ArithmeticException in doubles

[javi cervera]

This code :
try {
double a = 10/0;
}
catch(ArithmeticException e)
{
System.out.println("catch");
}
print catch
but this code:

[javi cervera]

this code:
try {
double a = 10.0/0.0;
}
catch(ArithmeticException e)
{
System.out.println("catch");
}
no print catch
can anybody explain me please ?
thanks in advanced

[Ron Newman]

Floating-point division by 0 doesn't cause an ArithmeticException. It produces an Infinity constant (or a NaN, if you divide 0 by 0).

[Thomas Paul]

And just to make it clear:
float f = 10/0; //is integer math even though the result is being moved to a float.
For example:
float f = 10/20; //f equals 0.0 because 10/20 = 0 in integer math

[Rahul Phadnis]

Try this to make it absolutely clear

 
public class TryMe {
public static void main(String[] args) {
try {
double a = 10/0;
}
catch(ArithmeticException e)
{
System.out.println("integer arithematic catch");
}

try {
double B = 10.0/0.0;
}
catch(ArithmeticException e)
{
System.out.println("double arithematic catch");

}

try {
int i =100;
int j =0;
double C = i/j;
}
catch(ArithmeticException e)
{
System.out.println("integer variable arithematic catch");

}



}
}
 

[Corey McGlone]

wasarta,
Welcome to Javaranch
We'd like you to read the Javaranch Naming Policy and change your publicly displayed name (change it here) to comply with our unique rule. Thank you.

[Ron Newman]

Since "10/0" is an integer division that produces a compile-time constant, why doesn't the compiler complain?