1
Lecture 7 - Arrays
Outline
4.6
4.7
4.8
4.9
4.10
Sorting Arrays
Case Study: Computing Mean, Median and Mode Using Arrays
Searching Arrays: Linear Search and Binary Search
Multiple-Subscripted Arrays
Thinking About Objects: Identifying a Class's Behaviors
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2
Sorting Arrays
• Sorting data
– Important computing application
– Virtually every organization must sort some data
• Massive amounts must be sorted
• Bubble sort (sinking sort)
– Several passes through the array
– Successive pairs of elements are compared
• If increasing order (or identical), no change
• If decreasing order, elements exchanged
– Repeat these steps for every element
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3
4.6
Sorting Arrays
• Example:
–
–
–
–
Original: 3 4 2 6 7
Pass 1: 3 2 4 6 7
Pass 2:
2 3 4 6 7
Small elements "bubble" to the top
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4
Case Study: Computing Mean, Median and
Mode Using Arrays
• Mean
– Average
• Median
– Number in middle of sorted list
– 1, 2, 3, 4, 5 (3 is median)
• Mode
– Number that occurs most often
– 1, 1, 1, 2, 3, 3, 4, 5 (1 is mode)
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// Fig. 4.17: fig04_17.cpp
// This program introduces the topic of survey data analysis.
// It computes the mean, median, and mode of the data.
#include <iostream>
5
Outline
1. Function prototypes
using std::cout;
using std::endl;
using std::ios;
1.1 Initialize array
#include <iomanip>
2. Call functions mean,
median, and mode
using std::setw;
using std::setiosflags;
using std::setprecision;
void
void
void
void
void
mean( const int [], int );
median( int [], int );
mode( int [], int [], int );
bubbleSort( int[], int );
printArray( const int[], int );
int main()
{
const int responseSize = 99;
int frequency[ 10 ] = { 0 },
response[ responseSize ] =
{ 6, 7, 8, 9, 8, 7, 8, 9,
7, 8, 9, 5, 9, 8, 7, 8,
6, 7, 8, 9, 3, 9, 8, 7,
7, 8, 9, 8, 9, 8, 9, 7,
6, 7, 8, 7, 8, 7, 9, 8,
7, 8, 9, 8, 9, 8, 9, 7,
 2000 Prentice Hall,
All rights
reserved.
5, Inc.
6, 7,
2, 5,
3, 9, 4,
8,
7,
8,
8,
9,
5,
6,
9,
8,
7,
9,
2,
3,
4,
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7, 8, 9, 6, 8, 7, 8, 9, 7, 8,
7, 4, 4, 2, 5, 3, 8, 7, 5, 6,
4, 5, 6, 1, 6, 5, 7, 8, 7 };
6
Outline
mean( response, responseSize );
median( response, responseSize );
mode( frequency, response, responseSize );
3. Define function
mean
return 0;
3.1 Define function
median
}
void mean( const int answer[], int arraySize )
{
int total = 0;
cout << "********\n
Mean\n********\n";
for ( int j = 0; j < arraySize; j++ )
total += answer[ j ];
cout <<
<<
<<
<<
<<
<<
<<
<<
<<
"The mean is the average value of the data\n"
"items. The mean is equal to the total of\n"
"all the data items divided by the number\n"
"of data items (" << arraySize
"). The mean value for\nthis run is: "
total << " / " << arraySize << " = "
setiosflags( ios::fixed | ios::showpoint )
setprecision( 4 )
static_cast< double >( total ) / arraySize << "\n\n";
}
void median( int answer[], int size )
{
 2000
Prentice
Hall, Inc. All rights reserved.
cout
<< "\n********\n
Median\n********\n"
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<< "The unsorted array of responses is";
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Outline
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printArray( answer, size );
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bubbleSort( answer, size );
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cout << "\n\nThe sorted array is";
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printArray( answer, size );
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cout << "\n\nThe median is element " << size / 2
3.1 Define function
median
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<< " of\nthe sorted " << size
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<< " element array.\nFor this run the median is "
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<< answer[ size / 2 ] << "\n\n";
3.1.1 Sort Array
3.1.2 Print middle
element
78 }
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80 void mode( int freq[], int answer[], int size )
81 {
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int rating, largest = 0, modeValue = 0;
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cout << "\n********\n
Mode\n********\n";
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for ( rating = 1; rating <= 9; rating++ )
freq[ rating ] = 0;
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for ( int j = 0; j < size; j++ )
++freq[ answer[ j ] ];
Notice how the subscript in
frequency[] is the value of an
element in response[]
(answer[]).
cout << "Response"<< setw( 11 ) << "Frequency"
<< setw( 19 ) << "Histogram\n\n" << setw( 55 )
<< "1
1
2
2\n" << setw( 56 )
Hall, Inc.0 All rights
95  2000 Prentice
<< "5
5 reserved.
0
5\n\n";
3.2 Define function
mode
3.2.1 Increase
frequency[]
depending on
response[]
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for ( rating = 1; rating <= 9; rating++ ) {
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cout << setw( 8 ) << rating << setw( 11 )
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<< freq[ rating ] << "
";
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if ( freq[ rating ] > largest ) {
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largest = freq[ rating ];
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modeValue = rating;
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}
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for ( int h = 1; h <= freq[ rating ]; h++ )
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cout << '*';
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Print stars depending on value of
frequency[]
cout << '\n';
}
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cout << "The mode is the most frequent value.\n"
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<< "For this run the mode is " << modeValue
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<< " which occurred " << largest << " times." << endl;
115 }
116
117 void bubbleSort( int a[], int size )
118 {
119
Outline
int hold;
120 2000 Prentice Hall, Inc. All rights reserved.
3.3 Define bubbleSort
121
for ( int pass = 1; pass < size; pass++ )
9
Outline
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for ( int j = 0; j < size - 1; j++ )
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if ( a[ j ] > a[ j + 1 ] ) {
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hold = a[ j ];
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a[ j ] = a[ j + 1 ];
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a[ j + 1 ] = hold;
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}
130 }
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132 void printArray( const int a[], int size )
133 {
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for ( int j = 0; j < size; j++ ) {
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if ( j % 20 == 0 )
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cout << endl;
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cout << setw( 2 ) << a[ j ];
}
141
} 2000 Prentice Hall, Inc. All rights reserved.
3.3 Define bubbleSort
3.3 Define printArray
Bubble sort: if elements out of order,
swap them.
********
Mean
********
The mean is the average value of the data
items. The mean is equal to the total of
all the data items divided by the number
of data items (99). The mean value for
this run is: 681 / 99 = 6.8788
********
Median
********
The unsorted
6 7 8 9 8 7
6 7 8 9 3 9
6 7 8 7 8 7
5 6 7 2 5 3
7 4 4 2 5 3
The sorted
1 2 2 2 3
5 6 6 6 6
7 7 7 7 7
8 8 8 8 8
9 9 9 9 9
array
8 9 8
8 7 8
9 8 9
9 4 6
8 7 5
array
3 3 3
6 6 6
7 7 7
8 8 8
9 9 9
of responses is
9 7 8 9 5 9 8 7
7 7 8 9 8 9 8 9
2 7 8 9 8 9 8 9
4 7 8 9 6 8 7 8
6 4 5 6 1 6 5 7
is
4 4
6 6
7 7
8 8
9 9
4
7
7
8
9
4
7
7
8
9
4
7
7
8
9
5
7
8
8
9
5
7
8
8
9
5
7
8
8
9
5
7
8
8
9
The median is element 49 of
the sorted 99 element array.
For this run the median is 7
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7
7
9
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7
8
5
7
7
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9
3
8
5
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Outline
4. Program Output
********
Mode
********
Response
11
Outline
Frequency
Histogram
5
1
0
1
5
2
0
2
5
1
1
*
2
3
***
3
4
****
4
5
*****
5
8
********
6
9
*********
7
23
***********************
8
27
***************************
9
19
*******************
The mode is the most frequent value.
For this run the mode is 8 which occurred 27 times.
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Program Output
12
Just Do It!
Write a program that asks a user to input as many
floating point numbers as they’d like (up to 100).
The program should ask each time whether the user
would like to enter another number (y/n). After the
user says no (“n”), the program should sort the
numbers from largest to smallest and print the sorted
list to the screen. Make sure you use a modified
version of the bubbleSort function from the prior
example.
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Pseudocode
While the user wants to enter another number
Get the user’s entry
Store the number
Keep track of how many elements have been entered
Sort the entered numbers
Display the sorted numbers
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Sample Solution
// arraysort.cpp
#include <iostream>
using std::cout; using std::cin; using std::endl;
void bubbleSort(float[], int);
//declarations
void swap(float& a, float& b);
//-------------------------------------------------------------int main()
{
const int SIZE = 100;
//maximum array elements
float nums[SIZE];
int count = 0;
//number of floats in array
char ans = 'y';
do {
cout << "Enter a number: ";
cin >> nums[ count++ ];
cout << "Would you like to enter another (y/n)? ";
cin >> ans;
} while ( ans != 'n' );
bubbleSort(nums, count);
for(int i=0; i<count; i++)
cout << nums[i] << endl
return 0;
}
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15
Sample Solution
void swap(float& a, float& b)
{
float temp;
temp = a;
a = b;
b = temp;
}
void bubbleSort(float f[], int c)
{
int hold;
for ( int j = 0; j < c - 1; j++ )
if ( f[ j ] < f[ j + 1 ] )
swap(f[ j ], f[ j + 1 ]);
}
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16
Searching Arrays: Linear Search and Binary
Search
• Search array for a key value
• Linear search
– Compare each element of array with key value
– Useful for small and unsorted arrays
• Binary search
– Can only be used on sorted arrays
– Compares middle element with key
• If equal, match found
• If key < middle, repeat search through the first half of the array
• If key > middle, repeat search through the last half of the array
– Very fast; at most n steps, where 2^nn > # of elements
• 30 element array takes at most 5 steps
– 52 > 30
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17
Just Do It!
Create a program that allows the user to input a number
of integers, and then stores them in an int array. Write
a function called maxint() that goes through the
array, element by element, looking for the largest one.
The function should take as arguments the address of
the array and the number of elements in it, and return
the index number of the largest element. The program
should call this function and then display the largest
element and its index number.
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18
Pseudocode
Main
While the user wants to enter another number
Get the user’s entry
Store the number
Keep track of how many elements have been entered
Determine the largest number entered (call the function)
Display the largest number
MaxInt function
Search all numbers
If the current number is larger than the previous largest
Save the index of the current number
Return the index of the largest number
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19
Sample Solution
// arraymax.cpp
#include <iostream>
using std::cout; using std::cin; using std::endl;
int maxint(int[], int);
//declaration
//-------------------------------------------------------------int main()
{
const int SIZE = 100;
//maximum array elements
int intarray[SIZE];
//array of ints
int count = 0;
//number of ints in array
int intval;
while(true)
{
cout << count << ". Enter an integer value (0 to end): ";
cin >> intval;
if(intval==0) break;
intarray[count++] = intval; //set array element to value
}
//find largest element
int bigdex = maxint(intarray, count);
cout << "\nLargest element is number " << bigdex;
cout << ", which contains " << intarray[bigdex] << endl;
return 0;
}
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20
Sample Solution (cont.)
//-------------------------------------------------------------int maxint(int ia[], int c)
//returns largest element
{
int bigdex = 0;
//assume it's element 0
for(int j=1; j<c; j++)
//if another is larger,
if( ia[j] > ia[bigdex] )
//put its index in bigdex
bigdex = j;
return bigdex;
}
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21
Multiple-Dimensional Arrays
• So far we’ve only looked at arrays that have a single
dimension – now we’ll look at multidimensional arrays
• Let’s start with a two-dimensional array:
1
District1
District2
District3
District4
Month
2
3
$235 $354
$123 $221
$421 $397
$801 $892
$547
$241
$401
$932
• A 2-dimensional array is similar to a table
• Example – salemon.cpp
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22
Multiple-Subscripted Arrays
• Multiple subscripts - tables with rows, columns
– Like matrices: specify row, then column.
Row 0
Column 0
a[ 0 ][ 0 ]
Column 1
a[ 0 ][ 1 ]
Column 2
a[ 0 ][ 2 ]
Column 3
a[ 0 ][ 3 ]
Row 1
a[ 1 ][ 0 ]
a[ 1 ][ 1 ]
a[ 1 ][ 2 ]
a[ 1 ][ 3 ]
Row 2
a[ 2 ][ 0 ]
a[ 2 ][ 1 ]
a[ 2 ][ 2 ]
a[ 2 ][ 3 ]
Column subscript
Array name
Row subscript
• Initialize
int b[ 2 ][ 2 ] = { { 1, 2 }, { 3, 4 } };
1
2
3
4
1
0
3
4
– Initializers grouped by row in braces
int b[ 2 ][ 2 ] = { { 1 }, { 3, 4 } };
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23
Multiple-Dimensional Arrays (cont.)
• Defining
double sales[DISTRICTS][MONTHS];
double sales[YEARS][MONTHS][DISTRICTS];
• Accessing
sales[d][m];
• Initializing
double sales[4][3] =
{ {x, x, x}, {x, x, x},
{x, x, x}, {x, x, x} }
• Example – saleinit.cpp
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24
Multiple-Subscripted Arrays
• Referenced like normal
cout << b[ 0 ][ 1 ];
– Will output the value of 0
– Cannot reference with commas
cout << b( 0, 1 );
• Will try to call function b, causing a syntax error
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// Fig. 4.23: fig04_23.cpp
// Double-subscripted array example
#include <iostream>
25
Outline
1. Initialize variables
using std::cout;
using std::endl;
using std::ios;
1.1 Define functions to
take double scripted
arrays
#include <iomanip>
using std::setw;
using std::setiosflags;
using std::setprecision;
const int students = 3;
const int exams = 4;
1.2 Initialize
studentgrades[][]
// number of students
// number of exams
int minimum( int [][ exams ], int, int );
int maximum( int [][ exams ], int, int );
double average( int [], int );
void printArray( int [][ exams ], int, int );
int main()
{
int studentGrades[
{ { 77, 68,
{ 96, 87,
{ 70, 90,
2. Call functions
minimum, maximum,
and average
Each row is a particular student,
each column is the grades on the
exam.
students ][ exams ] =
86, 73 },
89, 78 },
86, 81 } };
cout << "The array is:\n";
printArray( studentGrades, students, exams );
cout << "\n\nLowest grade: "
 2000 Prentice
Hall, Inc. AllstudentGrades,
rights reserved.
<< minimum(
students, exams )
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<< "\nHighest grade: "
<< maximum( studentGrades, students, exams ) << '\n';
for ( int person = 0; person < students; person++ )
cout << "The average grade for student " << person << " is "
<< setiosflags( ios::fixed | ios::showpoint )
<< setprecision( 2 )
<< average( studentGrades[ person ], exams ) << endl;
return 0;
}
// Find the minimum grade
int minimum( int grades[][ exams ], int pupils, int tests )
{
int lowGrade = 100;
for ( int i = 0; i < pupils; i++ )
for ( int j = 0; j < tests; j++ )
if ( grades[ i ][ j ] < lowGrade )
lowGrade = grades[ i ][ j ];
return lowGrade;
}
// Find the maximum grade
int maximum( int grades[][ exams ], int pupils, int tests )
{
int highGrade = 0;
 2000
Allirights
reserved. i++ )
forPrentice
( intHall,
i Inc.
= 0;
< pupils;
26
Outline
2. Call functions
minimum, maximum,
and average
3. Define functions
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for ( int j = 0; j < tests; j++ )
if ( grades[ i ][ j ] > highGrade )
highGrade = grades[ i ][ j ];
return highGrade;
}
// Determine the average grade for a particular student
double average( int setOfGrades[], int tests )
{
int total = 0;
for ( int i = 0; i < tests; i++ )
total += setOfGrades[ i ];
return static_cast< double >( total ) / tests;
}
// Print the array
void printArray( int grades[][ exams ], int pupils, int tests )
{
cout << "
[0] [1] [2] [3]";
for ( int i = 0; i < pupils; i++ ) {
cout << "\nstudentGrades[" << i << "] ";
for ( int j = 0; j < tests; j++ )
cout << setiosflags( ios::left ) << setw( 5 )
<< grades[ i ][ j ];
}

} 2000 Prentice Hall, Inc. All rights reserved.
Outline
3. Define functions
The array is:
[0]
studentGrades[0] 77
studentGrades[1] 96
studentGrades[2] 70
[1]
68
87
90
[2]
86
89
86
[3]
73
78
81
Lowest grade: 68
Highest grade: 96
The average grade for student 0 is 76.00
The average grade for student 1 is 87.50
The average grade for student 2 is 81.75
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28
Outline
Program Output
29
Just Do It!
Create a program that creates a deck of cards, shuffles the deck,
and deals the 52 cards to each of 4 players. Be sure to deal the
cards one at a time to each player – DON’T give one player 13,
then the next 13, etc.
This problem may end up being more complicated than it initially
sounds. Think carefully about how to store the deck of cards and
then have a separate storage for the cards in the player’s hands.
Feel free to use functions!
Hints: Think multi-dimensional arrays.
Use rand to shuffle the deck.
Use ints to represent the suits and values of the cards.
A single card requires two values to represent it.
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30
Pseudocode
Create a deck of cards
Shuffle the deck
For each card in the deck
Give the card to the next player
For each player
For each card in the player’s hand
Display the card
Go to the next line before displaying the next hand
 2000 Prentice Hall, Inc. All rights reserved.
31
Sample Solution
// bridge.cpp – deals four bridge hands
#include <iostream>
#include <cstdlib>
//for srand(), rand()
#include <ctime>
//for time for srand()
using std::cout; using std::cin; using std::endl;
// card suits
const int clubs = 0;
const int hearts = 1;
const int spades = 2;
const int diamonds = 3;
//from 2 to 10 are integers without names
const int jack = 11;
const int queen = 12;
const int king = 13;
const int ace = 14;
const int PLAYERS = 4;
const int CARDS_IN_HAND = 13;
const int CARDS_IN_DECK = 52;
const int VALUE = 0;
const int SUIT = 1;
//-------------------------------------------------------------// function display_card prototype
void display_card(int[]);
 2000 Prentice Hall, Inc. All rights reserved.
32
Sample Solution (cont.)
////////////////////////////////////////////////////////////////
int main()
{
int count, j, k, tempVal, tempSuit;
card deck[CARDS_IN_DECK][2];
card player_hands[PLAYERS][CARDS_IN_HAND][2];
cout << endl;
for(j=0; j<CARDS_IN_DECK; j++)
//make an ordered deck
{
int num = (j % CARDS_IN_HAND) + 2; //cycles through 2 to 14, 4 times
int su = j / CARDS_IN_HAND; //cycles through 0 to 3, 13 times
deck[j][VALUE] = num;
//set card
deck[j][SUIT] = su;
}
//shuffle the deck
srand( time(NULL) );
//seed random numbers with time
for(j=0; j<CARDS_IN_DECK; j++)
//for each card in the deck,
{
k = rand() % CARDS_IN_DECK;
//pick another card at random and swap
tempVal = deck[j][VALUE]; tempSuit = deck[j][SUIT];
deck[j][VALUE] = deck[k][VALUE]; deck[j][SUIT] = deck[k][SUIT];
deck[k][VALUE] = tempVal; deck[k][SUIT] = tempSuit;
}
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33
Sample Solution (cont.)
count = 0;
for(j=0; j<CARDS_IN_DECK; j++)
//deal four hands of 13 cards
{
player_hands[j%4][j/4][VALUE] = deck[j][VALUE];
player_hands[j%4][j/4][SUIT] = deck[j][SUIT];
}
for(j=0; j<PLAYERS; j++)
//display the player hands
{
cout << "\nPlayer " << j+1 << " holds: ";
for(k=0; k<CARDS_IN_HAND; k++)
{
display_card(player_hands[j][k]);
cout << ", ";
}
cout << endl;
}
return 0;
} //end main
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34
Sample Solution (cont.)
//-------------------------------------------------------------// display_card function definition
void display_card(int aCard[])
//display the card
{
if( aCard[VALUE] >= 2 && aCard[VALUE] <= 10 )
cout << aCard[VALUE];
else
switch(aCard[VALUE])
{
case jack: cout << "J"; break;
case queen: cout << "Q"; break;
case king: cout << "K"; break;
case ace:
cout << "A"; break;
}
switch(aCard[SUIT])
{
case clubs:
cout << static_cast<char>(5); break;
case hearts:
cout << static_cast<char>(3); break;
case spades:
cout << static_cast<char>(6); break;
case diamonds: cout << static_cast<char>(4); break;
}
}
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35
Thinking (some more) About Objects
• Previous chapters
– Chapter 2 - identified classes
– Chapter 3 - determined many attributes of our classes
• This chapter
– Determine operations (behaviors)
• Chapter 5
– Focus on interactions between classes
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36
Class Operations
• Class operations
– Service class provides to clients
• Radio - setting station, volume
– Objects usually do not perform operations spontaneously
• Operations invoked by sending object (client object)
• Sends message to receiving object (server object)
• Requests object perform specific operation
• Deriving operations
– Examine problem statement for verbs and verb phrases
– Relate phrases to particular classes
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37
Class Operations (II)
Cla ss
Verb p hra ses
Elevator
moves, arrives at a floor, resets the elevator button, sounds
the elevator bell, signals its arrival to a floor, opens its
door, closes its door
Clock
ticks every second
Scheduler
randomly schedules times, creates a person, tells a person
to step onto a floor, verifies that a floor is unoccupied,
delays creating a person by one second
Person
steps onto floor, presses floor button, presses elevator
button, enters elevator, exits elevator
Floor
resets floor button, turns off light, turns on light
FloorButton
summons elevator
ElevatorButton signals elevator to move
Door
(opening of door) signals person to exit elevator, (opening
of door) signals person to enter elevator
Bell
none in problem statement
Light
none in problem statement
Building
none in problem statement
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38
Creating Class Operations
• Creating operations
– Examine verb phrase
"moves" verb is listed with Elevator
–
–
Should "moves" be an operation?
No message tells elevator to move
• Moves in response to a button, on condition that door is closed
• "Moves" should not be an operation
– "Arrives at floor" should not be an operation
• Elevator itself decides when to arrive, based on time
– "resets elevator button" - implies elevator sends message to
elevator button, telling it to reset
• ElevatorButton needs an operation to provide this service
to the elevator
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39
Creating Class Operations (II)
• Format
– Write operation name in bottom compartment in class
diagram
– Write operations as function names, include return type,
parameters
resetButton() : void
• Function takes no parameters, returns nothing
• Other verbs
– Bell - provides service of ringing
– Floor - signal arrival of elevator
– Door - open and close
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40
Creating Class Operations (III)
Elevator
currentFloor : int = 1
direction : enum = up
capacity : int = 1
arrivalTime : int
moving : bool = false
processTime( time : int ) : void
personEnters( ) : void
personExits( ) : void
summonElevator( ) : void
prepareToLeave( ) : void
Scheduler
Clock
Door
time : int = 0
open : bool = false
getTime( ) : int
tick( ) : void
openDoor( ) : void
closeDoor( ) : void
Floor
Bell
occupied : bool = false
elevatorArrived( ) : void
isOccupied( ) : bool
<none yet>
ringBell( ) : void
FloorButton
floor1ArrivalTime : int
floor2ArrivalTime : int
pressed : bool = false
processTime( time : int ) : void
pressButton( ) : void
resetButton( ) : void
Light
on : bool = false
turnOff( ) : void
turnOn( ) : void
Person
ID : int
ElevatorButton
pressed : bool = false
stepOntoFloor( ) : void
exitElevator( ) : void
enterElevator( ) : void
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resetButton( ) : void
pressButton( ) : void
Building
<none yet>
runSimulation( ) : void
41
Class Clock, Building
• Class Clock
– Has phrase "ticks every second"
– "Getting the time" is an operation clock provides
• Is the ticking also an operation?
• Look at how simulation works
• Building, once per second, will
–
–
–
–
Get time from clock
Give time to scheduler
Give time to elevator
Building has full responsibility for running simulation,
therefore it must increment clock
– getTime and tick therefore operations of Clock
– processTime operation of Scheduler and
Elevator
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42
Class Scheduler
• Class Scheduler
– "randomly schedules times", "delays creating a person by
one second"
– Scheduler performs these actions itself, does not provide
service to clients
– "create person" - special case
• A Person object cannot respond to a create message because
it does not yet exist
• Creation left to implementation details, not an operation of a
class
• More Chapter 7
– "tells a person to step onto a floor" - class Person should
have a function that the scheduler can invoke
• stepOntoFloor - operation of class Person
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43
Class Person
• More verb phrases
– "verifies a floor is unoccupied" - class Floor needs a
service to let other objects know if a floor is occupied or not
• isOccupied returns true or false
• Class Person
– "presses floor button", "presses elevator button"
• Place operation pressButton under classes
FloorButton and ElevatorButton
– "enter elevator", "exit elevator" - suggests class Elevator
needs operations to correspond to these actions
• Discover what, if any, actions operations perform when
concentrate on implementation
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44
Class Floor, FloorButton, ElevatorButton
• Class Floor
– "resets floor button" - resetButton operation
– "turns off light", "turns on light" - turnOff and turnOn
in class Light
• Class FloorButton and ElevatorButton
– "summons elevator" - summonElevator operation in
class Elevator
– "signals elevator to move" - elevator needs to provide a
"move" service
• Before it can move, must close door
• prepareToLeave operation (performs necessary actions
before moving) in class Elevator
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45
Class Door
• Class Door
– Phrases imply door sends message to person to tell it to exit
or enter elevator
– exitElevator and enterElevator in class Person
• Notes
– We do not overly concern ourselves with parameters or
return types
– Only want a basic understanding of each class
– As we continue design, number of operations may vary
• New operations needed
• Some current operations unnecessary
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46
Sequence Diagrams
• Sequence Diagram
– Model our "Simulation loop"
– Focuses on how messages are sent between objects over time
• Format
– Each object represented by a rectangle at top of diagram
• Name inside rectangle (objectName)
– Lifeline - dashed vertical line, represents progression of time
• Actions occur along lifeline in chronological order, top to
bottom
– Line with arrowhead - message between objects
•
•
•
•
Invokes corresponding operation in receiving object
Arrowhead points to lifeline of receiving object
Name of message above message line, includes parameters
Parameter name followed by colon and parameter type
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47
Sequence Diagrams (II)
: Building
: Clock
: Scheduler
tick( )
{currentTime < totalTime}
getTime( )
time
processTime( currentTime : int )
processTime( currentTime : int )
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: Elevator
48
Sequence Diagrams (III)
• Flow of control
– If object returns flow of control or returns a value, return
message (dashed line with arrowhead) goes back to original
object
• Clock object returns time in response to getTime message
received from Building object
• Activations
– Rectangles along lifelines - represent duration of an activity
• Height corresponds to duration
• Timing constraint
– In our diagram, text to far left
– While currentTime < totalTime objects keep
sending messages as in the diagram
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building : Building
49
scheduler : Scheduler
processTime( time )
floor1 : Floor
floor2 : Floor
isOccupied( ) : bool
bool converge
The two lifelines
[occupied = true]
scheduler
second floor same way as the first
[occupied =handles
false]
When
returns control to building
createfinished,: Person
personArrives( )
building sends processTime message to
scheduler delayArrival( floor1 )
scheduleArrival( floor1 )
scheduler decides whether to create a new person
Can only create a new person if floor unoccupied
Checks by
sending
isOccupied
floor
floor1
returns
true
or false message toisOccupied(
) : bool
object
scheduler's lifeline splits - conditional executionbool
of activities.
Condition supplied for each lifeline.
After new
Person
[occupied
= false] object created, it steps on first floor
create
Person
If true - scheduler calls Person
delayArrival
object :sends
personArrives
message to
personArrives( )
[occupied = true]
Not an operation - not invokedfloor1
by another
object
object
If false - scheduler
creates
new Person
delayArrival( floor2
) scheduler
schedules new arrival for floor1
scheduleArrival(
floor2 )
object
calls its own scheduleArrival
function (not an
When new objects created, rectangle
corresponds to
operation)
time. Message "create" sent from one object to
another (arrowhead points to new object).
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50
Conclusion
• Discussed operations of classes
– Introduced sequence diagrams to illustrate operations
• Chapter 5
– Examine how objects interact with each other
 2000 Prentice Hall, Inc. All rights reserved.