1((

• Usually same value if expressions have no side effects• Consider ))lambda)x y( )+ x 2(( 1 )/ 1 0((• Scheme uses Applicative order

Why Applicative order?

• Think of the optimization)define )f x( )* x x((

)define )g x( )sqrt x((

Normal order evaluates )sqrt 2( twice

)f )sqrt 2(( )* )sqrt 2( )sqrt 2((

• What would happen if sqrt produced a side effect )e.g. prints on a display(?

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What is the Scheme order?(define (square x) (* x x))(define (sum-of-squares x y) (+ (square x) (square y)))(define (foo a) (sum-of-squares (+ a 1) (* a 2)))(foo 5)

With DrScheme stepper*:

*works for Beginning Student Language level only

Reminder: Special Forms

Evaluation not according to standard rule or/and may have side effects 32

Form Effect Example

)DEFINE name expr( Changes environment table )DEFINE x )+ 2 1(()DEFINE )f x( )+ x 1((

)IF a expr1 expr2( Only one out of expr1 and expr2 is evaluated depending on the value of a

)IF )= x 0( )myprint 1( )myprint 2((

)AND exp1 exp2( If exp1 is false ,exp2 is not evaluated

)AND )> 0 a( )> 2 )/ 1 a((

)OR exp1 exp2 exp3( If exp1 is true , exp2 is not evaluated, if exp2 is true, exp3 will not be evaluated

)OR )= 0 a( )> 1 )/ 2 a((

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דוגמא – ערך מוחלט

(define (abs x)

(if (< x 0)

(- x)

x))

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עוד דוגמא

(define (foo a b)

((if (> b 0) + -) a b))

(foo 2 6)

(foo 5 -4)