Array index

[Heba Mahmoud]

 

public class TestArray {
	
	public static void main(String[] args) {
		
		int [] x = {1,2,3,4};
		
		for (int i : x ){
			x[i]= 0;
		}
		for (int i : x ){
			System.out.println(i);
		}
		
		
	}
	

}

i don't understand why this code ouput 0030?
 :rolleyes:
   

[Heba Mahmoud]

the source is

inquistion john meyers's SCJP 5 mock exam

but really i have a question regarding quoting the questions?

i prepared for the exam for along time and each time when i found any diffcult question for me , i copied it in some word file till i can ask about it
so i have a lot questions that i don't remember from where i had got it and what its source??

so that matter may make me don't quote some of the questions i send.

[Neha Daga]

your code executes this way:

first iteration:
i = 1, x[1] = 0, array becomes: x = (1, 0 , 3, 4)
second iteration:
i = 0, x[0] = 0, array becomes: x = (0, 0, 3, 4)
third iteration:
i = 3, x[3] = 0, array becomes: x = (0, 0, 3, 0)
fourth iteration:
i = 0, x[0] = 0, array becomes: x = (0, 0, 3, 0)

final answer: x = (0, 0, 3, 0)

[Abimaran Kugathasan]

Neha Daga, I couldn't get that point. How it is happening? Since second enhanced for loop works correctly (traversed through according to index position), so why is that for first?

[Neha Daga]

ok , I hope you know how for each loop works it iterates through elements in the loop.

when first iteration occurs it sets 'i' to the first value in array that is 1. Now the next statement sets the element at index represented by 'i' to 0 so the element at index 1 is set to 0. Now when the second iteration occurs 'i' is set to the second element that is the element at index 1 which is 0, so now the next statement will set the element at index 0 will be changed to 0. the third iteration sets 'i' to 3 value at the index 2 and hence the value at index 3 will be changed to 0. and in the last iteration the i will be set to 0 beacuase now index 3 has been set to 0 in the last iteration and hence the next statement will change the first value to 0.

[KrishnaPrasad raghavan]

This indeed is a good question and the explanation ever better.

[Abimaran Kugathasan]

Yea.. Me too.

class TestArrayIndex	{
	public static void main(String[] args)	{
		//int[] x = {1,2,3,4,5,6};

		int[] x = new int[]{1,2,8,4,5,6};

		for(int i : x)	{
			x[i] = 0;
		}

		for(int i : x)	{
			System.out.println(i);
		}
	}
}

This gives me RTE(ArrayIndexOutOfBoundsException), At line 8. Please help!

[Henry Wong]

Abimaran Kugathasan wrote:
This gives me RTE(ArrayIndexOutOfBoundsException), At line 8. Please help!

If you were able to follow the explanation for the first example, you should be able to figure out this example -- it is basically doing the same thing with different numbers.

Henry

[Ankit Garg]

@Abimaran What will happen when the first for loop will iterate for the third time?? At that time i will be 8, so the statement inside the loop will be: x[8] = 0; and that will result in an ArrayIndexOutOfBoundsException...

[Abimaran Kugathasan]

Oh! I confused.... Shit! That i refers the elements of that array, not the index. Soooo Sorry. What a stupid mistake! I start to forget all.......
Sorry heba mahmoud.
Sorry Neha Daga.
Sorry KrishnaPrasad raghavan.

I may confused All.

Thanks a lot Ankit Garg! Thanks.........