Question

Give the output of the following program snippets
Rewrite the program snippets in to while loop
for(j=2;j<=20;j=j+2)
{
if (j==10)
break;
System.out.println(" " +j);
}​

Answer 1

$\textsf{\large{\underline{Corrected C{o}de}:}}$

for(int j=2;j<=20;j=j+2){

   if (j==10)

       break;

   System.out.println(" "+j);

}

After converting it into while loop:

int j=2;

while(j<=20){

   if(j==10)

       break;

   System.out.println(" "+j);

   j=j+2;

}

$\textsf{\large{\underline{O{u}tput}:}}$

2

4

6

8

$\textsf{\large{\underline{Explanation}:}}$

Let us dry run our program so as to predict the output.

When j = 2:

> j == 10 returns false.

> if block is not executed.

2 is printed and a new line is inserted.

The value of j is incremented by 2.

——————————————————————————

When j = 4:

> j == 10 returns false.

> if block is not executed.

4 is printed and a new line is inserted.

The value of j is incremented by 2.

——————————————————————————

When j = 6:

> j == 10 returns false.

> if block is not executed.

6 is printed and a new line is inserted.

The value of j in incremented by 2.

——————————————————————————

When j = 8:

> j == 10 returns false.

> if block is not executed.

8 is printed and a new line is inserted.

The value of j in incremented by 2.

——————————————————————————

When j = 10:

> j == 10 returns true.

> if block is executed.

> loop is terminated using break statement.

——————————————————————————

So, the final output of the program is:

2

4

6

8

See the attachment for verification.

Answer 2

Answer:

you can see it's JavaDoc. it's variability

is based on method overloading.

(Multiple methods with the same

name, but with different parameters)

This print stream is standing output

so called standard output.