Answer:
The correction is as follows:
(a)
small = arr[0];
for (int j = 0; j < arr.length; j++){
if (small > arr[j]) {
small = arr[j]; } }
(b)
double sum = 0;
int n = arr.length;
for (int k = arr.length-1; k >= 0; k--)
sum = sum + arr[k];
double avg = sum /n;
(c)
int[] a = {1,2,3,4,5};
int[] b = new int[5];
for (int i=0; i<a.length; i++)
b[i] = a[a.length-i-1];
Explanation:
Required
Correct each of the given code
a.
This line is correct; it initializes the smallest to the first element
small = arr[0];
This iterates through the array; however, the last element is left out.
for (int j = 0; j < arr.length-1; j++)
So, the correct code is: <em>for (int j = 0; j < arr.length; j++){</em>
The expected remaining part of the program which compares the elements of the array is missing
I've completed the code [See the answer section]
b.
By syntax this line is correct. However, for the average to be calculated as double; sum has to be declared as double
int sum = 0;
So, the correct code is: <em>double sum = 0; </em>
This line correctly calculates the length of the array
int n = arr.length;
This iteration will create an endless loop
for (int k = arr.length; k <= 0; k--)
So, the correct code is: <em>for (int k = arr.length-1; k >= 0; k--) </em>
This correctly add up the elements of the array
sum = sum + arr[k];
This correctly calculate the average of the array elements
double avg = sum /n;
(c)
The array a is incorrectly initialized because there is no matching end curly brace
int[] a = {1,2,3,4,5);
So, the correct code is: <em>int[] a = {1,2,3,4,5};</em>
This correctly create array b with 5 elements
int[] b = new int[5];
This iterates through a; however, the last element is left out
for (int i=0; i<=a.length; i++)
So, the correct code is: <em>for (int i=0; i<a.length; i++)</em>
This will create an out of bound error
b[i-1] = a[a.length-i];
So, the correct code is: <em>b[i] = a[a.length-i-1];</em>
