Doubt In foreach Loop

[srinibash udayasingh]

Hello Ranchers
can any one explain the following code snipets

int []arr = {1,2,3,4};
for ( int i : arr )
{
arr[i] = 0;
}
for ( int i : arr )
{
System.out.println(i);
}

out put here is :0030

[Abdullah Mamun]

int []arr = {1,2,3,4};
for ( int i : arr )
{
    arr[i] = 0;
}

After executing this code the arr[] will contain {0,0,3,0}
How??? -- lets see

executing for loop and retrieving the first element which is 1 or i = 1
set arr[1] = 0;
now arr = {1,0,3,4} // Look the second element became 0 now

looping again and retrieving second element which is 0 or i = 0
set arr[0] = 0;
now arr = {0,0,3,4}

looping again and retrieving third element which is 3 or i = 3
set arr[3] = 0; // Look the fourth element became 0 now
now arr = {0,0,3,0}

looping again and retrieving fourth and last element which is 0 or i = 0
set arr[0] = 0;
now arr = {0,0,3,0}

So the final array is {0,0,3,0} and thats why you are getting this output.

Hope this helps
[ August 22, 2007: Message edited by: Al Mamun ]

[srinibash udayasingh]

Thank You very Much for your good explanation.Now my doubt cleared

[Mohammed shareef]

Instead Of Anonymous array use normal array declaration &initialization,u will get expected output.There is some mystery with enhance for with anonymous array & if we increase length of Anonymous array to 5 it is giving ArrayIndexOutOfBound Exception .Let's wait &see who willl solve this mystery.

[Mohammed shareef]

Excellent Al Mamun ,but you have not explain why control is going to i+1 position,then coming back to i position.

[Manfred Klug]

Originally posted by Mohammed shareef:
if we increase length of Anonymous array to 5 it is giving ArrayIndexOutOfBound Exception .Let's wait &see who willl solve this mystery.

There is no mystery about the exception. Just follow the explanation from Al Mamun and add the following:

looping again and retrieving fifth element which is ... or i = ...
set arr[...] = 0;

I think it should be clear why the ArrayIndexOutOfBoundException occurs.

[Abdullah Mamun]

Originally posted by Mohammed shareef:

Excellent Al Mamun ,but you have not explain why control is going to i+1 position,then coming back to i position.

Actually, there is no pattern or any other thing like going to i+1 or coming back to i position in this example. The for loop is traversing through the array and the value of each index of the array is determining the value of i.

[Rishiraj Bayerd]

Hi Al Mamun,
Excellent explanation. Simply superb� if we take int [] arr={10,20,30,40} instead of int [] arr={1,2,3,4} we got ArrayOutofBound exception , because at for(i:arr), the value of is 10 i.e value of first element(value 0th place) and that becomes use as index of arr at line arr[i]=0;

with regards
Rishi

[Amitji Sharma]

Thanks Al Mamun
for such a Good Explanation.

[Neha Bhattad]

Thanks Al Mamun.

Instead Of Anonymous array use normal array declaration &initialization,u will get expected output.

My question is: Why would the answer be any diffent if it wasnt an Anonymous array? I tried and im getting the same response 0030.

int [] arr1 = new int[4];
arr1[0] = 1;
arr1[1] = 2;
arr1[2] = 3;
arr1[3] = 4;
for(int i: arr1)
arr1[i] = 0;
for(int i:arr1)
System.out.println(i);

[Rishiraj Bayerd]

Hi Neha,
i think it does not depends on array type(i.e. Anonymous,Simple). All mistry hide at first foreach loop. I try to draw the picture.

Original code:
for(int i: arr1) arr1[i] = 0;

int i:arr1 means the value of the i is the first element of arr1(0th place of arr1) i.e. the value of the i is 1.

For fisrt time... Then arr1[i]=0 becomes arr1[1]=0 .
So,(Original array) {1,2,3,4}=>{1,0,3,4}

Again for second time...
The value of the i is the second element of arr1(1st place of arr1) i.e. the value of the i is 0. Then arr1[i]=0 becomes arr1[0]=0
So,(Original array) {1,0,3,4}=>{0,0,3,4}


Again for Third time...
The value of the i is the third element of arr1(2nd place of arr1) i.e. the value of the i is 3. Then arr1[i]=0 becomes arr1[3]=0
So,(Original array) {0,0,3,4}=>{0,0,3,0}

Again for last time...(length of arr1 is 4)
The value of the i is the fourth element of arr1(3rd place of arr1) i.e. the value of the i is 0. Then arr1[i]=0 becomes arr1[0]=0 So,(Original array) {0,0,3,0}=>{0,0,3,0}

So the final output will be 0030

thanks....

with regards

Rishi

[Abdullah Mamun]

Hi Neha

Originally posted by Mohammed shareef:
...Instead Of Anonymous array use normal array declaration &initialization,u will get expected output....

This statement is not correct here. The output is totally depends on the first for-each loop and the assignment inside the loop. So don't worry