Explanation of Solution
a. The given code:
// creating variable s and initializing by “1”.
int s = 1;
// iterate the for loop up to “n<=5”.
for (int n = 1; n <= 5; n++)
{
// variable “s” is added with value of “n”
s = s + n;
//prompting the value of “s”
System.out.print(s + " ");
}
Explanation:
The above snippet of code is used to demonstrate “for” loop. In the code,
- An integer variable “s” is created and assigned a value “1”.
- Iterate “for” loop up to the condition “n <= 5”. The loop variable was initialized with “n=1” and every iteration “n” is incremented by “1”.
- The variable “s” is added with value of “n”.
- The value of “s” is printed on the screen.
Trace table for the code snippet:
s = 1 + 1 = 2
s = 2 + 2 = 4
s = 4+ 3 = 7
s = 7+ 4 = 11
s = 11 +5 = 16
Expected output:
2 4 7 11 16
b. The given code:
// creating variable s and initializing by “1”.
int s = 1;
// iterate the for loop up to “s<=10”.
for (int n = 1; s <= 10;System.out.print(s + " "))
{
//loop variable n is added with 2
n = n + 2;
/*variable “s” added with value of “n” and the value is assigned to s*/
s = s + n;
}
Explanation:
The above snippet of code is used to demonstrate “for” loop. In the code,
- An integer variable “s” is created and assigned a value “1”.
- Iterate “for” loop up to the condition “s <= 10”...
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