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2 dimensional Array!

 
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Hi!
int[][] ar2=new int[1][3];
System.out.println(ar2[1].length);
The above given ArrayIndexOutOfBound Exception.I know that the length of array is 2.Is it not possible to take the length of each dimension?Can anybody explain please?Thanks.
 
Greenhorn
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Hi sridhar,
int[][] ar2=new int[1][3];//Line1
System.out.println(ar2[1].length);//Line2

Array Indexes Start from 0 not 1,so there is no ar2[1] since in line 1 u declared an array of int[1][3]
Line 2 can be replaced by
System.out.pritnln(ar2[0].length);//prints 3
 
Sridhar Srinivasan
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Hi!
If I say
int[][] ar2={{1,2,3},{4,5,6}};//Line1
I can say that ar2[0][1] has value 2 and ar2[1][1] has value 5.
In this case can I find the length of ar2[1]?Am I rite?R something wrong?Please explain.Thanks
 
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Hi Sridhar,
Yes. You are right in your prediction about the length but as earlier noted you probably missed that 0 and 1 indexing logic ....
I guess, if you can refer to the JLS Chapter 10 Arrays then it might make things more clear to you. I as myself didnt refer to JLS much when I took the test but many people have it almost by Heart, to remember from top of my mind -Marlene Miller name comes up...moderators would know it...
Regards
Maulin
 
Sridhar Srinivasan
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In that case, I can say that int[][] ar2=new int[2][2]
The above is similiar to int[][] ar2={{1,2},{3,4}}- rite?
If I am able to get the length of ar2[1].length of the second statement, why can't I get the length of the first one.
 
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Hi Sridhar,
I dont think u will have any problem with the first statement.
Even for tht , u can get the length of array using arr2[1].length.It prints 2.
vineela
 
Sridhar Srinivasan
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Hi Vineela!
Then, how about the following:
int[][] ar2=new int[1][3];//Line1
System.out.println(ar2[1].length);//Line2
This shouldn't give any error.rite?Thanks.
 
Vineela Devi
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Hi Sridhar,
U r trying to create a multidimensional array.It usually means an array of arrays.

int[][] ar2=new int[1][3];//Line1
System.out.println(ar2[1].length);//Line2

so, here the length of the main array is only 1and length of subarrays is 3. so, as the index starts from 0, there will be only ar[0]. ar[1] doesnot exist at all. so, u will definitely get ArrayIndexOutOfBoundsException.
when u try to create a two dimensional array, say like this:
int array[][] = new int[2][2];
This means tht the length of main array is 2 and the length of each subarray is 2.
in java, since non rectangular arrays can also be defined, u can simply omit the size of the subarrays.
i.e. int array[][] = new int[2][];
is also valid.
The length of subarrays can declared later when we use.
i.e. we can declare as array[0] = new int[2];//Length of first subarray(referred by array[0]) is 2
array[1] = new int [3];
//Length of Second subarray(referred by array[1]) is 3
Hope it is clear to u now,why u r getting the exception.

Vineela
 
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You should get an ArrayIndexOutOfBoundsException on line 2
had you written ar2[0] it would have worked. the array is 1 long, counting from 0, therefore indexing [1] is indexing the 2nd element which doesn't exist.
 
lowercase baba
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int[][] ar2=new int[1][3];//Line1
System.out.println(ar2[1].length);//Line2


when you do this, you have a two dimensional array. we all agree on that.
your 2d array contains 1 1d array. your 1d array contains 3 ints.
in other words ar2 has 1 element (which is an array) in it, at index position 0 -> ar2[0]. that is the ONLY thing in ar2.
so when you say ar2[1], you are past the end of ar2, so you get the error.
 
Sridhar Srinivasan
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Thanks a lot for all ur explaination.I understood well.
 
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