Arrays 06.ppt

 2000 Prentice Hall, Inc.
All rights reserved.
1
Chapter 6 - Arrays
Outline
6.1 Introduction
6.2 Arrays
6.3 Declaring Arrays
6.4 Examples Using Arrays
6.5 Passing Arrays to Functions
6.6 Sorting Arrays
6.7 Case Study: Computing Mean, Median and Mode Using Arrays
6.8 Searching Arrays
6.9 Multiple-Subscripted Arrays

2
6.1 Introduction
• Arrays
– Structures of related data items
– Static entity – same size throughout program
– Dynamic data structures discussed in Chapter 12

3
6.2 Arrays
• Array
– Group of consecutive memory locations
– Same name and type
• To refer to an element, specify
– Array name
– Position number
• Format:
arrayname[ position number ]
– First element at position 0
– n element array named c:
• c[ 0 ], c[ 1 ]...c[ n – 1 ]
Name of array
(Note that all
elements of this
array have the
same name, c)
Position number
of the element
within array c
c[6]
-45
6
0
72
1543
-89
0
62
-3
1
6453
78
c[0]
c[1]
c[2]
c[3]
c[11]
c[10]
c[9]
c[8]
c[7]
c[5]
c[4]

4
6.2 Arrays
• Array elements are like normal variables
c[ 0 ] = 3;
printf( "%d", c[ 0 ] );
– Perform operations in subscript. If x equals 3
c[ 5 - 2 ] == c[ 3 ] == c[ x ]

5
6.3 Declaring Arrays
• When declaring arrays, specify
– Name
– Type of array
– Number of elements
arrayType arrayName[ numberOfElements ];
– Examples:
int c[ 10 ];
float myArray[ 3284 ];
• Declaring multiple arrays of same type
– Format similar to regular variables
– Example:
int b[ 100 ], x[ 27 ];

6
• Initializers
int n[ 5 ] = { 1, 2, 3, 4, 5 };
– If not enough initializers, rightmost elements become 0
int n[ 5 ] = { 0 }
• All elements 0
– If too many a syntax error is produced syntax error
– C arrays have no bounds checking
• If size omitted, initializers determine it
int n[ ] = { 1, 2, 3, 4, 5 };
– 5 initializers, therefore 5 element array

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7
1. Initialize array
2. Loop
3. Print
1 /* Fig. 6.8: fig06_08.c
2 Histogram printing program */
3 #include <stdio.h>
4 #define SIZE 10
5
6 int main()
7 {
8 int n[ SIZE ] = { 19, 3, 15, 7, 11, 9, 13, 5, 17, 1 };
9 int i, j;
10
11 printf( "%s%13s%17sn", "Element", "Value", "Histogram" );
12
13 for ( i = 0; i <= SIZE - 1; i++ ) {
14 printf( "%7d%13d ", i, n[ i ]) ;
15
16 for ( j = 1; j <= n[ i ]; j++ ) /* print one bar */
17 printf( "%c", '*' );
18
19 printf( "n" );
20 }
21
22 return 0;
23 }

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8
Program Output
Element Value Histogram
0 19 *******************
1 3 ***
2 15 ***************
3 7 *******
4 11 ***********
5 9 *********
6 13 *************
7 5 *****
8 17 *****************
9 1 *

9
• Character arrays
– String “first” is really a static array of characters
– Character arrays can be initialized using string literals
char string1[] = "first";
• Null character '0' terminates strings
• string1 actually has 6 elements
– It is equivalent to
char string1[] = { 'f', 'i', 'r', 's', 't', '0' };
– Can access individual characters
string1[ 3 ] is character ‘s’
– Array name is address of array, so & not needed for scanf
scanf( "%s", string2 );
• Reads characters until whitespace encountered
• Can write beyond end of array, be careful

Outline
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10
1. Initialize strings
2. Print strings
2.1 Define loop
2.2 Print characters
individually
2.3 Input string
3. Print string
Program Output
1 /* Fig. 6.10: fig06_10.c
2 Treating character arrays as strings */
4
5 int main()
6 {
7 char string1[ 20 ], string2[] = "string literal";
8 int i;
9
10 printf(" Enter a string: ");
11 scanf( "%s", string1 );
12 printf( "string1 is: %snstring2: is %sn"
13 "string1 with spaces between characters is:n",
14 string1, string2 );
15
16 for ( i = 0; string1[ i ] != '0'; i++ )
17 printf( "%c ", string1[ i ] );
18
19 printf( "n" );
20 return 0;
21 }
Enter a string: Hello there
string1 is: Hello
string2 is: string literal
string1 with spaces between characters is:
H e l l o

11
• Passing arrays
– To pass an array argument to a function, specify the name of
the array without any brackets
int myArray[ 24 ];
myFunction( myArray, 24 );
• Array size usually passed to function
– Arrays passed call-by-reference
– Name of array is address of first element
– Function knows where the array is stored
• Modifies original memory locations
• Passing array elements
– Passed by call-by-value
– Pass subscripted name (i.e., myArray[ 3 ]) to function

12
• Function prototype
void modifyArray( int b[], int arraySize );
– Parameter names optional in prototype
• int b[] could be written int []
• int arraySize could be simply int

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13
1. Function definitions
2. Pass array to a
function
2.1 Pass array element
to a function
3. Print
1 /* Fig. 6.13: fig06_13.c
2 Passing arrays and individual array elements to functions */
4 #define SIZE 5
5
6 void modifyArray( int [], int ); /* appears strange */
7 void modifyElement( int );
8
9 int main()
10 {
11 int a[ SIZE ] = { 0, 1, 2, 3, 4 }, i;
12
13 printf( "Effects of passing entire array call "
14 "by reference:nnThe values of the "
15 "original array are:n" );
16
17 for ( i = 0; i <= SIZE - 1; i++ )
18 printf( "%3d", a[ i ] );
19
20 printf( "n" );
21 modifyArray( a, SIZE ); /* passed call by reference */
22 printf( "The values of the modified array are:n" );
23
24 for ( i = 0; i <= SIZE - 1; i++ )
25 printf( "%3d", a[ i ] );
26
27 printf( "nnnEffects of passing array element call "
28 "by value:nnThe value of a[3] is %dn", a[ 3 ] );
29 modifyElement( a[ 3 ] );
30 printf( "The value of a[ 3 ] is %dn", a[ 3 ] );
31 return 0;
32 }
Entire arrays passed call-by-
reference, and can be modified
Array elements passed call-by-
value, and cannot be modified

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14
3.1 Function
definitions
Program Output
33
34 void modifyArray( int b[], int size )
35 {
36 int j;
37
38 for ( j = 0; j <= size - 1; j++ )
39 b[ j ] *= 2;
40 }
41
42 void modifyElement( int e )
43 {
44 printf( "Value in modifyElement is %dn", e *= 2 );
45 }
Effects of passing entire array call by reference:
The values of the original array are:
0 1 2 3 4
The values of the modified array are:
0 2 4 6 8
Effects of passing array element call by value:
The value of a[3] is 6
Value in modifyElement is 12
The value of a[3] is 6

15
6.6 Sorting Arrays
• Sorting data
– Important computing application
– Virtually every organization must sort some data
• 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
• 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

16
6.7 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 the median
• Mode – number that occurs most often
– 1, 1, 1, 2, 3, 3, 4, 5
– 1 is the mode

Outline
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17
1. Function prototypes
1.1 Initialize array
2. Call functions mean,
median, and mode
1 /* Fig. 6.16: fig06_16.c
2 This program introduces the topic of survey data analysis.
3 It computes the mean, median, and mode of the data */
5 #define SIZE 99
6
7 void mean( const int [] );
8 void median( int [] );
9 void mode( int [], const int [] ) ;
10 void bubbleSort( int [] );
11 void printArray( const int [] );
12
13 int main()
14 {
15 int frequency[ 10 ] = { 0 };
16 int response[ SIZE ] =
17 { 6, 7, 8, 9, 8, 7, 8, 9, 8, 9,
18 7, 8, 9, 5, 9, 8, 7, 8, 7, 8,
19 6, 7, 8, 9, 3, 9, 8, 7, 8, 7,
20 7, 8, 9, 8, 9, 8, 9, 7, 8, 9,
21 6, 7, 8, 7, 8, 7, 9, 8, 9, 2,
22 7, 8, 9, 8, 9, 8, 9, 7, 5, 3,
23 5, 6, 7, 2, 5, 3, 9, 4, 6, 4,
24 7, 8, 9, 6, 8, 7, 8, 9, 7, 8,
25 7, 4, 4, 2, 5, 3, 8, 7, 5, 6,
26 4, 5, 6, 1, 6, 5, 7, 8, 7 };
27
28 mean( response );
29 median( response );
30 mode( frequency, response );
31 return 0;
32 }

Outline
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18
3. Define function
mean
3.1 Define function
median
3.1.1 Sort Array
3.1.2 Print middle
element
33
34 void mean( const int answer[] )
35 {
36 int j, total = 0;
37
38 printf( "%sn%sn%sn", "********", " Mean", "********" );
39
40 for ( j = 0; j <= SIZE - 1; j++ )
41 total += answer[ j ];
42
43 printf( "The mean is the average value of the datan"
44 "items. The mean is equal to the total ofn"
45 "all the data items divided by the numbern"
46 "of data items ( %d ). The mean value forn"
47 "this run is: %d / %d = %.4fnn",
48 SIZE, total, SIZE, ( double ) total / SIZE );
49 }
50
51 void median( int answer[] )
52 {
53 printf( "n%sn%sn%sn%s",
54 "********", " Median", "********",
55 "The unsorted array of responses is" );
56
57 printArray( answer );
58 bubbleSort( answer );
59 printf( "nnThe sorted array is" );
60 printArray( answer );
61 printf( "nnThe median is element %d ofn"
62 "the sorted %d element array.n"
63 "For this run the median is %dnn",
64 SIZE / 2, SIZE, answer[ SIZE / 2 ] );

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19
65 }
66
67 void mode( int freq[], const int answer[] )
68 {
69 int rating, j, h, largest = 0, modeValue = 0;
70
71 printf( "n%sn%sn%sn",
72 "********", " Mode", "********" );
73
74 for ( rating = 1; rating <= 9; rating++ )
75 freq[ rating ] = 0;
76
77 for ( j = 0; j <= SIZE - 1; j++ )
78 ++freq[ answer[ j ] ];
79
80 printf( "%s%11s%19snn%54sn%54snn",
81 "Response", "Frequency", "Histogram",
82 "1 1 2 2", "5 0 5 0 5" );
83
84 for ( rating = 1; rating <= 9; rating++ ) {
85 printf( "%8d%11d ", rating, freq[ rating ] );
86
87 if ( freq[ rating ] > largest ) {
88 largest = freq[ rating ];
89 modeValue = rating;
90 }
91
92 for ( h = 1; h <= freq[ rating ]; h++ )
93 printf( "*" );
94
3.2 Define function
mode
3.2.1 Increase
frequency[]
depending on
response[]
Notice how the subscript in
frequency[] is the value of an
element in response[]
(answer[])
Print stars depending on value of
frequency[]

Outline
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20
3.3 Define bubbleSort
3.3 Define printArray
95 printf( "n" );
96 }
97
98 printf( "The mode is the most frequent value.n"
99 "For this run the mode is %d which occurred"
100 " %d times.n", modeValue, largest );
101}
102
103 void bubbleSort( int a[] )
104 {
105 int pass, j, hold;
106
107 for ( pass = 1; pass <= SIZE - 1; pass++ )
108
109 for ( j = 0; j <= SIZE - 2; j++ )
110
111 if ( a[ j ] > a[ j + 1 ] ) {
112 hold = a[ j ];
113 a[ j ] = a[ j + 1 ];
114 a[ j + 1 ] = hold;
115 }
116 }
117
118 void printArray( const int a[] )
119 {
120 int j;
121
122 for ( j = 0; j <= SIZE - 1; j++ ) {
123
124 if ( j % 20 == 0 )
125 printf( "n" );
Bubble sort: if elements out of order,
swap them.

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21
Program Output
126
127 printf( "%2d", a[ j ] );
128 }
129 }
********
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 array of responses is
7 8 9 8 7 8 9 8 9 7 8 9 5 9 8 7 8 7 8
6 7 8 9 3 9 8 7 8 7 7 8 9 8 9 8 9 7 8 9
6 7 8 7 8 7 9 8 9 2 7 8 9 8 9 8 9 7 5 3
5 6 7 2 5 3 9 4 6 4 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
The sorted array is
1 2 2 2 3 3 3 3 4 4 4 4 4 5 5 5 5 5 5 5
5 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7
7 7 7 7 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8
8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8
9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
The median is element 49 of
the sorted 99 element array.
For this run the median is 7

Outline
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22
Program Output
********
Mode
********
Response Frequency Histogram
1 1 2 2
5 0 5 0 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.

23
6.8 Searching Arrays: Linear Search and
Binary Search
• Search an array for a key value
• Linear search
– Simple
– Compare each element of array with key value
– Useful for small and unsorted arrays

24
6.8 Searching Arrays: Linear Search and
Binary Search
• Binary search
– For sorted arrays
– Compares middle element with key
• If equal, match found
• If key < middle, looks in first half of array
• If key > middle, looks in last half
• Repeat
– Very fast; at most n steps, where 2n > number of elements
• 30 element array takes at most 5 steps
– 25 > 30 so at most 5 steps
5

25
• Multiple subscripted arrays
– Tables with rows and columns (m by n array)
– Like matrices: specify row, then column
Row 0
Row 1
Row 2
Column 0 Column 1 Column 2 Column 3
a[ 0 ][ 0 ]
a[ 1 ][ 0 ]
a[ 2 ][ 0 ]
a[ 0 ][ 1 ]
a[ 1 ][ 1 ]
a[ 2 ][ 1 ]
a[ 0 ][ 2 ]
a[ 1 ][ 2 ]
a[ 2 ][ 2 ]
a[ 0 ][ 3 ]
a[ 1 ][ 3 ]
a[ 2 ][ 3 ]
Row subscript
Array name
Column subscript

26
• Initialization
– int b[ 2 ][ 2 ] = { { 1, 2 }, { 3, 4 } };
– Initializers grouped by row in braces
– If not enough, unspecified elements set to zero
int b[ 2 ][ 2 ] = { { 1 }, { 3, 4 } };
• Referencing elements
– Specify row, then column
printf( "%d", b[ 0 ][ 1 ] );
1 2
3 4
1 0
3 4

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27
1. Initialize variables
1.1 Define functions to
take double scripted
arrays
1.2 Initialize
studentgrades[][]
2. Call functions
minimum, maximum,
and average
1 /* Fig. 6.22: fig06_22.c
2 Double-subscripted array example */
4 #define STUDENTS 3
5 #define EXAMS 4
6
7 int minimum( const int [][ EXAMS ], int, int );
8 int maximum( const int [][ EXAMS ], int, int );
9 double average( const int [], int );
10 void printArray( const int [][ EXAMS ], int, int );
11
12 int main()
13 {
14 int student;
15 const int studentGrades[ STUDENTS ][ EXAMS ] =
16 { { 77, 68, 86, 73 },
17 { 96, 87, 89, 78 },
18 { 70, 90, 86, 81 } };
19
20 printf( "The array is:n" );
21 printArray( studentGrades, STUDENTS, EXAMS );
22 printf( "nnLowest grade: %dnHighest grade: %dn",
23 minimum( studentGrades, STUDENTS, EXAMS ),
24 maximum( studentGrades, STUDENTS, EXAMS ) );
25
26 for ( student = 0; student <= STUDENTS - 1; student++ )
27 printf( "The average grade for student %d is %.2fn",
28 student,
29 average( studentGrades[ student ], EXAMS ) );
30
31 return 0;
32 }
Each row is a particular student,
each column is the grades on the
exam.

Outline
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28
3. Define functions
33
34 /* Find the minimum grade */
35 int minimum( const int grades[][ EXAMS ],
36 int pupils, int tests )
37 {
38 int i, j, lowGrade = 100;
39
40 for ( i = 0; i <= pupils - 1; i++ )
41 for ( j = 0; j <= tests - 1; j++ )
42 if ( grades[ i ][ j ] < lowGrade )
43 lowGrade = grades[ i ][ j ];
44
45 return lowGrade;
46 }
47
48 /* Find the maximum grade */
49 int maximum( const int grades[][ EXAMS ],
51 {
52 int i, j, highGrade = 0;
53
54 for ( i = 0; i <= pupils - 1; i++ )
55 for ( j = 0; j <= tests - 1; j++ )
56 if ( grades[ i ][ j ] > highGrade )
57 highGrade = grades[ i ][ j ];
58
59 return highGrade;
60 }
61
62 /* Determine the average grade for a particular exam */
63 double average( const int setOfGrades[], int tests )
64 {

Outline
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29
3. Define functions
65 int i, total = 0;
66
67 for ( i = 0; i <= tests - 1; i++ )
68 total += setOfGrades[ i ];
69
70 return ( double ) total / tests;
71 }
72
73 /* Print the array */
74 void printArray( const int grades[][ EXAMS ],
76 {
77 int i, j;
78
79 printf( " [0] [1] [2] [3]" );
80
81 for ( i = 0; i <= pupils - 1; i++ ) {
82 printf( "nstudentGrades[%d] ", i );
83
84 for ( j = 0; j <= tests - 1; j++ )
85 printf( "%-5d", grades[ i ][ j ] );
86 }
87 }

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30
Program Output
The array is:
[0] [1] [2] [3]
studentGrades[0] 77 68 86 73
Lowest grade: 68
Highest grade: 96
The average grade for student 0 is 76.00

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