Question on multi dimensional array

[Veena Pointi]

The following code

class arrconv{
public static void main(String args[]){
byte b2[][][][]=new byte[2][3][1][2];
//int b4[][][][]={{{{1,2}},{{3,4}},{{5,6}}},{{{7,8}},{{9,10}},{{11,12}},{{13,14}}}};
byte b[][]={{1,2},{3,4}};
b2[0][0]=b;
for(int i=0;i<2;i++)
for(int j=0;j<3;j++)
for(int k=0;k<1;k++)
for(int l=0;l<2;l++)
System.out.println("b2["+i+"]["+j+"]["+k+"]["+l+"] "+b2[i][j][k][l]);
Object[][] obj = (Object[][])b2.clone();
System.out.println(obj.length);
      for(int x = 0;x<1; x++) {
 for(int y=0;y<2;y++){
        byte[][] ia = (byte[][])obj[x][y];
        System.out.println("ia["+x+"]["+y+"]"+ia[x][y]);
}
} 
}
}

The o/p when u compile the above code is

b2[0][0][0][0] 1
b2[0][0][0][1] 2
b2[0][1][0][0] 0
b2[0][1][0][1] 0
b2[0][2][0][0] 0
b2[0][2][0][1] 0
b2[1][0][0][0] 0
b2[1][0][0][1] 0
b2[1][1][0][0] 0
b2[1][1][0][1] 0
b2[1][2][0][0] 0
b2[1][2][0][1] 0
2
ia[0][0]1
ia[0][1]0

In the above code ,the line b2[0][0]=b; assigns byte values 1,2 of b[][] array to b2[][][][] array.
But I wanna assign byte values 3,4 of b[][] array to b2[][][][] array.I am unable to do it....Can anybody try the code & tell me how to do it?
Thanks
Veena

[Dan Chisholm]

Veena,
I'm not sure that I understand your question, but it sounds like you are saying that you can not access all of the elements of array b after it is assigned to array b2. In other words, it sounds like you can not access all of the elements of array b using the reference b2. I have not tried to run the code, but I think that your declaration of b2 is the problem. While the dimensions of array b are 2 X 2, the dimensions of array b2 are 2 X 3 X 1 X 2. You might have more luck accessing b[1][0] and b[1][2] using the reference b2 if array b2 were declared as a 2 X 3 X 2 X 2 array.
I hope that I correctly understood your question.

[Veena Pointi]

Dan,
I am sorry if my question was not clear.In the above code byte values 1,2 of b array are assigned to b2 array.What should I do to assign byte values 3,4 of b array to b2 array.The above code produces following o/p.
b2[0][0][0][0] 1
b2[0][0][0][1] 2
b2[0][1][0][0] 0
b2[0][1][0][1] 0
b2[0][2][0][0] 0
b2[0][2][0][1] 0
b2[1][0][0][0] 0
b2[1][0][0][1] 0
b2[1][1][0][0] 0
b2[1][1][0][1] 0
b2[1][2][0][0] 0
b2[1][2][0][1] 0
2
ia[0][0]1
ia[0][1]0
What changes I should make in the above code to get the following o/p
b2[0][0][0][0] 3
b2[0][0][0][1] 4
b2[0][1][0][0] 0
b2[0][1][0][1] 0
b2[0][2][0][0] 0
b2[0][2][0][1] 0
b2[1][0][0][0] 0
b2[1][0][0][1] 0
b2[1][1][0][0] 0
b2[1][1][0][1] 0
b2[1][2][0][0] 0
b2[1][2][0][1] 0
2
ia[0][0]1
ia[0][1]0
I hope I made my doubt clear.

[Dan Chisholm]

Originally posted by Veena Point:

What changes I should make in the above code to get the following o/p
b2[0][0][0][0] 3
b2[0][0][0][1] 4

I have not tried to run the code, but my assumptions are as follows.
The values 3 and 4 are stored at
b2[0][0][1][0] 3
b2[0][0][1][1] 4
Since you have declared b2 as
byte b2[][][][]=new byte[2][3][1][2];
you won't have access to the elements where 3 and 4 are stored. You could solve the access problem by defining b2 as follows.
byte b2[][][][]=new byte[2][3][2][2];
If you want the declaration of b2 to remain unchanged and if you don't need access to the first two elements of b then you could try the following new assignment statement.
b2[0][0][0]=b[1];
As I mentioned earlier, I have not tried the code but I think that it will work.
I wish you luck.

[Veena Pointi]

Originally posted by Dan Chisholm:

If you want the declaration of b2 to remain unchanged and if you don't need access to the first two elements of b then you could try the following new assignment statement.
b2[0][0][0]=b[1];

Wow!It worked out.Thank you Dan.You really think very deep.This multi dimensional array problem was bugging me from many days.Now it is clear.Thanks alot.
Veena