Kathy Book Question - Array References(Urgent)

[peter chong]

Q12 Chap 1 Pg49
public class test{
public static void main(String []args){
byte [][] big = new byte[7][7];
byte [][]b=new byte[2][1];
byte b3=5;
byte b2[][][][]=new byte [2][3][1][2];
//line of code to be inserted.Answer is A,B,E,F
}
}
A.b2[0][1]=b;
B.b[0][0]=b3;
C.b2[1][1][0]=b[0][0];
D.b2[1][2][0]=b;
E.b2[0][1][0][0]=b[0][0];
F.b2[0][1]=big;
I have go through the answer 's explaination,
but can't understand how to get the solution.
Apart from that,I have also refer to diagram in Fig 1-7 Pg31,but still confused.
Can someone please enlighten me,many thanks.

[Barry Gaunt]

A little trick I use:
Consider the array byte[][][][] b2; and look at A) b2[0][1]. We have only used two indices [][] from the four [][][][] that have been declared, that leaves two [][] left over. So to b2[0][1] we can assign a byte[][]. Now b is a byte[][], it's declared as one, so b2[0][1] = b; is valid.
Look now at C) b2[1][1][0]. We have used three indices [][][] from the four [][][][] that have been declared, and that leaves one [] left over. So to b2[1][1][0] we can assign a byte[]. But b[0][0] is a primitive byte not an array byte[], so we cannot assign it to b2[1][1][0]
Similar arguments for the rest.
[ March 25, 2003: Message edited by: Barry Gaunt ]

[Cor Lieftink]

Hi,
I have a remark about this same question. I am confused that answer F is correct.
B2 is assigned an created array of the dimensions [2][3][1][2]. I would think that in the second dimension you can put only references to arrays of dimensions [1][2]. I would think that corresponds with the amount of memory that is allocated on the heap. How can answer F be good, if one puts in the element b2[0][1] a reference to a [7][7] array?
Thanks for the answer in advance!
Cor
Update: tested the code
I tested the code and it was possible to do it.

  
public static void main(String[] args) {
byte [][] big = new byte [7][7];
big[6][6]=9;
byte b2 [][][][] = new byte [2][3][1][2];
b2[0][1] = big;

byte testByte = b2[0][1][6][6];
System.out.println(Byte.toString(testByte)); // produces 9
byte testByte2 = b2[0][2][6][6];
System.out.println(Byte.toString(testByte2)); //produces ArrayIndexOutOfBoundsException. 
 
}

I added a testByte2 to see if the overall structure of the array is changed, but that is not the case. So apparently the original dimensions do not impose limits on the way the elements of the array are filled in with existing, and therefore already memory-allocated, arrays. The only limit is the number of dimensions. If you increase for example the number of dimensions of big from 2 to 3, you get a compiler error "cannot convert from byte[][][] to byte [][]".
[ January 20, 2004: Message edited by: Cor Lieftink ]

[sanjana narayanan]

Hi Berry,

Your explanation was very good. THis is the only area i was not comfortable. After going thru u'r explanation , it was crystal clear.
Thanks,
Sanjana