Floating Point Number and dividebyzero arithmatic exception

[veer jain]

Why does the Floating Point numbers does NOT produce the dividebyzero arithmatic exception ??

    
double x;
x = 24.0/0;

The above code runs fine, But Why ?? Why is it designed like that ?? It should through an exception...

Any Thgoughts ?

Regards
Veer

[Jelle Klap]

Because IEEE 754 specifies that division by zero for floating point numbers should result in either positive or negative infinity depending on the sign bit.

[Gitesh Ramchandani]

* if the value of the divisor in floating-point division is 0 no exception is thrown; the value of the results are as follows:
o division of a positive floating-point value: POSITIVE_INFINITY
o division of a negative floating-point value: NEGATIVE_INFINITY
o division of a floating-point value by -0: POSITIVE_INFINITY

    10.34 /  0           // result:  Infinity
   -10.34 /  0           // result: -Infinity    
    10.34 / -0           // result:  Infinity
     0    /  0           // result:  NaN (Not a number) 

[veer jain]

Thanks a ton Guys !!