Exception Question.

[Amisha Shah]

Question:

The following class will print '2' when compiled and run.

class Test
{
public static int[ ] getArray() { return null; }
public static void main(String[] args)
{
int index = 1;
try
{
getArray()[index=2]++;
}
catch (Exception e){ } //empty catch
System.out.println("index = " + index);
}
}
A. true
B. flase

output of this code is 2. i thought output is 1. can anyone eplain me how program work?

[Naseem Khan]

  
class Test
{
public static int[ ] getArray() { return null; }
public static void main(String[] args)
{
int index = 1;
try
{
getArray()[index=2]++; // at line 1
}
catch (Exception e){ } //empty catch
System.out.println("index = " + index);
}
}

at line 1, the value of 2 is assigned to index.
So basically line 1 will be like this....
getArray()[2]++;
getArray method returns null. Since u r retrieving third element of int array which is null. U will get get NullPointer exception, if u print it in catch by printStackTrace() mthod.
But since ur catch is empty, it does nothing and after catch normal execution follows.... and u will get the of value of index 2.

regards

Naseem.K

[wise owen]

class Test
{
public static int[ ] getArray() { return null; }
public static void main(String[] args)
{
int index = 1;
try
{
getArray()[index=2]++;
}
catch (Exception e){ } //empty catch
System.out.println("index = " + index);
}
}
Steps:
1. getArray() called and return a reference X
2. index =2 executed and index is 2 now.
3. It runs X[2]++, an exception is thrown because X is null
4. catach clause do nothing
5. print out index

[Uma Vinodh]

Its a very clear explanation.

Thanks.