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Math.sqrt

kundan varma
Ranch Hand

Joined: Mar 08, 2004
Posts: 322
HI All
can somebody tell me what is the logic behind such output:
System.out.println("Math.sqrt( -9.0 ) \t" + Math.sqrt( -9.0 ) );
System.out.println("Math.sqrt( -0.0 ) \t" + Math.sqrt( -0.0 ) );
System.out.println("Math.sqrt( 0.0 ) \t" + Math.sqrt( 0.0 ) );

OUT PUT IS

Math.sqrt( -9.0 ) NaN
Math.sqrt( -0.0 ) -0.0 //doubt
Math.sqrt( 0.0 ) 0.0

Thanks
kundan


SCJP1.4,SCBCD,SCEA,CNA
Failures are practice shoots for success.
kundan varma
Ranch Hand

Joined: Mar 08, 2004
Posts: 322
Hi All
I got the answer from api which states
"
public static double sqrt(double a)Returns the correctly rounded positive square root of a double value. Special cases:
If the argument is NaN or less than zero, then the result is NaN.
If the argument is positive infinity, then the result is positive infinity.
If the argument is positive zero or negative zero, then the result is the same as the argument.
Otherwise, the result is the double value closest to the true mathetmatical square root of the argument value.
Parameters:
a - a double value.
Returns:
the positive square root of a. If the argument is NaN or less than zero, the result is NaN.
"
THanks
kundan
 
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