// Math.sqrt() question

[Meg Adal]

As per API

public static double sqrt(double a)
Returns the square root of a double value.
Parameters:
a - a double value.
Returns:
the square root of a. If the argument is NaN or less than zero, the result is NaN.

So I wrote a code

  
class A {
public static void main(String s[]){
System.out.println( Math.sqrt(Double.MIN_VALUE)); 
// it is printing 2.2227587494850775E-162
System.out.println(Math.sqrt(Integer.MIN_VALUE)); 
// it gives NaN
}
}

Why it is not giving me NaN for double
Can anybody explain me or give the link which explains diff. behavior for Double's Min and Max values
Thanks in advance
Megan

[Ron Newman]

Double.MIN_VALUE is not a negative number. It is the smallest positive number representable as a double.
In my opinion, the designers of Java made a mistake in using the same symbol (MIN_VALUE) to mean two quite different things for Integer vs. Double.

[Jessica Sant]

Check out the definitions for Double.MIN_VALUE and Integer.MIN_VALUE:
static double MIN_VALUE: The smallest positive value of type double.
static int MIN_VALUE: The smallest value of type int.
So, by def'n Double.MIN_VALUE is positive, so you'll get an answer. Integer.MIN_VALUE is negative, so you'll get NaN