finding values in two dimensional loop

[Jason Attin]

Hi guys, I'm reading the OCA book and there is an example that I'm not really understanding. I run the code myself, it works but I still have trouble understanding the whole logic. So, here is the code:
package loops;

public class loop_test {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		int[][] list = {{1,13,5},{1,2,5},{2,7,2}};
		int searchValue = 2;
		int positionX = -1;
		int positionY = -1;
		PARENT_LOOP: for(int i=0; i<list.length; i++){
			System.out.println("list.length is " + list.length + " at index " + i);
			for(int j=0; j<list[i].length; j++){
				System.out.println("list[i].length is  " + list[i].length + " at index " + j);
				//SEARCHING VALUE FINDS THE FIRST MATCHING ONE AND COMES OUT OF BOTH LOOPS
				/*if(list[i][j] == searchValue){
					System.out.println( " i is " + i + " j is " + j + " list[i][j] " + list[i][j]);
					positionX = i;
					positionY = j;
					break PARENT_LOOP;
				}*/
				//SEARCHING VALUE FINDS THE SECOND MATCHING AND COMES OUT OF BOTH LOOPS
				if(list[i][j] == searchValue){
					System.out.println( " i is " + i + " j is " + j + " list[i][j] " + list[i][j]);
					positionX = i;
					positionY = j;
					break;
				}
				//SEARCHING VALUE FINDS THE LAST MATCHING ONE AND COMES OUT OF BOTH LOOPS
				/*if(list[i][j] == searchValue){
					System.out.println( " i is " + i + " j is " + j + " list[i][j] " + list[i][j]);
					positionX = i;
					positionY = j;					
				}*/
				
			}
		}
		if(positionX == -1 || positionY == -1){
			System.out.println("Value " + searchValue + " not found");
			
		}
		else{
			System.out.println("Value " + searchValue + " found at: " + positionX + " " + positionY );
		}

	}

}

So, let me explain. The commented out code

//SEARCHING VALUE FINDS THE FIRST MATCHING ONE AND COMES OUT OF BOTH LOOPS
				/*if(list[i][j] == searchValue){
					System.out.println( " i is " + i + " j is " + j + " list[i][j] " + list[i][j]);
					positionX = i;
					positionY = j;
					break PARENT_LOOP;
				}*/

finds the first matching and prints out the result: 1,1. OK that's clear.
The second matching snippet instead:

if(list[i][j] == searchValue){
					System.out.println( " i is " + i + " j is " + j + " list[i][j] " + list[i][j]);
					positionX = i;
					positionY = j;
					break;
				}

prints the second matching 2,0. Here I would have expected it to print the last matching instead, 2,2. So what I don't get is why it prints the second matching and comes out of both loops. Here the break statement should allow the application to come out of the inner loop only not the outer. What is it that makes it break from the outer loop as well?

Third snippet:

//SEARCHING VALUE FINDS THE LAST MATCHING ONE AND COMES OUT OF BOTH LOOPS
				/*if(list[i][j] == searchValue){
					System.out.println( " i is " + i + " j is " + j + " list[i][j] " + list[i][j]);
					positionX = i;
					positionY = j;					
				}*/

I think this is clear,  no break statement the loop continue executing. ANy idea?
thanks

[Henry Wong]

Jason Attin wrote:
The second matching snippet instead:

if(list[i][j] == searchValue){
					System.out.println( " i is " + i + " j is " + j + " list[i][j] " + list[i][j]);
					positionX = i;
					positionY = j;
					break;
				}

prints the second matching 2,0. Here I would have expected it to print the last matching instead, 2,2. So what I don't get is why it prints the second matching and comes out of both loops. Here the break statement should allow the application to come out of the inner loop only not the outer. What is it that makes it break from the outer loop as well?

Basically, the code finds the first one... just like in the first snippet. However, since the break statement doesn't break out of both loops (it just breaks out of the inner one), the code continues (with the next outer loop element) and finds the second matching case. And again, the break statement doesn't break out of both loops (it just breaks out of the inner one). However, in this second case, the outer loop is already at the last element. So, the outer loops merely completes, and the code after the loop reports that it found the second matching case.

Henry

[Jason Attin]

Right, thanks. Just so I understand it correctly though...when the value 2 in position 2,0 is found and the break executes, control goes back to the outer loop but the value of i is still 2 and i<list.length is true. Does i++ gets executed as soon as control comes back to the outer loop? that's the only way I can explain i<list.length being false. Hope that makes sense

[Henry Wong]

Jason Attin wrote:when the value 2 in position 2,0 is found and the break executes, control goes back to the outer loop but the value of i is still 2 and i<list.length is true. Does i++ gets executed as soon as control comes back to the outer loop? that's the only way I can explain i<list.length being false. Hope that makes sense

It doesn't matter whether the "i<list.length" expression is true or false, upon completing the inner loop. Expressions are only checked at the top of the loop, and only after initialization (i=0) or reinitialization (i++) has executed. So, the inner loop completes, then the code runs to complete the current iteration of the outer loop. Upon that, the reinitialization code is executed, and then the expression is checked, to see if another iteration should run.

Henry

[Jason Attin]

OK thanks, I think I understand now