CSCI 1101B {101B} 09/19/13 The Golden Lecture Chapter 2 Mohammad - - PDF document

9/21/2013 1

CSCI 1101B {101B} 09/19/13 The Golden Lecture

Mohammad T. Irfan

Chapter 2

What’s in a name? Variables

• TYPE

NAME;

• int i;
• int i, j, anotherVar;
• double i, j = 10.37, anotherVar;
• Same as the following two lines:
• double i, j, anotherVar;
• j = 10.37;
• boolean isFound = false;
• char c = ‘x’;
• Why are they called variables?

int num = 10; num = 20;

Constants

• final int NUM = 10;
• Later on, you can’t change the value of NUM

NUM = 20; //Error

Objects

• int num = 5;
• FilledOval circle = new FilledOval (10, 50,100, 100,canvas);

Type (Class): FilledOval Name (Object): circle Value: constructor Type: int Name: num Value: 5

Chapter 3

Working with Numbers

9/21/2013 2 Arithmetic Operations

• double radius = 10.0;
• double sideLength = 2.0;
• double circleArea, squareArea, totalArea;
• circleArea = 3.14159 * radius * radius;
• squareArea = sideLength * sideLength;
• totalArea = circleArea + squareArea;

Printing numbers and text

• double numMiles;
• numMiles = 10.50;
• System.out.println (“I ran ” + numMiles + “ miles!”);

Chapter 4

Making Choices if Statement

• if ( condition )

{ if-part; //one or more statements } else if (another condition) { else-if-part; //one or more statements } else { else-part; //one or more statements } //End of if-else: no semicolon is necessary here

• //Next statements start here... (outside of if-else)

Optional

Example: Hannaford Finding the maximum shortage

• Input
• # of items n, stocks s1, s2, …, sn, demands d1, d2, …, dn
• Steps

For item# i ranging from 1 to n Set its shortage ti = di – si EndFor maxShortage = t1 maxShortageItem = 1 For item# i ranging from 2 to n if ti > maxShortage then maxShortage = ti maxShortageItem = i endif EndFor

if (ti > maxShortage) { maxShortage = ti; maxShortageItem = i; }

Problem

• Roll a die twice.
• Both numbers are odd  You win
• One is odd, the other even  It’s a tie
• None is odd  Computer wins
• Pseudocode
• firstRoll = outcome of rolling the die
• secondRoll = outcome of rolling the die again
• if (firstRoll is odd AND secondRoll is also odd)
• utput “I win ”;

else if (firstRoll is odd OR secondRoll is odd)

• utput “It’s a tie!”

else

• utput “Computer wins ”

9/21/2013 3 Revisiting the Hannaford Example

• Finding the 2nd maximum shortage

if (ti > maxShortage2 && ti != maxShortage) { maxShortage2 = ti; maxShortageItem2 = i; } if (ti > maxShortage2 and ti ≠ maxShortage) maxShortage2 = ti maxShortageItem2 = i Endif

Chapter 14

Arrays Array Definition

• int [] numberList = new int[5];
• numberList[0] = 10;
• numberList[1] = 15;
• numberList[2] = 5;
• numberList[3] = 30;
• numberList[4] = 50;

numberList

Position#: 0 1 2 3 4

numberList

10 15 5 30 50 1 2 3 4

Another way

• int [] numberList = {10, 15, 5, 30, 50};

numberList

10 15 5 30 50 1 2 3 4

Operations with an Array

• int b = numberList[0];
• numberList[3] = numberList[1] + 2 * numberList[4];
• numberList[2] = b * 4;

numberList

10 15 5 30 50 1 2 3 4 115 40

9/21/2013 4 Revisiting Hannaford Example

• Finding the 2nd maximum shortage

if (t[i] > maxShortage2 && t[i] != maxShortage) { maxShortage2 = t[i]; maxShortageItem2 = i; } if (ti > maxShortage2 and ti ≠ maxShortage) maxShortage2 = ti maxShortageItem2 = i Endif

Chapter 7, 13

Iterative Statements for Loop

• for (initialization; condition; update)

{ statements to be repeated ... }

If condition is true If condition is false, exit the loop

for Loop Example

• int i;
• for (i = 0;

i < 5; i = i + 1) { System.out.println (numberList[i]); }

If condition is true If condition is false, exit the loop

Java Reference

• docs.oracle.com/javase/7/docs/api/

Lab Assignments

Due next week in lab (No penalty for errors)

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Bring (partial) solutions to the next lab

• 1. Calculate the roots of a given quadratic equation.
• 2. For the rolling die problem, exchange the if and the else

if part. What goes wrong?

• 3. Solve the problem of finding the maximum shortage

using Java.

• Input:
• int n = 10;
• int [] stock = {20, 5, 30, 90, 60, 15, 25, 75, 150, 10};
• int [] demand = {14, 6, 2, 33, 25, 46, 277, 129, 182, 199};
• //Test with different numbers
• 4. Solve the 2nd maximum shortage problem.