Chapter 6, Problem 6RE

Explanation of Solution

a.

Given statements:

//declare the variables and assign the values

int i = 0;

int j = 10;

int n = 0;

//check "i" is less than "j"

while(i < j)

{   

    //increment the value

    i++;

    //decrement the value

    j--;

    //increment the value

    n++;

}

Trace table for the above loops:

i	j	n
0	10	0
1	9	1
2	8	2
3	7	3
4	6	4
5	5	5

Explanation:

Here, the “i” value is “0”, “j” value is “10” and “n” value is “0”.

• In first iteration of “while” loop checks “i < j” which means “0 < 10”. The condition becomes true, so it increment the “i” value, i = 1, decrement the “j” value, j = 9 and also increment the “n”, n = 1...

Explanation of Solution

b.

Given statements:

//declare the variables and assign the values

int i = 0;

int j = 0;

int n = 0;

//check "i" is less than "10"

while(i < 10)

{   

    //increment the value

    i++;

    //calculate the "n" value

    n = n + i + j;

    //increment the value

    j++;

}

Trace table for the above loops:

i	j	n
0	0	1
1	1	4
2	2	9
3	3	16
4	4	25
5	5	36
6	6	49
7	7	64
8	8	81
9	9	100

Explanation:

Here, the “i” value is 0, “j” value is 0 and “n” value is 0.

• In first iteration of “while” loop checks “i < 10” which means “0 < 10”. The condition becomes true, so it increment the “i” value, i = 1, calculate “n” value which means (0 + 1 + 0 = 1) and also increment the “j”, j = 1.
• In second iteration of “while” loop checks “i < 10” which means “1 < 10”. The condition becomes true, so it increment the “i” value, i = 2, calculate “n” value which means (1+ 2 + 1 = 4) and also increment the “j”, j = 2.
• In third iteration of “while” loop checks “i < 10” which means “2 < 10”. The condition becomes true, so it increment the “i” value, i = 3, calculate “n” value which means (4+ 3 + 2 = 9) and also increment the “j”, j = 3.
• In fourth iteration of “while” loop checks “i < 10” which means “3 < 10”...

Explanation of Solution

c.

Given statements:

//declare the variables and assign the values

int i = 10;

int j = 0;

int n = 0;

//check "i" is greater than "0"

while(i > 0)

{   

    //decrement the value

    i--;

  //increment the value

    j++;

    //calculate the "n" value

    n = n + i - j;

}

Trace table for the above loops:

i	j	n
10	0	8
9	1	14
8	2	18
7	3	20
6	4	20
5	5	18
4	6	14
3	7	8
2	8	0
1	9	-10

Explanation:

Here, the “i” value is 10, “j” value is 0 and “n” value is 0.

• In first iteration of “while” loop checks “i > 0” which means “10 > 0”. The condition becomes true, so it decrement the “i” value, i = 9, increment the “j”, j = 1 and calculate “n” value which means (0 + 9 - 1 = 8).
• In second iteration of “while” loop checks “i > 0” which means “9 > 0”. The condition becomes true, so it decrement the “i” value, i = 8, increment the “j”, j = 2 and calculate “n” value which means (8 + 8 - 2 = 14).
• In third iteration of “while” loop checks “i > 0” which means “8 > 0”. The condition becomes true, so it decrement the “i” value, i = 7, increment the “j”, j = 3 and calculate “n” value which means (14 + 7 - 3 = 18).
• In fourth iteration of “while” loop checks “i > 0” which means “7 > 0”...

Explanation of Solution

d.

Given statements:

//declare the variables and assign the values

int i = 0;

int j = 10;

int n = 0;

//check "i" is greater than "0"

while(i != j)

{   

    //increment the "i" by 2

    i = i + 2;

  //decrement the "j" by 2

    j = j – 2;

    //increment the value

    n++;

}

Trace table for the above loops:

i	j	n
0	10	0
2	8	1
4	6	2
6	4	3
8	2	4
...