CS 152: Programming Language Paradigms March 12 Class Meeting Department of Computer Science San Jose State University Spring 2014 Instructor: Ron Mak

SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 2 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 3 Screenshot of the original Smalltalk environment.

SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 4 Smalltalk Messages  Unary message: A message with no arguments.  Keyword message: A message that expect arguments. The message selector (method name) ends in a colon. Example:  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. Example: at:put: is the complete message selector. Set new includes: 'Hello' a at: i+1 put: (a at: i).

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

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

SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 7 Smalltalk Variables, cont’d  Statements are separated by periods.  A sequence of messages to the same object are separated by semicolons: Example:  Smalltalk variables use reference semantics, not value semantics. 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 8 Smalltalk Equality and Identity Operators  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 9 Smalltalk Statements  The to:do message creates a loop. Even control statements are expressed with message passing.  Enclose a code block with square brackets [ ] 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 10 Smalltalk Statements, cont’d  An ifTrue:ifFalse message expresses a choice of action.  Print the contents of an array: array do: [:element | Transcript nextPutAll: element printString; cr] "Print to the transcript each element followed by a carriage return."

SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 11 Review for the Midterm  A test of understanding, not memorization.  Open book, open notes, open laptop, open Internet. 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 12 Review for Midterm, cont’d  Historical overview von Neumann architecture Machine and assembly languages FORTRAN COBOL Algol Pascal  Abstractions Data Control Basic Structured Unit

SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 13 Review for Midterm, cont’d  Paradigms 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 14 Review for Midterm, cont’d  Good language design Efficiency Regularity Generality Orthogonality Uniformity Security Extensibility _

SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 15 Review for Midterm, cont’d  Functional programming Programs as functions Recursion Referential transparency Lack of state First-class objects Higher-order functions  Scheme 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 16 Review for Midterm, cont’d  Scheme, cont’d Procedures and predicates Conditional expressions Recursion  flat  mutual  deep  tail recursion Symbol manipulation Symbolic differentiation Scope  local bindings  closure

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

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

SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 19 Review for Midterm, cont’d  Object-oriented programming 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 20 Review for Midterm, cont’d  Smalltalk Purely object-oriented Objects Messages IDE _

SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 21 Sample Midterm Question: Scheme  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. If there is a tie, it moves only one of the minimum elements. Example: (define nums '( )) (min-to-head nums)  ( )

SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 22 Sample Midterm Question: Scheme, cont’d  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. Example: (sort nums)  ( )

SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 23 Sample Midterm Question: Scheme, cont’d 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. This is a selection sort. (define sort (lambda (lst) (if (null? lst) '() (let ((mth-lst (min-to-head lst))) (cons (car mth-lst) (sort (cdr mth-lst)))) )))

SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 24 Sample Midterm Question: Scheme, cont’d  Write a recursive Scheme procedure min-to-head. 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.  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.  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 Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 25 Sample Midterm Question: Scheme, cont’d  Assume that min-to-head has correctly worked on the cdr of the list. 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. If necessary, swap them. 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 26 Sample Midterm Question: Prolog  Given this Prolog database and the query: Explain the result of the query in terms of resolution, unification, and backtracking. /*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).

SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 27 Sample Midterm Question: Prolog, cont’d  Answer:  Resolution: The query launch(10) matches Rule 5.  Unification: Rule 5: Variable Start is instantiated and bound to 10.  Rule 5: First subgoal: nasa(10) matches Rule 4. _ /*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).

SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 28 Sample Midterm Question: Prolog, cont’d  Rule 5: First subgoal: nasa(10) matches Rule 4.  Variable Start is bound to 10.  Rule 4: First subgoal: Match Fact 2.  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. ?- launch(10). /*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!').

SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 29 Sample Midterm Question: Prolog, cont’d  Rule 4’s failure causes a backtrack to match Rule 3.  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. ?- launch(10). /*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!').

SJSU Dept. of Computer Science Spring 2014: March 12 CS 152: Programming Language Paradigms © R. Mak 30 Sample Midterm Question: Prolog, cont’d  Rule 4: Fourth subgoal causes the countdown to end in failure.  Rule 5: First subgoal has a not, and therefore that subgoal succeeds.  Rule 5: Second subgoal prints “We have liftoff!” _ /*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!').