CMPS 135 Introduction to Programming

Instructor:

Dr. Cong-Cong Xing

Part I

Overview of Computers and Programming

0. Java

High level programming language Developed by James Gosling (and his

team) at Sun Microsystems in 1991. Formally born in 1995 Naming: oak (tree) Java (coffee) High popularity in industry and academia

1. Computer Hardware System

Basic architecture

CPUInput devices Output devices

Storage devices

Input devices– Enter data into computers– Ex: keyboard, scanner

Output devices– Observe computation results– Ex: monitor, printer

CPU (Central Processing Unit)– Control unit: coordinate all computation tasks– ALU (arithmetic-logic unit): carry out computations

Storage devices– Major storage device: memory (also called primary

memory, RAM, 1st storage device) which stores info temporarily

2nd storage devices store info permanently. Ex: hard disks, floppy disk, optical disks (CDs, DVDs), U-drive

Motherboard (main board): host memory, CPU, etc.

2. Problem Solving and Programming

Problem

Implementation

Specifications Testing

Design

Requirements

Problem: problem statement Requirements: complete understanding of

the problem (what needs to be done) Specs: explicit list of input(s), output(s),

relevant formulas (if any), data structures, and methodology

Design: algorithm (a step-by-step procedure showing how to solve the problem)

Implementation: coding of design in the chosen language (ex: Java)

Testing: try the best to ensure the program works correctly

3. Overview of Programming Languages

Machine language (lowest level)– Native language for computers– Binary numbers– The only language that computers can understand– Ex: 00011100111100010 (meaning?)

Assembly language (low level)– 2-to-4 letter commands– Ex: ADD R2 R4 (meaning?)– Assembly programs needs to be translated into

machine programs by assembler

High-level languages– English-like constructs– Ex: x = a+b (meaning?)– Examples of high-level languages

• Pascal, C, C++, Fortran, COBOL, Java

4. Processing a High-level Language

editor Sourcefile

CompileSource file

Objectfile

Linker/loader

executablein RAM

Fix err err

MoreObj files

Part II

Welcome to Java

1. The First Program

// my first Java program

public class Greeting

{ public static void main (String args[])

{ System.out.println(“Welcome to Java”);

}

}

2. Dissection of the First Program

What is the output of the program?– Welcome to Java (is displayed on monitor)

How to get the output?– Compile and execute the program

Java is case-sensitive. Main, main, and MaIn are different.

Line by line interpretation (informally)

// my first Java program

public class Greeting

{ public static void main (String args[])

{ System.out.println(“Welcome to Java”);

}

}

Comments. Ignored by compiler.Starts the def of class Greeting

“public” and “class” are key words“Greeting” can be changed

Starts the def of method mainNecessary for every Java program

Every “word” is fixedProg execution starts here

Printout “Welcome to Java”On monitor

Signifies the begin and end Class Greeting and

Method main

3. Compilation and Execution (GUI Dr. Java)

Compilation and Execution (command line)

Every java program must be save in a file ended with .java– Ex: Greeting.java

Compilation: javac <filename>– Ex: javac Greeting.java

Execution: java <filename (w/o .java)>– Ex: java Greeting

(Why command line then?)

4. Java Packages

Many predefined classes are grouped together to form various packages. Ex: class JOptionPane is in package javax.swing

How to use packages?– Syntax: import <name of package>

Ex: print “welcome to java” again

import javax.swing.JOptionPane;public class GreetingAgain{ public static void main (String args[]) { JOptionPane.showMessageDialog(null, “Welcome to Java”); System.exit(0); } }

Direct compiler to load JOptionPane

Terminate progSuccessfully.

Required for GUI.

Print “welcom to Java” in a

Pop-up window(See next slide)

Cont’d

Part III

Basic Elements

1. Identifiers

Names that users make up for describing data and programs

Ex: Greeting (a class name) Rule to make identifiers

– A letter (a – z or A – Z) followed by a combination of letters and digits (0 – 9)

More examples:

A, a, Max, MAX, daysOfWeek (legal) %right, 30days, #missing (illegal)

2. Data Types and Values

Name Type Ex of values

int Integer 1, 2, -1, -2, 3

double Real number 1.0, 2.3, -4.5

char Character ‘a’, ‘m’, ‘1’

String (S is capital)

Strings “abc”, “123d”

boolean Logical true, false

3. Variables

Special identifiers that hold certain type of data Each variable corresponds to a memory location

(or memory cell)– Ex: a 2 variable a holds 2– b 1.23 variable b holds 1.23

Value of variables may be changed (by programmers) at any time

Each variable must have a data type

4. Declaration of variables

Syntax– <type> var1, var2, …, varn;

Ex: – int a, b, c; --------------(a) – double A; --------------(b)

Meaning: (a) declares variables a, b, and c are of type int. (b) declares variable A is of type double

Every variable must be declared first before it can be used

5. Arithmetic Expressions

Math formulas expressed in Java Addition operator: +

– In math: a+b– In Java: a+b

Subtraction operator: -– In math: a-c– In Java: a-c

Multiplication operator: *– In math: b×c– In Java: b*c

Cont’d

Division operator: /– In math: a÷b– In Java: a/b

Modulo operator: %– In math: a mod b (the remainder of a÷b)– In Java: a%b– Ex: 7%4 = 3

Cont’d

Note: a/b, when both a and b are integers, the result of a/b is also an integer (the integer part of the quotient, no rounding). Otherwise, the result is a real.

Ex: 7/4 = 1 2/4 = 0 2.0/4 = 0.5 2/4.0 = 0.5 2.0/4.0 = 0.5

Cont’d

3/2 hours = how many seconds ?

60*60*3/2 or 3/2*60*60 or neither

Cont’d

More examples

math Java

(a+b+c)÷3 (a+b+c)/3

a+b-c÷2 a+b-c/2

(b2-4ac) ÷2a (b*b-4*a*c)/(2*a)

(a+b)(a-b) ÷(a2-b2) fill in

6. Precedence of Operators

From highest to lowest:

( )

* / %

+ - For operator of the same level: innermost

first, from left to right Consistent with math

high

low

Cont’d

Ex: evaluation tree of (a+b+c)/3

(a + b + c) / 3

+

/

1

2

3

result

+

Ex: evaluation tree of (b*b-4*a*c)/(2*a)

(b * b – 4 * a * c) / (2 * a)

/

*

-

*

**1

2

3

4

5

6

result

7. Assignment Statements

Syntax: <var> = <expression>; Ex: a = 1; b = 2; b = a+b-2; a = a+1; b = b – 1; Meaning: evaluates the expression on the right

first and stores the result of evaluation into the variable on the left (a memory cell).

Note: = does not mean “equal”! Types on both sides of = MUST match!

8. Keywords

Reserved words by Java. They have special meanings to Java and should be avoided by users.

Examples: int, double, boolean, char,

while, if, for, void, static

9. Input and Output

String type: another data type in Java. Any double quoted sequence of symbols are of type String.

Ex: “123”, “ab”, “ab45H$%”,

“1”, “_”, “ “

Cont’d

Input statement: reads data into programs– Syntax:

import javax.swing.JOptionPane;

<var> = JOptionPane.showInputDialog(“___”);

note: <var> must be of type String– Meaning: prompts and reads data into variable var– Convert string type variable vstr to int or double type

• <var> = Integer.parseInt(vstr);• <var> = Double.parseDouble(vstr);

Cont’d

Ex: the following statements

String astr; int a; astr=JOptionPane.showInputDialog(“enter a value for a”);

a = Interger.parseInt(astr);

pops up a window and prompts for input and converts astr (String type) to a (int type)

Cont’d

Output statement: display computation results on monitor.– Syntax: System.out.println(“msg”+var); or

System.out.print(“msg”+var);– Meaning: print msg and the value of variable

var on the monitor. – with ln, line is changed. w/o ln, line is not

changed.

Cont’d

– Ex: the statements

a = 1; b =2;

System.out.println(“value of a is: ”+a);

System.out.println(“value of b is: ”+b);

Produces the following on monitor:

value of a is: 1

value of a is: 2

Cont’d

– Ex: the statements

a = 1; b =2;

System.out.print(“value of a is: ”+a);

System.out.println(“value of b is: ”+b);

Produces the following on monitor:

value of a is: 1value of b is: 2

10. Case Study

Problem: write a Java program that prompts and inputs a radius of a circle and computes the area of the circle.

Requirements: users enter the radius, program outputs the corresponding area.

Specification: – Input(s): radius of a circle– Output(s): area of the circle– Relevant formula: area = pi*radius*radius

Cont’d

– Data structure (dictionary of variables)

Name Type Usage

r double Input, used to hold radius

rs String String variable for r

area double Output, used to hold the result

Cont’d Design (structure chart) (How to read the chart? from left to right, from top to

bottom)

START

Declare r, area, rs

double r,areaString rs

Prompt and read r

rs=JOptionPane.showInputDialog(“enter a radius”);

r=Double.parseDouble(rs);

Echor

System.out.println(“radius is”+r);

Computes area

area=pi*r*r

Display area

System.out.println(“area is “+area);

Cont’d

Test Cases: hand-compute some problem instances and use them to run/test your program

r area

1.0 3.14

2.3 16.6

11.2 393.9

Cont’d

Implementation (coding): translate the structure chart into Java code. (next page)

Cont’d

// Name: your name // NSU#: your N-# // date: today's date // course: course # // description: this program prompts a radius of a circle and outputs // the area of the circle

import javax.swing.JOptionPane;

public class CompArea { public static void main(String args[]) { // declaration of variable String rs; double r, area;

Cont’d

// prompts and reads radius rs = JOptionPane.showInputDialog("Enter a radius"); // convert rs to a real number r = Double.parseDouble(rs);

// echo the radius(for printout purpose) System.out.println("The radius is: "+r);

// compute the area area = Math.PI*r*r;

// display the result System.out.println("The area is: "+area);

// terminate the program System.exit(0); } } // end of program

Part IV

Selection Structures

1. Relational and Logical Operators

Relation expressions

Operator Meaning Example Value

< < 1<2, x<1 true, depending on x

> > 3>4 false

<= ≤ 3.1<=3.2 true

>= ≥ 100>=100 true

== = (equal) 3 == 3 true

!= ≠ 3 != 3 false

Cont’d

Logic expressions– Logic “and” operator (&&) (a, b are any

expressions of type boolean)

a b a && b

true true true

true false false

false true false

false false false

Cont’d

Ex: true && false false

(1>3) && (3>1) false

(1>=1) && (1<=1) && (1==1) true

Note : true and false are values (of type boolean). Just like numbers can be used to build arithmetic expressions, true and false can be used to build relational and logic expressions.

– Logic “or” (||) (a, b are any expressions of type boolean)

a b a || b

true true true

true false true

false true true

false false false

Cont’d ex: true || false true

(1>2) || (3>4) false

((1>3)&&(3>1)) || (2>1) true

– Logic “not” (!) (a is any expression of type boolean)

a !a

true false

false true

Cont’d

ex: !true false

!(1>2) true

!(1>3) && (3>1) true

((1>2) && (1==1)) || (!(2>3)) true

(true || (4>=4)) && !(!(4<4)) false

Cont’d

Comparing relational expressions, logic expressions and arithmetic expressions. They are all expressions and each has a type and a value.

Relational exp

Logic exp Arithmetic exp

type boolean boolean int or double

example

3>4 5==5 (1>2) || (3==5) 3+4 (5+4)*2

value false true false 7 18

2. Precedence of Operators

From highest to lowest: ( ) ! * / % + -< <= > >=== !=&&||

Cont’d

Ex: evaluation tree

1 > 2 || 2 > 1

>

||

>1 2

3

true

false true

Cont’d

(1 >= 2) || (3 < 4) && (5 != 5)

>=

||

<

&&

!=12 3

4

5

false true false

false

false

3. if Statement

Syntax:

if (condition) { statement 1; statement 2; …..

statement n; }note: condition must be a

boolean expression Meaning: (in flow chart)

cond

stmnt1

stmnt2

stmntn

falsetrue

note: this is not a structure chart

Cont’d

Ex:

if ( grade >= 60 )

{ System.out.println(“you passed”);

System.out.println(“good work”); }

Its structure chart:grade>=60

print “you passed”

T

print “goodwork”

4. if-else Statement

Syntax: if (condition) { stmnt1; ….. stmntn; } else { stmnt1’; ….. stmntn’; } Meaning: (in flow chart)

cond

stmnt1

stmntn stmntn’

stmnt1’

truefalse

note: this is not a structure chart

Cont’d

Ex: if (grade >=60) { System.out.println(“you passed”); System.out.println(“congratulations!”); } else { System.out.println(“you failed”);

System.out.println(“work harder!”); }

Its structure chart

grade>=60

print “youpassed”

print “congratulations”

print “work harder”

print “youfailed”

T F

Cont’d Note:

– In if and if-else statements, if there is only one statement after if (or else), the pair of { } can be omitted. Conversely, if there is no { } after if (or else) in if and if-else statements, the system will take the first statement (whatever it is) after if (or else) as the “true” (or “false”) body. The rule of thumb is: always put a pair of { } around the “true” and “false” bodies.

Cont’d

Ex: what’s the output of the following code fragment?

Temp = 56;

if (Temp>55)

{ System.out.println(“HD temp is over 55 degree!”);

System.out.println(“Shut down system immediately!”);

}

else

{ System.out.println(“HD temp is OK”);

System.out.println(“Keep monitoring HD”);

}

Cont’d

Ex: what’s the output of the following code fragment?

Temp = 56;

if (Temp>55)

{ System.out.println(“HD temp is over 55 degree!”);

System.out.println(“Shut down system immediately!”);

}

else

System.out.println(“HD temp is OK”);

System.out.println(“Keep monitoring HD”);

Cont’d

Ex: what’s the output of the following code fragment?

Temp = 56;

if (Temp>55)

System.out.println(“HD temp is over 55 degree!”);

System.out.println(“Shut down system immediately!”);

else

{ System.out.println(“HD temp is OK”);

System.out.println(“Keep monitoring HD”);

}

Cont’d

Ex: what’s the output of the following code fragment?

Temp = 56;

if (Temp>55)

System.out.println(“HD temp is over 55 degree!”);

System.out.println(“Shut down system immediately!”);

else

System.out.println(“HD temp is OK”);

System.out.println(“Keep monitoring HD”);

Cont’d

Ex: Given x, write a code fragment s.t. if x=0, print zero; if

x>0, print positive; if x<0, print negative.

Idea? (design?)

X=0?

X>0?zero

pos neg

T F

T F

Cont’d

if (x==0){ System.out.println(“zero”);}else{ if (x>0) { System.out.println(“positive”); } else { System.out.println(“negative”); }}

Cont’d

An alternative design:

if (x==0) System.out.println(“zero”);if (x>0) System.out.println(“positive”);if (x<0) System.out.println(“negative”);

X>0?X=0? X<0?

zero pos neg

T T T

Cont’d

Case study

Find the max among 3 integers.– Problem: Given 3 integers, find the largest

one.– Requirements: write a Java program which

prompts and reads 3 integers, compare them, and outputs the largest integer. E.g., given 3,4,10, then 10 should be selected.

Cont’d

– Specifications: • Input(s): 3 integers• Output(s): the largest one among the 3 integers• Relevant formula: n/a. The major computation is

comparison.• Data structure: (next pg)

Cont’d

Name Type Usage

a int input, hold the 1st integer

b int input, hold the 2nd integer

c int input, hold the 3rd integer

sa String string for a

sb String string for b

sc String string for c

largest int output, hold the largest integer

Cont’d

– Design

start

P & Ra, b, c

Find the max

output end

Read sa Read sb Read sc

largest =a largest=b nothing largest=c

displaylargest

Convertsa to a

Echo ofa

Convert sb to b

Echo ofb

Convert sc to c

Echo of c

a>=b largest>=c

T F T F

Cont’d

– Test cases

inputs output

a b c largest

1 2 3 3

10 24 -5

24

100 34 90 100

-1 -34 -3 -1

Cont’d

An alternative solution – Design (partial)

a>=b

a>=cb>=c

display c display b display c display a

f t

f t f t

5. Switch Statement

Syntax: switch (expression) { case label1: stmnt1; break; case label2: stmnt2; break; …… case labeln: stmntn; break; default: default stmnt; break; }

Note: * break cannot be missed *if two or more cases have the same action, we can combine them: case label1; case label2; stmnts; beak; *type of expression must be int or char. Type of label must be of the same type as that of expression.

Cont’d

Meaning (flow chart)

exp=l1

exp=l2

exp=ln

default

breakstmntn

breakstmnt2

breakstmnt1t

f

t

f

t

f

Cont’d

Structure chartSwitch()

stmnt1 stmnt2 stmnt3 stmntn

exp=l1 exp=l2 exp=l3exp=ln

Cont’d

Ex: (num is a variable of type int) switch (num) { case 1: System.out.println(“It’s 1”); break; case 2: System.out.println(“It’s 2”); break; case 3: case 4: case 5: System.out.println(“It’s 3 or 4 or 5”); break; default: System.out.println(“Invalid data”); break; }

Cont’d

Case Study– Problem: computes a student’s letter grade by the following

chart. (assume score is an integer between 0 and 100) score grade 90+ A 80 – 89 B 70 – 79 C 60 – 69 D 59- F

Cont’d

– Requirements:• Users enter a student’s number grade (e.g. 85), the program

outputs the corresponding letter grade (e.g. B).

– Specifications:• Input(s): a number grade (integer)• Output(s): the corresponding letter grade• Formula: no specific formula. Job is done through

comparisons.• Data structure:

Cont’d

Name Type Usage

score int hold input (number grade)

sscore String String for score

grd int Scaled-down of score

Cont’d

– Design– Q: how to “map” 99, 98, 97,…,90 all to 9 so that we

can output an A? (89, …,80 all to 8, 79,..,70 all to 7, …..)

– A: (trick)type casting reals to integer: coerce reals to integers by dropping everything after decimal point.

example: (int) (8.9) = 8 (int) (99/10.0) = 9 (int) (4.0) = 4 (int) (98/10.0) = 9 (int) (-4.3) = -4 (int) (97/10.0) = 9

Cont’d

(structure chart)

START

P&Rsscore

Decide lettergrade

end

Convetsscore

to score

Echoscore

grd=(int)(score/10.0)

switch(grd)

print “A” print”B” print “C” print “D” print “F”

10,9 8 6 5,4,3,2,1,0switch(grd)

7

cont’d

– Test cases

– Implementation: (next page)

score 100 95 81 89 75 70 66 51 0

grade A A B B C C D F F

Cont’d

– Implementation (partial)– switch(grd)– { case 10: case 9:– System.out.println("A"); break; – case 8:– System.out.println("B"); break;– case 7:– System.out.println("C"); break;– case 6:– System.out.println("D"); break;– case 5: case 4: case 3: case 2: case 1: case 0:– System.out.println(“F"); break;– default:– System.out.println(“invalid data"); break;– }

Part V

Loops

1. while loop

Syntax: while (cond)

{ stmnt1;

…..

stmntn;

}

note: cond must be a boolean expression

loopbody

Cont’d

Meaning (in flow chart)

cond

stm1

stm2

stmn

f

t

In English:(1) cond is evaluated first. As long as cond is true, all statements in the loop body will be executed one by one (probably many times). (2) at the time cond becomes false, the first statement out of loop body will be executed. Loop is exited.

other stmnt

Cont’d

Meaning (in structure chart)

stmnt1 stmnt2 stmntn

cond

Cont’d

Ex: what’s the output of the code fragment?

counter = 1;

while (counter <= 5)

{ System.out.print(counter);

counter = counter + 1;

}

Cont’d

Output is : 12345

How many times has the body of loop been executed? 5 (not 6). There is an attempt for the 6th loop, but it is aborted.

Loop Val of counter at entry of loop

Counter <=5?

Output so far

1 1 true 1

2 2 true 12

3 3 true 123

4 4 true 1234

5 5 true 12345

6 6 false (exit loop)

Cont’d

Ex:

counter = 1; while (counter < 5) { System.out.print(counter); counter = counter + 1; }

How many times will the body be executed?

Cont’d

Ex:

counter = 1; while (counter > 0) { System.out.print(counter); counter = counter + 1; }

How many times will the body be executed?

Cont’d

Ex:

counter = 1; while (counter < 0) { System.out.print(counter); counter = counter + 1; }

How many times will the body be executed?

Cont’d

Ex: output of the following code fragment?

c=10; while (c>=1) { System.out.print(c); c = c-1; }

10987654321

Loop printed Val of c

1 10 9

2 9 8

3 8 7

4 7 6

5 6 5

6 5 4

7 4 3

8 3 2

9 2 1

10 1 0

Cont’d

Ex: a typical use of while loop – control the validity of input data. Suppose we want to read a radius (r) of a circle, we can use the following code to ensure the entered radius must be nonnegative.

while (r<0) { // print some warning message System.out.println(“Invalid data, try again”);

// use JOptionPane to read radius again rs = JOptionPane.showInputDialog("Enter a radius"); r = Double.parseDouble(rs); }

Cont’d

Ex: another typical use of while loop – controls the stop/continuation of programs.

….. answer = 1;

while(answer ==1) { // do whatever computation, e.g. compute area // output results

// prompts to stop/continue the program answerstr = JOptionPane.showInputDialog(“Do you want to

continue? 1-yes,0-no”); answer = Integer.parseInt(answerstr); }

System.out.println(“bye”); …………

Cont’d

Ex: compute 2100

idea: 2100 = 2*2*…*2 (100 times). Set up a loop running 100 times. In each time, a 2 is multiplied to the current result.

code (fragment):

prod = 1; counter=1; while(counter <= 100) { prod = prod * 2.0 ; // why not 2? counter = counter+1; } // prod holds the result. Print prod

Cont’d

How does it work?

loop counter at entry of loop

counter<= 100?

prod at end of loop

1 1 true 21 (2)

2 2 true 22 (2*2)

3 3 true 23 (2*2*2)

…. … … ….

100 100 true 2100 (2*…*2)

101 101 false

Cont’d

Ex: compute 1+2+…+100

c=1; sum=0; while (c<=100) { sum = sum +c; c = c + 1; } // sum holds the result, print sum

Cont’d

Ex: compute n! for any natural number n. Note: n! is called the factorial of n and is defined as: n! = 1*2*….*n e.g., 3! = 1*2*3 = 6 4! = 1*2*3*4 = 24 1! = 1 0! = 1 (defined)

Cont’d

Code (partial)

p=1; c=1; // initialize p and c

while (c<=n) // n acquires a value before this point

{ p = p*c;

c = c + 1;

}

// p holds n!, we can printout p

Cont’d

How does the loop work?

loop c at entry of loop

c <= n? (e.g. n=100)

p at end of loop

1 1 true 1! (1)

2 2 true 2! (1*2)

3 3 true 3! (1*2*3)

…. … … ….

100 100 true 100! (1*2…*100)

101 101 false

Cont’d

Ex: compute 1!+2!+…+n! for any natural number n>0. e.g.:

if n=1, 1!=1 if n=3, 1!+2!+3!=9

s=0; p=1; y=1; while (y <=n) // whatever n is { p = p*y; s = s+p; y = y+1; } // s holds the result

Cont’d

Case Study– Problem: compute the average of your computer

science class– Requirements: Users enter each student’s grade into

computer and enters 9999 to signify the end. The program takes these grades and computes the average. (Note: users do not need to input the number of students into computer, the program should be able to detect the number of students automatically.)

– Specifications:• Inputs: each student’s grade• Output: the average of those grade

Cont’d

• Relevant formula:

avg = total grade/number of students• major programming construct: loops• Data structure:

Name Type Usage

grd double holds each student’s grade. Input

sum double holds the total grade

n int holds the number of students

avg double holds average. Ouput

grdstr String associated with grade

Cont’d

– Design: (note how the– loop is represented) START

P & R first grade

grdstr

Convertgrdstr to grd

echo

sum=0n=0

accumulate total grade

add curren

tgrd to

total

sum=sum+grd

increase #of students

n=n+1

P &Rgrdstr

convertgrdstrto grd

echo

computeavgerage

avg=sum/n

displayavg end

whilegrd !=9999

Cont’d

– Test Case:

– Implementation:

avg

grd 10 55 69 100 98 76 58 85 73 75 69

grd 100 99 98 95 76 83 92 100 91 51 66 95

// Name: your name// ssn: your ssn// date: today's date// course: cmps120// description: fill out by yourself

import javax.swing.JOptionPane;

public class CompAvg{ public static void main(String args[]) { // declaration of variables String grdstr; double grd, avg, sum; int n;

// prompts and reads the first grade grdstr = JOptionPane.showInputDialog("Enter a grade");

// convert grdstr to grd grd = Double.parseDouble(grdstr);

// echo System.out.println("The grade is: "+grd);

// initialization sum =0; n=0;

// accumulate total grade while (grd != 9999) { sum = sum + grd; n = n+1; grdstr = JOptionPane.showInputDialog("Enter a grade"); grd = Double.parseDouble(grdstr); System.out.println("The grade is: "+grd); }

// compute average avg = sum/n;

// dispaly resultSystem.out.println("The average

is:"+avg); // termination System.exit(0); }}// end of program

2. For-Loop

Syntax: for (<init>;<cond>;<incre>) { body } Ex: for (i=1;i<=10;i++) { System.out.println(i); }

Cont’d

Meaning: special counter-controlled while

loop. The initialization is made first, then condition is checked. If false, the body is skipped. If true, body of the loop is executed and the increment is made and the condition is checked again. Just like the first time, what to do next strictly depends on the value of the condition. In another word, the loop will keep running until (sooner or later) the condition (value) becomes false. (see flow chart)

init

body

incre

cond

t

f

Cont’d

Meaning of the example code:

i=1

Print i

i++

i<=10

t

f

For i=1 to 10,i++

Print i

Cont’d

ex: what is the output of the following code?

for (i=1; i<=10; i++)

{ System.out.println(“this is”+i+”pass”);

System.out.println(“i is “+i);

}

Output: this is 1 pass

i is 1

this is 2 pass

i is 2

…..

this is 10 pass

i is 10

Cont’d

ex: what is the output of the following code?

for (i=100; i>=1; i--)

{ System.out.println(i);

}

Output:

100

99

98

….

2

1

Cont’d

ex: what’s the output of the code?

for (i=2; i<=10; i=i+2)

{ System.out.println(i);

}

Output:

2

4

6

8

10

Cont’d

Ex: write a for-loop to printout

2,5,8,11,14,17,20 each at a line.

for (i=2; i<=20; i=i+3)

{ System.out.println(i); }

Cont’d

Nested loops: loops (while-loops and/or for-loops) can be embedded inside another loop to make (complicated) nested loops.

Ex: what’s the output of the code?

for (i=1; i<=5; i++)

{ for (j=1; j<=4; j++)

{ System.out.print(“*”); }

System.out.println();

}

Body ofOut loop

Body ofin loop

Cont’d

****

****

****

****

****

i controls # of

rows, j controls

# of columns

j=1 j=2 j=3 j=4

i=1 * ** *** ****

i=2 *****

******

*******

********

i=3 *********

**********

***********

goes on in the same way

i=4

i=5

Cont’d

Ex: write a program fragment to printout

***************

for (i=1; i<=5; i++)

{ for (j=1; j<=?; j++)

{ System.out.print(“*”); }

System.out.prrintln();

}

analysis (to decide ?)

1st row i=1 ?=1

2nd row i=2 ?=2

3rd row i=3 ?=3

4th row i=4 ?=4

5th row i=5 ?=5

cont’d

from the analysis we see: ?=i, so the code is:

for (i=1; i<=5; i++)

{ for (j=1; j<=i; j++)

{ System.out.print(“*”); }

System.out.prrintln();

}

Cont’d

ex: write a program fragment to printout

*****

****

***

**

*

for (i=1; i<=5; i++)

{ for (j=1; j<=?; j++)

{ System.out.print(“*”); }

System.out.prrintln();

}

analysis (to decide ?)

1st row i=1 ?=5

2nd row i=2 ?=4

3rd row i=3 ?=3

4th row i=4 ?=2

5th row i=5 ?=1

Cont’d

from the analysis we see:

?=6-i, so the code is as follows (structure chart on the right)

for (i=1; i<=5; i++)

{

for (j=1; j<=6-i; j++)

{ System.out.print(“*”); }

System.out.prrintln();

}

Println()

Print”*”

For j=1;j<=6-i;j++

For i=1,i<=5,i++

Cont’d

ex: write a program fragment to printout

*

**

***

****

*****

for (i=1; i<=5; i++)

{ // print leading blanks

for (k=1; k<=?;k++)

{System.out.print(“ “);}

// print *’s

for (j=1; j<=??; j++)

{ System.out.print(“*”); }

System.out.prrintln();

}

Cont’dRow i ?

(blanks)

?? (*’s)

1st 1 4 1

2nd 2 3 2

3rd 3 2 3

4th 4 1 4

5th 5 0 5

Cont’d It is easy to see: ?=5-i and ??

=i, so the code is:

for (i=1; i<=5; i++){ // print leading blanks for (k=1; k<=5-i; k++) {System.out.print(“ “);}

// print *’s for (j=1; j<=i; j++) { System.out.print(“*”); }

System.out.prrintln(); }

Row i ? (blanks)

?? (*’s)

1st 1 4 1

2nd 2 3 2

3rd 3 2 3

4th 4 1 4

5th 5 0 5

Cont’d

Structure chart of previous code

Println()

Print “ “ Print “*”

For i=1;i<=5;i++

For k=1;k<=5-i;k++ For j=1;j<=i;j++

Cont’d

Ex: what about this diagram?

*************************

for (i=1; i<=4; i++){ for (j=1;j<=i; j++) { System.out.print(“*”);} System.out.println(); }

for (i=1; i<=5; i++) { System.out.print(“*”);} System.out.println();

for (i=4; i>=1; i--){ for (j=1; j<=i; j++) { System.out.print(“*”);} System.out.println();}

Part VI

Modular Programming

1. Methods

Basic idea of modular programming: break down large and complicated programs into smaller and easy-to-handle programs (called modules).

1. Methods

In Java, modules are represented as methods: there are two types of methods in Java: –value-producing methods: called

functions–valueless methods: called

procedures

1. Methods

Syntax of methods:

public static type method-name (parameter list)

{ body of method }

Cont’d

ex: a function that computes the square of an integer.

public static int sq (int x) { return x*x; }

sq takes an integer x and returns the square of x.

return valtype name

para name

paratype

what isreturned

Cont’d

to use (to call) this function:

a = sq(1); or

b = sq(3);

the definition and usage of functions in Java are similar to that in math.

2. Program Structure

public class class-name { public static … method1(…) { … } public static …. method2(…) { …. } ………………. pubic static void main(String args[]) { ….. } }

call

call

3. Pre-defined Math functions

usage: Math.fun-name(…) ex:

– Math.sqrt(9.0); square root of 9.0– Math.max(2,3); the max of 2 and 3– Math.PI; the pi

(3.1415926535897932384626433…)

4. Examples

Computes the square of a real number.

import javax.swing.*; public class CompSq { // method sq public static double sq(double x) { return x*x; } // main method public static void main(String args[]) { double a,b; // use JOptionPane, read a real into a b = sq(a); System.out.println(“the sq root of “+a+”is “+b); System.exit(0); } }

Cont’d

compute the area of a circle. import javax.swing.*; public class Area { // module A: get input public static double getR() { double r; String sr; sr = JOptionPane.showInputDialog(“enter a radius”); r = Double.parseDouble(sr); return r; }

// module B: compute area public static double doComp(double x) { return Math.PI*x*x; }

Cont’d

// module C: display resultpublic static void display(double y){ System.out.println(“the area is “+y); }

// main methodpublic static void main(String args[]) { double radius, area; radius = getR(); area = doComp(radius); display(area); System.exit(0); }} // end of clas

5. Local variables, variable lifetime

Local variables: variables defined inside methods and method (formal) parameters.

Variable lifetime: the time period in which variables exist in memory with respect to program execution. Some variables exist for a short period of time, others may exist for the entire execution of the program.

(Local) variable management: Each time a method is called, a memory area is created for local variables. When the call is returned, this memory area will be erased and freed up.

Cont’d

method f(…)

{…..}

.. main(…)

{….. f (..) }

… class

call

create

Memory area

erase Memory area

Memory area for local variables in f

6. Global variables

Variables defined in classes and out of all methods, with static modifier are called global variables. Also called class variables. They are visible to all methods.

Note: global variable is dangerous, and is not recommended generally.

……… class T{ static int x;

…. m1 {… } ….. main() {…. }}

visible

x is global

visible

7. Example

public class Ex{ static int id =1; // class var, global public static double average(double x,

double y) { double avg; // local var avg = (x+y)/2; System.out.println(“id is “+id); return avg; } public static void main(String args[]) { double a, b, avg; // local var a=1; b=2; avg = average(a,b); System.out.println(“id is “+id); System.out.println(“average is “+avg); System.exit(0); }}

1

2

1

2

1.5

a

b

avg

x

y

avg

1.5M area erased upon return

8. Case Study

Problem: Given a positive integer n, compute 1+2+…+n Requirements: Users input a positive integer n, the program computes

the sum from 1 to n. E.g., if n=1, ouput = 1 if n=2, output = 1+2=3 if n=10, output = 1+2+…+10=55 Specification:

– Input: positive integer n– Output: 1+2+…+n– Relevant formula/methodology: modules, loops– Method getN:

• purpose: It takes nothing and returns the positive integer entered by the user• type: () int• data structure

Name Type usage

n Int Input, positive integer

Sn String String for n

Cont’d

– Method doComp:• purpose: computes 1+…+n and returns the result.• type: int int• data structure

Name Type Usage

n int formal parameter

i int loop controller

sum int hold 1+…+n

Cont’d

– Method display:• purpose: display the

result• type: int ()• data structure

Name Type Usage

r int formal parameter

Cont’d

– Method main• Purpose: default• Type: default• Data structure:

name type usage

num int hold input integer

result int hold 1+…+n. Output

Cont’d

Design:

Main(String args[])

result=doComp(num)

Num=getN()

display(result)

end

Cont’d

getN()

sn=JOpt… convert sn to n return n

display(int r)

print r

doComp(int n)

sum=0

sum=sum+i

return sumfor i=1 to n

Cont’d

Implementation: left as a lab.

Part VII

Arrays

1. 1-D Arrays

Concept: – 1-d array is a single list (with order

embedded) of collection of data of the same type.

– It corresponds to the notion of finite tuples in math. (x1,x2, …., xn)

1. 1-D Arrays

– ex:

an int array of size =5

an boolean array of size =3

1 2 3 4 5

true false true

Cont’d

Syntax: type var-name[] = new type [n];

Meaning: – Indicates that var-name is an array variable

(not a conventional variable)– Each element in the array is of type type.– Allocates n cells in memory for this array

which are indexed from 0 to n-1. (not from 1 to n).

An alternative way to define arrays:

type[] var-name = new type [n];

Cont’d

Ex: (typing this) int a[] = new int[10];Creates the following in memory

The 1st element is referenced by a[0]The last element is referenced by a[9]All elements (a[0],…,a[9]) are of type int

a

a[0] a[1] a[2] a[8] a[9]

Cont’d

Setting and getting values in an array– Each array element can be treated as an

individual variable.– Ex: int b[] = new int[5];

b[0] =1; b[1]=2;

b[2]=5; b[4]=5; b[3]=6;

results in

1 2 5 6 5b

b[0] b[1] b[2] b[3] b[4]

Cont’d

and

System.out.println(b[3]);

prints 5.

c=b[0]+b[4];

puts 6 into variable c

b[0]=b[1]*b[2];

puts 10 into b[0]

Cont’d– Ex: combine loops with arrays.

int a[] = new int[10];

for (i=0; i<10; i++)

{ a[i]=0; }results in

for (j=0; j<10; j++)

{ a[j] = j; } results in

0 0 0 0 0 0 0 0 0 0a

[0] [1] ……… [8] [9]

0 1 2 3 4 5 6 7 8 9

[0] [1] ……… [8] [9]

a

to printout the contents of array a,

for (k=0; k<10; k++)

{ System.out.println(a[k]); }

Cont’dQ: how to reverse a[]? i.e., make a[] like

will for (i=0; i<10; i++) a[9-i] = a[i];work? if not, what does it do?

will for (i=0; i<10; i++) { t = a[9-i]; a[9-i] =a[i]; a[i]=t; }work? if not, what does it do?

9 8 7 6 5 4 3 2 1 0a

[0] [1] ……… [8] [9]

Cont’d

solution:

// reverse a[] into b[] for (i=0; i<10; i++) { b[9-i] = a[i] ; } // copy b[] back to a[] for (i=0; i<10; i++) { a[i] = b[i]; }

b[] is an auxiliary array.

Pass arrays as parameters

// def array a

int[] a = new int a[10];

…..

// call method m

// passing array a to it

…m(a);

}

public static void m(int x[])

{ ……….

// body of method m

//Note:

// operations performed

// on array x will affect

// array a in the caller

// no array “return” needed.

}

Cont’d

Examples of arrays– Compute the average of a class.

• Input: # of students and grade of each student.• Output: average of the class• Analysis: (data structure) n: type int, holds the # of students g[]: type double array, holds the grade of each student avg: type double, holds the average of class sum: type double, holds the summation of all grades• relevant formula: sum = ∑g[i], avg = sum/n

Cont’d

• Design:

START

P & Rn

read allgrades

cal sum of grade cal averagedisplay

avg

read each grade

into g[i]

sum=sum+g[i]

avg = sum/n

End

for i=1 ton

for i=1 to n

Cont’d

Implementation:

• import javax.swing.*;

• public class Array• {

• public static void main(String args[])• { int n, i;• String sn;• double avg,sum, g[] = new double[100];

• // read # of students• sn=JOptionPane.showInputDialog("How many students?");• n=Integer.parseInt(sn);

Cont’d

– // read grade of each student– for (i=1;i<=n;i++)– g[i]=Double.parseDouble(JOptionPane.showInputDialog("grade?"));

– // totoal grades– sum=0;– for (i=1;i<=n;i++)– sum = sum + g[i];

– // compute avgerage– avg = sum/n;– – // output– System.out.println("average is "+avg);– – // terminate– System.exit(0);– }– }

Cont’d

– ex: sorting of numbers• given a set of numbers, sort them into ascending

order (i.e., from small to large). e.g.:

2 4 3 1 5 will be sorted to

1 2 3 4 5• analysis: the strategy of bubble sort. We use an

example to illustrate the idea of bubble sort. Suppose we want to sort the set of integers:

2 4 3 1 5

Cont’d

2 4 3 1 5

2 4 3 1 5

1 4 3 2 5

2 4 3 1 5

1 2 3 4 5

1 4 3 2 5

1 4 3 2 5

1 3 4 2 5

1 2 4 3 5

1 2 4 3 5

1 2 4 3 5

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

Cont’d

Java code:public class BSort{ public static void main(String args[]) { int n, j, i, temp; int a[] = new int[100]; String sn;

sn = JOptionPane.showInputDialog(“how many numbers?”); n = Integer.parseInt(sn);

for (i=0; i<n; i++) // read data into array a, you can do it by yourself

Cont’d

for (i=0; i<=(n-2); i++) { for (j=i+1; j<=(n-1); j++) { if (a[i]>a[j]) { temp = a[i]; a[i] = a[j]; a[j] = temp; } } } // output array a[] for (i=0; i<n; i++) System.out.println(a[i]); System.exit(0); } }

2. 2-D Arrays

Concept: – a table of collection of data of the same type– ex:

size: 2 x 4 (# of rows x # of columns); type of element: int Syntax:

– type var-name[][] = new type[n][m];– ex: int t[][] = new int[10][20];

Meaning: – indicates that var-name is a 2-d array– each element in the array is of type type– size of array is n rows and m columns, total n x m cells– index of cells starts form 0 (not 1!)

4 5 21 4

88 7 6 5

Cont’d

ex: int a[][] = new int[3][4];

size : 3 x 4; 3 rows, 4 columnsEach element in the array is identified by its row # and

col #, e.g., the element at 2nd row and 3rd column is referenced by a[1][2].

a[0][0] a[0[1] a[0][2] a[0][3]

a[1][0] a[1][1] a[1][2] a[1][3]

a[2][0] a[2][1] a[2][2] a[2][3]

Cont’d

Setting and getting values in a 2-d array– each array element can be treated just like an individual

variable– ex: a[1][2]=5; b = a[1][2]+a[1][3]; System.out.println(a[0][0]);– loops and arrays// a[][] can be filled up in the following way for (i=0; i<3; i++) { for (j=0; j<4; j++) { a[i][j] = 0; } // assign 0 to each cell }

0 0 0 0

0 0 0 0

0 0 0 0

result

Cont’d

– ex:

// fill row-by-row

for (i=0; i<3; i++)

{ for (j=0; j<4; j++)

a[i][j]=i*j;

}

// fill col-by-col

for (j=0; j<4; j++)

{ for (i=0; i<3; i++)

a[i][j]=i*j;

}

0 0 0 0

0 1 2 3

0 2 4 6

0 0 0 0

0 1 2 3

0 2 4 6

Cont’d

– printout:// printout row-by-rowfor (i=0; i<3; i++) { for (j=0; j<4; j++) {System.out.print(a[i][j]);} System.out.println(); } // printout col-by-colfor (j=0; j<4; j++){ for (i=0; i<3; i++) {System.out.print(a[i][j]);} System.out.println();}

0 0 0 0 1st row 0 1 2 3 2nd row 0 2 4 6 3rd row

0 0 0 1st col 0 1 2 2nd col 0 2 4 3rd col 0 3 6 4th col

Cont’d

Call-by-value (and call-by-reference)– call-by-value: one of the parameter passing

modes. When a method is called, a (separate) copy of the argument is passed to the parameter and any changes made to this copy is local. All single-valued variables and individual array element of base types are passed by call-by-value.

Cont’d

– ex:…..main(String …)

{ int a;

a=3;

System.out.println(a);

AddTwo(a);

System.out.println(a);

}

public static void AddTwo(int x)

{

x = x+2;

}

- x is local, changes made to x will not affect variable a in main

3

a

x

3

5

Cont’d

– Call-by-reference: one of the parameter passing modes. When a method is called, a reference of the argument is passed to the parameter. Any changes made to the parameter will affect the original argument. Arrays are passed by call-by-reference. (different perspective: call-by-value since b is a ref variable and its value is a reference already.)

– ex: ….main(String…) { int b[] = new int[10]; // fill up array b b[1]=1; AddTwo(b); Sytem.out.println(b[1]); }

public static void AddTwo(int x[]){ x[1] = x[1]+2; }

b and x contains the same reference to the array

x 1 3 x x x

b

x

Cont’d

Ex of 2-d arrays: suppose

compute the average of each person.

Basic idea: use a 2-d array to store the grades of the 3 students. Write a method which handles the computation of average.

cmps math eng avg

John 80 90 85

Mary 70 75 76

Al 100 95 98

Cont’d

– Java code:public class Ex{ public static void main(String

args[]) { int i; double grade[][] =

{{80,90,85,0},{70,75,76,0},{100,95,98,0}};

compAvg(grade); for (i=0; i<3;i++) System.out.print(grade[i][3]); System.exit(0); } // end of main

public static void compAvg(double g[][])

{ double sum=0; i, j; for (i=0; i<3; i++) { for (j=0; j<3; j++) { sum=sum+g[i][j];} g[i][3] = sum/3.0; sum=0; } } // end of compAvg

}// end of class