Why not ArithmeticException thrown?

[sandhi mridul]

Hi,

class T1{
static void test(int i){
int j=i/2;  //line1
int k=i>>>1;
assert j==k:i;
}
public static void main(String[] args) 
{
test(0);//line 2
test(2);
test(-2);
test(1001);
test(-1001);
 
}}

ans=An assertionEROR with -2 as message is thrown.

My question is when line 2 is called ,why its not throwing assertion error at line 1?

Thanks
Sandhya
[ March 13, 2008: Message edited by: sandhi mridul ]

[Sheryl Pinto]

Hi Sandhi,

You will get an ArithmeticException if you try to divide by 0 at line 1.
0/2 = 0, no ArithmeticException.

Thanks,
Sheryl.
[ March 13, 2008: Message edited by: Sheryl Pinto ]

[Jelle Klap]

When line 2 is executed and T1.test() is called with int value 0 the first statement attemps to divide 0 by 2, which is perfectly legal.
Dividing 0 by an arbitry value is legal and will always result in 0.
Dividing an arbitrary value by 0 is not legal and will always result in an java.lang.ArithmeticException being thrown.
I believe you have the two mixed up.

Edit: Sheryl beat me to it
[ March 13, 2008: Message edited by: Jelle Klap ]

[sandhi mridul]

Ohh yaa.
sorry for overlooking.
Thanks for all your reply.