Switch statement inside a nested loop. Dealing with multi dimensional array

Question

import java.util.Scanner;

public class Main {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int x[][] = {
{1, 3, 5, 7, 9},
{2, 4, 6, 8, 0},
{12, 23, 34, 45, 56}
};
for (int i = 0; i <= 2; i++) {
for (int j = 0; j <= 4; j++) {
switch (i) {

case 0:
System.out.print(x[i][j] + "\t");
if (j == 4) {
System.out.println();
}
break;
case 1:
System.out.print(x[i][j] + "\t");
if (j == 4) {
System.out.println();
}
break;
}//endSwitch
}
}//end i
int a[] = new int[2];
System.out.print("\nColumn (Horizontal): ");
a[0] = input.nextInt();
System.out.print("Row (Vertical): ");
a[1] = input.nextInt();
System.out.println("Number: " + x[a[0]][a[1]]);
}
}

Can someone explain on why does every time i remove the "break" on case's body, the output will be doubled?

Like, if i remove the break in "case 0" the out put will be 1 1 3 3 5 5 7 7 9 9 instead of 1 3 5 7 9.

thank you.

Answer 1

Kakai

Since there is no break in case 0 so your code will execute like this

System.out.print(x[i][j] + "\t");
System.out.print(x[i][j] + "\t");

if (j == 4) {
System.out.println();
}

break;

So according to this

1 1 then break
3 3 then break
5 5 then break

And so on.

Answer 2

Kakai

Break use to stop program there and move on for next checking.

In switch case if you don't use break then next case will execute by default. If you don't use break in case 0 then case 1 will execute and so on.

Answer 3

Kakai

It's working fine. What you get error?

Answer 4

Kakai

No error but because of no break in case 0 output would be different

https://code.sololearn.com/cojmY3INDUBF/?ref=app

Answer 5

Kakai

Just take a paper and pen and see why this happens.

Answer 6

Do you know why we use break statement in switch ..?
Break statement is for control the flow of execution.

if you remove break statement at case 0 , then it doesn't stop, it continuing execution remaining statements until break appears.

Answer 7

Kakai Answer for your question why it's printing double times means...
You write exactly same code in case 0 and case 1 to read array elements. So , when you didn't use break in case 0 then it will execute case1 too ..which has same code as case 0.
it's printed whatever case 0 printed.

Answer 8

Kakai I am girl. by the way pleasure to help you.

Answer 9

Navya Sri and I Am AJ ! Thank you guys, ❤️. I got it now. Been tryin' to solve this for 1h hahaha.

Answer 10

Anyways, i got what i want to create but i just want to understand that kind of error. Thank you.

Answer 11

Yeah i know but why does when i remove the break in case's 0 the out put will be doubled? I want to understand that badly.
Out put with break (fine) : 1 3 5 7 9
Out put without break: 1 1 3 3 5 5 7 7 9 9

Answer 12

Try to run it without break in case 0, it'll get some logical errors.

Answer 13

I Am AJ ! Look at the out put, you got 1 1 3 3 5 5 7 7 9 9. I want to understand that. Supposed to be if it will run two times, u must get 1 3 5 7 9 1 3 5 7 9. But in that case, why 1 1 3 3 5 5 7 7 9 9?