Array Array Array Array Array Array Arra

Chapter 7
Arrays
PROGRAMMING IN ANSI C

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Question
 Questions:
How do we read in 100 scores and decide the every
grade of the score according to the average score?
 If a score is 10 or above greater than the average,
it is classified as grade 'A';
 If a score is 10 or above less than the average, it
is classified as grade 'C';
 Others are classified as grade 'B'.

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Question
main()
{ float s1, s2, …, s100, ave;
char grade1, grade2, ..., grade100;
scanf ( "%f", &s1 );
scanf ( "%f", &s2 );
……
ave = (s1 + s2 + … + s100) /100;
if ( s1 >= ave + 10 ) grade1 = 'A';
else if ( s1 < ave – 10 ) grade1 = 'C';
else grade1 = 'B';
if ( s2 >= ave + 10 ) grade2 = 'A';
……
}
 This program uses 100
variables s1, s2, ..., s100 to
deal with similar data and
do similar operations.
 We can use a structure to
organize those similar
variables – Array.
Array is a derived data type.
An array is a fixed-size sequenced collection
of elements of the same data type.

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Question
main()
{ float s[100], ave = 0; int i;
char grade[100];
for ( i = 0; i < 100; i ++ )
{ scanf ( "%f", &s[i] );
ave = ave + s[i];
}
ave = ave / 100;
for ( i = 0; i < 100; i++ )
if ( s[i] >= ave + 10 ) grade[i] = 'A';
else if ( s[i] < ave – 10 ) grade[i] = 'C';
else grade[i] = 'B';
}

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Chapter 6
 In this chapter, we will learn:
 One-dimensional arrays
 Two-dimensional arrays

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valid identifier
the type of the elements
One-dimensional Arrays - Declaration
 The general form of array declaration:
type array_name [size];
e.g. int a[6]; An integer constant,
the maximum number of
elements can be stored
inside the array
a
a[5]
5
a[4]
4
a[3]
3
a[2]
2
a[1]
1
a[0]
0
Array name is the start address
of this array in memory, and its
value is an address constant.
The spaces of elements
are continuous.
The first element is a[0].
The subscripts begin with 0.
The array "a" contains 6 elements.
The last element is a[5].
The subscripts of array a[n]
end with n-1.
Don't use
a[6] !!!

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main()
{ int i, s = 10;
float f[s];
f[0] = 0;
for ( i = 1; i < s; i ++ )
f[i] = f[i - 1] + 1;
……
}
main()
{ int i, s = 10;
float f[10];
f[0] = 0;
for ( i = 1; i < s; i ++ )
f[i] = f[i - 1] + 1;
……
}
#define SIZE 10
main()
{ int i;
float f[SIZE];
f[0] = 0;
for ( i = 1; i < SIZE; i ++ )
f[i] = f[i - 1] + 1;
……
}
One-dimensional Arrays - Declaration
 When the elements of
an array are used, the
subscripts of an array
can be integer
constants or variables
or expressions.
 However, in the
declaration of an array,
the size can be only an
integer constant
expression.
Error … : Constant expression
required in function ...

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 Like primary variables, an array must be declared
before it is used.
 We can only use the elements one by one, and can't
use the whole array.
(Notice: The array name is the start address of this
array, but it can't represent the overall elements.)
e.g. int a[6];
……
printf ( "%d", a );
One-dimensional Arrays - Usage of
Elements

e.g. int a[6], i;
……
for ( i = 0; i <= 5; i++ )
printf( "%d", a[i] );

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Elements
#define SIZE 10
main()
{ int i;
float f[SIZE];
f[0] = 0;
for ( i = 1; i < SIZE; i ++ )
f[i] = f[i - 1] + 0.1;
for ( i = 0; i < SIZE; i = i + 2 )
printf ( "%4.1f", f[i] );
}
0.0 0.2 0.4 0.6 0.8
The declaration of array
The usage of array elements

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Elements
e.g. int a[6];
a[6] = 10;

a
a[5]
5
a[4]
4
a[3]
3
a[2]
2
a[1]
1
a[0]
0
The elements are from a[0] to a[5].
a[6] is invalid.
 The array name "a" is the start address of this array,
which is the address of the first element a[0].
 a[i] is the (i+1)th element, the subscript i represents the
offset from the start address of the array.
 When a element is used, the compiler calculate its
address according to the start address and the
subscript, and then take out the data from the address.

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Elements
main()
{
int a[2];
a[0] = 1;
a[1] = 2;
printf ( "%dn", a[0] );
printf ( "%dn", a[1] );
printf ( "%d %x", a, a );
}
1
2
- 42 ffd6

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Elements
main()
{
int a[2];
scanf ( "%d", &a[0] );
a[1] = a[0] * 2;
printf ( "%dn", a[0] );
printf ( "%dn", a[1] );
}
5
5
10
scanf ( "%d", a );
scanf ( "%d", a + 1 );

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One-dimensional Arrays - Initialization
1. Like primary variables, we can initialize the elements
of an array when the array is declared.
 The general form of initialization of array is:
type array-name[size] = { list of values }
The values in the list are separated by commas.
e.g. int a[5] = { 1, 2, 3, 4, 5 };
Equivalent to:
int a[5];
a[0] = 1; a[1] = 2; a[2] = 3; a[3] = 4; a[4] = 5;

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2. Initialization may be partial.
 The number of initializers may be less than the
declared size. In such cases, the remaining elements
are initialized to zero.
e.g. int a[5] = { 1, 2, 3 };
Equivalent to:
int a[5];
a[0] = 1; a[1] = 2; a[2] = 3; a[3] = a[4] = 0;

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2. If the number of initializers is more than the
declared size, the complier will produce an error.
e.g. int a[3] = { 1, 2, 3, 4, 5 };
Error … : Too many initializers in function ...

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3. The size may be omitted.
 In such cases, the compiler allocates enough space
for all initialized elements. If the number of
initializers is n, the declared size of the array is n.
e.g. int a[ ] = { 1, 2, 3, 4, 5 };
Equivalent to:
int a[5] = { 1, 2, 3, 4, 5 };

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4. If the elements are not initialized before they are
used in an expression, the result of this expression is
unexpected.
 Before the elements are used, they must be
initialized, whether in compile time or in run time.
e.g. int a[3];
printf ( "%d, %d, %d", a[0], a[1], a[2] );
0, 64, 3117

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One-dimensional Arrays - Program 1
 Read in 10 integers, and find out the maximum and
the minimum and the average.
 Step 1: Read 10 numbers into array "a".
 Step 2: max = min = sum = a[0];
 Step3: for ( i = 1; i < 10; i++ )
{ if ( a[i] > max ) max = a[i];
if ( a[i] < min ) min = a[i];
sum = sum + a[i]; }
 Step 4: ave = sum / 10;
 Step 5: Output the values of max, min and ave.
N-S Flow Chart
for i = 0 to 9
Read in a[i]
max = min = sum = a[0];
ave = (float) sum / 10
for i = 1 to 9
max = a[i]
Y N
max<a[i]
min = a[i]
Y N
min>a[i]
sum = sum + a[i]
Output: max, min, ave

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#define N 10
main()
{ int a[N], max, min, sum, i; float ave;
printf ( "Enter %d integers:n", N );
for ( i = 0; i < N; i++ ) scanf ( "%d", &a[i] );
for ( i = 1; i < N; i++)
{ if ( a[i] > max ) max = a[i];
sum = sum + a[i]; }
ave = (float) sum / N;
printf ("max = %d, min = %d, ave = %.1f", max, min, ave );
}
#define N 10
main()
{ int a[N], max, min, sum, i; float ave;
printf ( "Enter %d integers:n", N );
for ( i = 0; i < N; i++ ) scanf ( "%d", a + i );
for ( i = 1; i < N; i++)
{ if ( a[i] > max ) max = a[i];
sum = sum + a[i]; }
ave = (float) sum / N;
printf ("max = %d, min = %d, ave = %.1f", max, min, ave );
}
Enter 10 integers:
1 2 3 4 5 6 7 8 9 10
max = 10, min = 1, ave = 5.5

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 Read in 10 scores, and decide the every grade of the
score according to the average score.
 If a score is 10 or above greater than the average,
it is classified as grade 'A';
 If a score is 10 or above less than the average, it
is classified as grade 'C';
 Others are classified as grade 'B'.
 Step 1: Read in 10 scores (float type).
 Step 2: Calculate the average score.
 Step 3: Compare every score with the average to
decide the every grade, and then output it.

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#define N 10
main()
{ float score[N], ave, sum = 0; int i; char grade;
for ( i = 0; i < N; i++ )
{ printf ( "Please input the score of student %d:", i+1 );
scanf ( "%f", &score[i] );
sum = sum + score[i]; }
printf("The average is : %.1fn", ave = sum / N);
for ( i = 0; i < N; i++ )
{ if ( score[i] - ave >= 10 ) grade = 'A';
else if ( score[i] - ave <= -10 ) grade = 'C';
else grade = 'B';
printf( "The grade of student %d is: %cn", i+1, grade); }
}

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 Output the former 40 numbers of the Fibonacci
sequence.
 Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, ……
F1 = 1 (n = 1)
F2 = 1 (n = 2)
Fn = Fn-1 + Fn-2 (n≥3)
f[39]
39
f[5]
5
f[4]
4
f[3]
3
f[2]
2
f[1]
1
1
f[0]
1
0
2
3
5
8
……

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main()
{
int i;
long f[40] = { 1, 1 };
for ( i = 2; i < 40; i++ )
f[i] = f[i-2] + f[i-1];
for ( i = 0; i < 40; i++ )
{ if ( i%4 == 0)
printf ( "n" );
printf ( "%12ld", f[i] );
}
}
main()
{
long f1 = 1, f2 = 1;
int i;
for ( i = 1; i <= 20; i++ )
{ printf ( "%12ld %12ld", f1,
f2 );
if ( i%2 == 0)
printf ( "n" );
f1 = f1 + f2;
f2 = f2 + f1;
}
}

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 Read in 20 characters, find out the character
specified by user, and output its position.
 Step 1: Read 20 characters into array "c".
 Step 2: Read in the character which the user wants
to search, and assign to variable ch.
 Step3: Compare ch with the array "c" from c[0] to
[9] to test whether the array contains this
character or not.

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main()
{ char c[20], ch; int i;
printf ( "Please input the 20 characters:n" );
for ( i = 0; i < 20; i++ ) scanf ( "%c", &c[i] );
printf("What do you want to search?"); scanf("%c", &ch);
for ( i=0; i<20; i++)
{ if ( c[i] == ch )
{ printf("'%c' at the position %d.", ch, i+1 );
break; }
else if ( i == 19 ) printf ( "No found!" ); }
}
main()
{ char c[20], ch; int i;
printf ( "Please input the 20 characters:n" );
for ( i = 0; i < 20; i++ ) scanf ( "%c", &c[i] );
scanf ( "%c", &ch );
printf("What do you want to search?"); scanf("%c", &ch);
for ( i=0; i<20; i++)
{ if ( c[i] == ch )
{ printf("'%c' at the position %d.", ch, i+1 );
break; }
else if ( i == 19 ) printf ( "No found!" ); }
}
Please input the 20 characters:
abcdefghijklmnopqrst
What do you want to search?No found!
Please input the 20 characters:
abcdefghijklmnopqrst
What do you want to search?
'd' at the positon 4.
d

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 Bubble sort: Arrange elements in the list according to
their values, in ascending order.
 Method: Compare the two adjacent numbers, and
put the larger number on the latter position.
 Step 1: Read n numbers into the array a.

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 Step 2: Compare the 1st and the 2nd number, and
put the larger on the latter position. Then compare
the 2nd and the 3rd number, and put the larger on
the latter position ... In the same way, until we have
compared and put the (n-1)th and the nth number –
we put the largest number on the last position.

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 Step 3: In the same way, we do with the former n-1
numbers, and the second largest number is put on
the last second position.
 Step 4: Repeat this process until all the numbers
are sorted.

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49 38 65 97 76 13 27 30
38 49 65 76 13 27 30 97
38
49
38 49 65
49
49 65 97
65
65 97 76
97
76 97 13
97
13 97 27
97
27 97 30
97
30 97
49
38
38 49 65
49
49 65 76
65
65 76 13
76
13 76 27
76
27 76 30
76
30 76
38 49 65 13 27 30 76 97
49
38
38 49 65
49
49 65 13
65
13 65 27
65
27 65 30
65
30 65
38 49 13 27 30 65 76 97
49
38
38 49 13
49
13 49 27
49
27 49 30
49
30 49
38 13 27 30 49 65 76 97
13
38
13 38 27
38
27 38 30
38
30 38
13 27 30 38 49 65 76 97
27
13
13 27 30
27
27 30
13 27 30 38 49 65 76 97
27
13
13 27
13 27 30 38 49 65 76 97
Comparison
7 times
6 times
5 times
4 times
3 times
2 times
1 time
a[0] a[1] a[2] a[3] a[4] a[5] a[6] a[7]
1st sorting
2nd sorting
3rd sorting
4th sorting
5th sorting
6th sorting
7th sorting
result
• To n numbers, altogether times of sorting.
• In ith sorting, altogether times of comparison.
n-1
"i" represents times of sorting: 1 <= i <= n-1
n-i
"j" represents the subscript of the compared
element : 1 <= j <= n-i
Every time, a[j-1] and a[j] are compared.

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#define N 10
main ( )
{ int a[N], i, j, t;
printf ( "Input %d numbers:n", N );
for ( i = 0; i <= N-1; i++) scanf ( "%d", &a[i] );
for ( i = 1; i <= N-1; i++ )
{ for ( j = 1; j <= N - i; j++ )
{ if ( a[j-1] > a[j] )
{ t = a[j-1];
a[j-1] = a[j];
a[j] = t; }
}
}
for ( i = 0; i < N; i++ ) printf ( "%d ",a[i] );
}

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 Selection sort: Arrange elements in the list according
to their values, in ascending order.
 Method: First select the smallest number and put it
on the first position. Then select the second
smallest number and put it on the second position...
In the same way, place all the numbers in ascending
order.

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49 38 65 97 76 13 27 30
a[0] a[1] a[2] a[3] a[4] a[5] a[6] a[7] Comparison
7 times
6 times
5 times
4 times
3 times
2 times
1 time
result
13 38 65 97 76 49 27 30
13 38 65 97 76 49 27 30
13 27 65 97 76 49 38 30
13 27 65 97 76 49 38 30
13 27 30 97 76 49 38 65
13 27 30 97 76 49 38 65
13 27 30 38 76 49 97 65
13 27 30 38 76 49 97 65
13 27 30 38 49 76 97 65
13 27 30 38 49 65 97 76
13 27 30 38 49 76 97 65
13 27 30 38 49 65 97 76
13 27 30 38 49 65 76 97
13 27 30 38 49 65 76 97
1th sorting
2nd sorting
3rd sorting
4th sorting
5th sorting
6th sorting
7th sorting
• To n numbers, altogether times of sorting.
• In ith sorting, altogether times of comparison.
n-1
"i" represents times of sorting: 1 <= i <= n-1
n-i
"j" represents the subscript of the compared
element : i <= j <= n-1
Every sorting, the smallest element in the compared
elements is exchange with the element a[i-1].

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#define N 10
main()
{ int a[N], i, j, t, k;
printf ( "Input %d numbers:n", N );
for ( i = 0; i < N; i++ ) scanf ( "%d", &a[i] );
for( i = 1; i < N; i++ )
{ k = i - 1;
for ( j = i; j <= N - 1; j++ )
if ( a[k] > a[j] ) k = j;
if(k!=i-1)
{ t = a[k]; a[k] = a[i-1]; a[i-1] = t; }
}
for ( i = 0; i < N; i++ ) printf ( "%d ", a[i] );
}

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 Read 10 integers into an array, and then rearrange
the elements of this array in the inverted order.
 Step 1: Read 10 integers into the array a.
 Step 2: Exchange a[0] and a[9], a[1] and a[8]....
a[4] and a[5].
 Step 3: Output this rearranged array.
90
80
70
60
50
40
30
20
10
0 0
80
70
60
50
40
30
20
10
90 0
10
70
60
50
40
30
20
80
90 0
10
20
60
50
40
30
70
80
90 0
10
20
30
50
40
60
70
80
90 0
10
20
30
40
50
60
70
80
90

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main()
{ int a[10], i, j, t;
printf ( "Please input the 10 integers:" );
for ( i = 0; i < 10; i++ ) scanf ( "%d", &a[i] );
printf ( "The old sequence is: " );
for ( i = 0; i < 10; i++ ) printf ( "%d ", a[i] );
printf("n");
for ( i = 0, j = 9; i < j; i++, j-- )
{ t = a[i]; a[i] = a[j]; a[j] = t; }
printf("The new sequence is: ");
for ( i = 0; i < 10; i++ ) printf ( "%d ", a[i] );
}
 Consider: How to rearrange
n numbers in the inverted order?
for ( i = 0; i < 5; i++ )
{ t = a[i]; a[i] = a[9-i]; a[9-i] = t; }

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 10 candidates participate in an election campaign, and
100 persons vote for them. Calculate the ballot of each
candidate, and output the result with the histogram
expressed by "*".
 Step 1: Declare and initialize the array:
int count[11] = {0};
 Step 2: Read in the vote. If one vote is for
candidate n, add 1 to count[n].
 Step 3: Output the result and the corresponding “*”.

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main()
{ int count[11] = {0}, i, j, n;
printf ( "Please input the 100 votes:" );
for ( i = 0; i < 100; i++ )
{ scanf ( "%d", &n );
count[n]++;
}
printf ( "Grade:tCounttGraphn" );
for ( i = 1; i <= 10; i++ )
{ printf ( "Candidate%3d: %d ballotst", i, count[i] );
for ( j = 0; j < count[i]; j++ ) printf ( "*" );
printf ( "n" );
}
}
Candidates : Count Graph
Candidate 1: 5 ballots *****
Candidate 2: 6 ballots ******
Candidate 3: 10 ballots **********
Candidate 4: 12 ballots ************
Candidate 5: 2 ballots **
Candidate 6: 7 ballots *******
Candidate 7: 18 ballots ******************
Candidate 8: 28 ballots ****************************
Candidate 9: 8 ballots ********
Candidate 10: 4 ballots ****

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Question
 Questions:
In maths, we often use the matrix, such as Am×n.
How do we express it in our program?

















25
19
2
7
5
13
0
6
3
5
3
1
3
4
A














43
42
41
33
32
31
23
22
21
13
12
11
3
4
a
a
a
a
a
a
a
a
a
a
a
a
A
Two-dimensional array!

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 The general form of array declaration:
type array_name [row_size] [column_size];
e.g. int a[4][3];
 The storage of the elements :
 The memory is one-dimensional.
 In memory: After all the elements of one line are
stored, another line is stored.
int a[3][2]
5
4
3
2
1
0










]
1
][
2
[
]
0
][
2
[
]
1
][
1
[
]
0
][
1
[
]
1
][
0
[
]
0
][
0
[
a
a
a
a
a
a
Two-dimensional Arrays
a[2][1]
a[2][0]
a[1][1]
a[1][0]
a[0][1]
a[0][0]
int c[2][3][4]
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23 c[1][2][3]
c[1][2][2]
c[1][2][1]
c[1][2][0]
c[1][1][3]
c[1][1][2]
c[1][1][1]
c[1][1][0]
c[1][0][3]
c[1][0][2]
c[1][0][1]
c[1][0][0]
c[0][2][3]
c[0][2][2]
c[0][2][1]
c[0][2][0]
c[0][1][3]
c[0][1][2]
c[0][1][1]
c[0][1][0]
c[0][0][3]
c[0][0][2]
c[0][0][1]
c[0][0][0]

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Two-dimensional Arrays
int a[3][4];
Each element a[i] is a
one-dimensional array
containing 4 elements.
0 a[0][0]
1 a[0][1]
2 a[0][2]
3 a[0][3]
4 a[1][0]
5 a[1][1]
6 a[1][2]
7 a[1][3]
8 a[2][0]
9 a[2][1]
10 a[2][2]
11 a[2][3]
a[0]
a[1]
a[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]
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]
The two-dimensional array a can
be regarded as a one-dimensional
array containing 3 elements.
a[2]
a[1]
a[0]

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Two-dimensional Arrays - Initialization
 A two-dimensional array can be initialized row by row.
int a[2][3] = { {1,2,0}, {4,0,0} };
Complete initialization
a[0][0] 1
a[0][1] 2
a[0][2] 0
a[1][0] 4
a[1][1] 0
a[1][2] 0
int a[2][3] = { {1,2}, {4} };
Partial initialization

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 A two-dimensional array can be initialized row by row.
a[0][0] 1
a[0][1] 2
a[0][2] 0
a[1][0] 4
a[1][1] 0
a[1][2] 0
int a[ ][3] = { {1,2}, {4} };
The size of the first
dimension can be omitted.
Notice: The size of the second dimension
can't be omitted!
e.g. int a[2][ ] = {{1,2}, {4}} is wrong!

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 A two-dimensional array can be initialized
with a list of initial values. a[0][0] 1
a[0][1] 2
a[0][2] 0
a[1][0] 4
a[1][1] 0
a[1][2] 0
int a[2][3] = { 1, 2, 0, 4, 0, 0 };
Complete initialization
int a[2][3] = { 1, 2, 0, 4 };
Partial initialization

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 A two-dimensional array can be initialized
with a list of initial values. a[0][0] 1
a[0][1] 2
a[0][2] 0
a[1][0] 4
a[1][1] 0
a[1][2] 0
int a[ ][3] = { 1, 2, 0, 4};
The size of the first
dimension can be omitted.
Notice: The size of the second dimension
can't be omitted!
e.g. int a[2][ ] = { 1, 2, 0, 4 }; is wrong!

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 Calculate each row sum of the matrix a[3][4].
 Step 1: Declare and initialize a two-dimensional
array a[3][4].
 Step 2: Declare a one-dimensional array s[3] to
store each row sum.
 Step 3: Calculate the sum of the ith row, and store
it to the element s[i].
 Step 4: Output the value of the array s.
Two-dimensional Arrays - Program 1

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main()
{ int a[3][4] = { {1,2,3,4}, {5,6,7}, {8,9,10} }, s[3];
int i, j, sum=0;
for ( i = 0; i < 3; i++, sum = 0)
{ for ( j = 0; j < 4; j++ )
sum = sum + a[i][j];
s[i] = sum;
}
for ( i = 0; i < 3; i++ )
printf("The row sum %d is: %dn",i+1,s[i]);
}
The row sum 1 is: 10
a[0][0] 1
a[0][1] 2
a[0][2] 3
a[0][3] 4
a[1][0] 5
a[1][1] 6
a[1][2] 7
a[1][3] 0
a[2][0] 8
a[2][1] 9
a[2][2] 10
a[2][3] 0

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Find the largest element from a matrix A3×4. Output
this element and its row number and column number.
 Step 1: Declare and initialize a two-dimensional
array a[3][4].
 Step 2: Declare and initialize the variable
max=a[0][0], row=0, column=0.
 Step 3: Find the largest element and use above 3
variables to store the largest element, the row
number and column number.
 Step 4: Output the values of max, row and column.

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main()
{ int a[3][4] = { {1,2,3,4}, {9,8,7,6}, {-10,10,-5,2} };
int max=a[0][0], row=0, colum=0, i, j;
for ( i = 0; i < 3; i++ )
for ( j = 0; j < 4; j++ )
if ( a[i][j] > max )
{ max = a[i][j];
row = i;
colum = j;
}
printf ( "max=%d, row=%d, column=%d",
max, row+1, colum+1 );
}
max=10, row=3, column=2
a[0][0] 1
a[0][1] 2
a[0][2] 3
a[0][3] 4
a[1][0] 9
a[1][1] 8
a[1][2] 7
a[1][3] 6
a[2][0] -10
a[2][1] 10
a[2][2] -5
a[2][3] 2

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Initialize a matrix with 2 rows and 3 columns and
output it. Then transpose this matrix and output it.








6
5
4
3
2
1
3
2
A












6
3
5
2
4
1
2
3
B
transpose

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Initialize a matrix with 2 rows and 3 columns and
output it. Then transpose this matrix and output it.
 Step 1: Declare and initialize 2 two-dimensional
arrays: a[2][3] and b[3][2].
 Step 2: Output the two-dimensional array a.
 Step 3: b[j][i] = a[i][j];
 Step 4: Output the two-dimensional array b.

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main()
{ int a[2][3] = { 1, 2, 3, 4, 5, 6 }, b[3][2], i, j;
for ( i = 0; i < 2; i++ )
{ for ( j = 0; j < 3; j++ )
{ printf ( "%5d", a[i][j] );
b[j][i] = a[i][j];
}
printf ( "n" );
} printf ( "n" );
for ( i = 0; i < 3; i++ )
{ for ( j = 0; j < 2; j++ )
printf ( "%5d", b[i][j] );
printf ( "n" ); }
}
1 2 3
4 5 6
1 4
2 5
3 6

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Homework
 Review Questions P212
 7.1, 7.3～7.8 (Write down in your exercise book)
 Programming Exercises

Array Array Array Array Array Array Arra

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