CS 152 Programming Language Paradigms March 12 Class

Smalltalk IDE Class Categories Class Names Message Categories Message Selectors Text Section SJSU Dept.

Screenshot of the original Smalltalk environment. SJSU Dept. of Computer Science Spring 2014: March

Smalltalk Messages o o Unary message: A message with no arguments. Keyword message: A

Smalltalk Messages, cont’d o Binary messages allow you to write arithmetic and comparison expressions

Smalltalk Variables o Use variables to refer to objects. n Example: The assignment operator

Smalltalk Variables, cont’d o o Statements are separated by periods. A sequence of messages

Smalltalk Equality and Identity Operators o o The equality operator is = The object

Smalltalk Statements o The to: do message creates a loop. n o Even control

Smalltalk Statements, cont’d o An if. True: if. False message expresses a choice of

Review for the Midterm o A test of understanding, not memorization. o Open book,

Review for Midterm, cont’d o Historical overview n n n o von Neumann architecture

Review for Midterm, cont’d o Paradigms n n n o o o Functional Logic

Review for Midterm, cont’d o Good language design n n n Efficiency Regularity Generality

Review for Midterm, cont’d o Functional programming n n n o Programs as functions

Review for Midterm, cont’d o Scheme, cont’d n Procedures and predicates Conditional expressions n

Review for Midterm, cont’d o Scheme, cont’d n Procedure as a first-class object o

Review for Midterm, cont’d o Logic n o Deductive reasoning Prolog n Declarative language

Review for Midterm, cont’d o Object-oriented programming n n n n Software reuse and

Review for Midterm, cont’d o Smalltalk n n Purely object-oriented Objects Messages IDE _

Sample Midterm Question: Scheme o Suppose you have a Scheme procedure min-to-head that takes

Sample Midterm Question: Scheme, cont’d o Write a recursive Scheme procedure sort that uses

Sample Midterm Question: Scheme, cont’d n n Base case: The empty list. Procedure sort

Sample Midterm Question: Scheme, cont’d o Write a recursive Scheme procedure min-to-head. n o

Sample Midterm Question: Scheme, cont’d o Assume that min-to-head has correctly worked on the

Sample Midterm Question: Prolog o Given this Prolog database and the query: /*1*/ go(0,

Sample Midterm Question: Prolog, cont’d /*1*/ go(0, 0) : - !. /*2*/ go(Control, Control).

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CS 152: Programming Language Paradigms March 12 Class Meeting Department of Computer Science San Jose State University Spring 2014 Instructor: Ron Mak www. cs. sjsu. edu/~mak

Smalltalk IDE Class Categories Class Names Message Categories Message Selectors Text Section SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 2

Screenshot of the original Smalltalk environment. SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 3

Smalltalk Messages o o Unary message: A message with no arguments. Keyword message: A message that expect arguments. n n The message selector (method name) ends in a colon. Example: Set new includes: 'Hello' o o o includes is a boolean method that has as an element argument and returns whether or not the element is in the set. Does the new set include the element 'Hello' ? If a message has more than one argument, a keyword and colon must precede each argument. n Example: a at: i+1 put: (a at: i). n at: put: is the complete message selector. SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 4

Smalltalk Messages, cont’d o Binary messages allow you to write arithmetic and comparison expressions with infix notation. n Examples: 3 + 4 3 < 4 SJSU Dept. of Computer Science Spring 2014: March 12 "Returns 7" "Returns true" CS 152: Programming Language Paradigms © R. Mak 5

Smalltalk Variables o Use variables to refer to objects. n Example: The assignment operator is : = or o Temporary variables are declared between vertical bars n o They are not capitalized. Smalltalk variables have no assigned data type n A variable can name any thing. SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 6

Smalltalk Variables, cont’d o o Statements are separated by periods. A sequence of messages to the same object are separated by semicolons: n o Example: Smalltalk variables use reference semantics, not value semantics. n A variable refers to an object. It does not contain an object. SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 7

Smalltalk Equality and Identity Operators o o The equality operator is = The object identity operator is == SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 8

Smalltalk Statements o The to: do message creates a loop. n o Even control statements are expressed with message passing. Enclose a code block with square brackets [ ] n n Similar to a Schema lambda form. A block can contain arguments as block variables. _ SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 9

Smalltalk Statements, cont’d o An if. True: if. False message expresses a choice of action. o Print the contents of an array: "Print to the transcript each element followed by a carriage return. " array do: [: element | Transcript next. Put. All: element print. String; cr] SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 10

Review for the Midterm o A test of understanding, not memorization. o Open book, open notes, open laptop, open Internet. n n But not open neighbor! Forbidden to communicate with anyone else during the exam. _ SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 11

Review for Midterm, cont’d o Historical overview n n n o von Neumann architecture Machine and assembly languages FORTRAN COBOL Algol Pascal Abstractions n n n Data Control Basic Structured Unit SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 12

Review for Midterm, cont’d o Paradigms n n n o o o Functional Logic Object-oriented von Neumann architecture von Neumann bottleneck Language syntax Language semantics Compilers Interpreters SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 13

Review for Midterm, cont’d o Good language design n n n Efficiency Regularity Generality Orthogonality Uniformity Security Extensibility _ SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 14

Review for Midterm, cont’d o Functional programming n n n o Programs as functions Recursion Referential transparency Lack of state First-class objects Higher-order functions Scheme n n Atoms and lists Prefix notation Binding List construction and destruction SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 15

Review for Midterm, cont’d o Scheme, cont’d n Procedures and predicates Conditional expressions n Recursion n o o flat mutual deep tail recursion n Symbol manipulation Symbolic differentiation n Scope n o o local bindings closure SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 16

Review for Midterm, cont’d o Scheme, cont’d n Procedure as a first-class object o o n as an argument as a return value Metalinguistic power o o eval apply _ SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 17

Review for Midterm, cont’d o Logic n o Deductive reasoning Prolog n Declarative language Objects and relations Fact, rules, and questions Conjuctions Variables n Backtracking n n o n n n place markers Resolution Unification Cut SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 18

Review for Midterm, cont’d o Object-oriented programming n n n n Software reuse and independence Extension Redefinition Abstraction Polymorphism Encapsulation Loose coupling Frameworks _ SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 19

Review for Midterm, cont’d o Smalltalk n n Purely object-oriented Objects Messages IDE _ SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 20

Sample Midterm Question: Scheme o Suppose you have a Scheme procedure min-to-head that takes an argument list of top-level integers and returns a list where the minimum element of the list is moved to the head. n n If there is a tie, it moves only one of the minimum elements. Example: (define nums '(4 3 2 1 0 1 8 0 2 9)) (min-to-head nums) (0 4 3 2 1 1 8 0 2 9) SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 21

Sample Midterm Question: Scheme, cont’d o Write a recursive Scheme procedure sort that uses procedure min-to-head as a helper to sort an argument list of top-level integers and return a list where the elements are sorted in ascending order. n Example: (sort nums) (0 0 1 1 2 2 3 4 8 9) SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 22

Sample Midterm Question: Scheme, cont’d n n Base case: The empty list. Procedure sort calls helper procedure min-to-head to move the smallest element to the head of the list. Call sort recursively on the remainder of the min-to-headed list in order to move the next smallest element to the second position. Continue until the base case. (define sort (lambda (lst) (if (null? lst) '() (let ((mth-lst (min-to-head lst))) (cons (car mth-lst) (sort (cdr mth-lst)))) n This is a selection sort. SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 23

Sample Midterm Question: Scheme, cont’d o Write a recursive Scheme procedure min-to-head. n o Base case: The empty list or a list containing only one element. Otherwise, assume that min-to-head was successfully applied to the cdr of the list. o o The procedure has moved the minimum value of the cdr of the list to the second position of the list. Compare the car of the list to the second position. o If necessary, swap the two values to put the lesser value at the head of the list. (define min-to-head (lambda (lst) (cond ((null? lst) '()) ((null? (cdr lst)) lst) (else (let ((mth-cdr (min-to-head (cdr lst)))) (if (> (car lst) (car mth-cdr)) (append (list (car mth-cdr) (car lst)) (cdr mth-cdr)) lst))) SJSU Dept. of Computer Science CS 152: Programming Language Paradigms 24 Spring 2014: March 12 © R. Mak )))

Sample Midterm Question: Scheme, cont’d o Assume that min-to-head has correctly worked on the cdr of the list. n o o Therefore, the second element of the list is the minimum within the cdr of the list. Compare the car of the list (the green element) against the red element. n If necessary, swap them. n Now the minimum of the entire list is at the head. Recursively call min-to-head on the cdr of the list. SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 25

Sample Midterm Question: Prolog o Given this Prolog database and the query: /*1*/ go(0, 0) : - !. /*2*/ go(Control, Control). /*3*/ go(Control, Start) : - Next is Start - 1, go(Control, Next). /*4*/ nasa(Start) : - go(Control, Start), write(Control), nl, fail. /*5*/ launch(Start) : - not(nasa(Start)), write('We have liftoff!'). ? - launch(10). n Explain the result of the query in terms of resolution, unification, and backtracking. SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 26

Sample Midterm Question: Prolog, cont’d /*1*/ go(0, 0) : - !. /*2*/ go(Control, Control). /*3*/ go(Control, Start) : - Next is Start - 1, go(Control, Next). /*4*/ nasa(Start) : - go(Control, Start), write(Control), nl, fail. /*5*/ launch(Start) : - not(nasa(Start)), write('We have liftoff!'). ? - launch(10). o Rule 5: First subgoal: nasa(10) matches Rule 4. o o Rule 4: First subgoal: Match Fact 2. o o o Variable Start is bound to 10. Variable Control is instantiated and shared with Start which is 10. Rule 4: First subgoal satisfied. Rule 4: Second subgoal: write(Control) print “ 10”. Rule 4: Third subgoal: Print a new line. Rule 4: Fourth subgoal: Fail. SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 28

Sample Midterm Question: Prolog, cont’d /*1*/ go(0, 0) : - !. /*2*/ go(Control, Control). /*3*/ go(Control, Start) : - Next is Start - 1, go(Control, Next). /*4*/ nasa(Start) : - go(Control, Start), write(Control), nl, fail. /*5*/ launch(Start) : - not(nasa(Start)), write('We have liftoff!'). ? - launch(10). o Rule 4’s failure causes a backtrack to match Rule 3. o o n o o o Rule 3: First subgoal: Set variable Next to Start – 1 which is 9. Rule 3: Second subgoal: Match Rule 2, set Control to 9. Rule 4: First subgoal satisfied. Rule 4: Second subgoal: write(Control) print “ 9”. Continue printing “ 8”, “ 7”, etc. down to “ 0”. When the countdown reaches 0, the cut operation in Rule 1 stops the countdown from backtracking further. SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 29