Chapter 6, Problem 22RE

Explanation of Solution

a. The given code:

// creating an integer variable n

int n = 1;

//iterate a for loop up to the condition i < 5.

for (int i = 2; i < 5; i++)

{

    //adding value of “i” to “n”

    n = n + i;

}

//printing the value of n

System.out.print(n);

Explanation:

The above snippet of code is used to demonstrate “for” loop. In the code,

• An integer variable “n” is declared and assigned a value “1”.
• Iterate “for” loop up to the condition “i < 5”. The loop variable was initialized with “i=1” and for every iteration “i” is incremented by “1”.
  • The variable “n” is added with value of “i”.
• The value of “n” is printed on the screen.

Trace table for the code snippet:

n = 1+2 = 3

n = 3+3 = 6;

n = 6+4 = 10;

Expected output:

10

b. The given code:

// creating an integer variable i.

int i;

//creating a double variable “n”

double n = 1 / 2;

//iterate a for loop up to the condition i <= 5.

for (i = 2; i <= 5; i++)

{

    //changing the value of “n”

    n = n + 1.0 / i;

}

//printing the value of i

System.out.print(i);

Explanation:

The above snippet of code is used to demonstrate “for” loop. In the code,

• An integer variable “i” is created.
• An integer variable “n” is created and assigned a value “1 / 2”.
• Iterate “for” loop up to the condition “i <= 5”. The loop variable was initialized with “i=2” and every iteration “i” is incremented by “1”.
  • The variable “n” is added with value of “n + 1.0 / i”.
• The value of “i” is printed on the screen.

Trace table for the code snippet:

i will be 6;

Expected output:

6

c. The given code:

// creating and initializing double variable “x”.

double x = 1;

// creating and initializing double variable “y”.

double y = 1;

// creating and initializing integer variable “i”.

int i = 0;

//iterate a do..while loop

do

{

    // the value of “y” is divided by “2”.

    y = y / 2;

    // the value of “x” is added with the value “x”.

    x = x + y;

    //the value of “i” is incremented by “1”...