Parsing a double from a string which holds a value greater than Double.MaxValue

Question

Consider the following java code:

String toParse = \"1.7976931348623157E308\"; //max value of a double in java        
double parsed = Double.parseDouble(toP         


        

Answer 1

The SE 7 version of the parseDouble documentation refers to the valueOf documentation which says:

Note that the round-to-nearest rule also implies overflow and underflow behaviour; if the exact value of s is large enough in magnitude (greater than or equal to (MAX_VALUE + ulp(MAX_VALUE)/2), rounding to double will result in an infinity and if the exact value of s is small enough in magnitude (less than or equal to MIN_VALUE/2), rounding to float will result in a zero.

This is consistent with the statement that the rounding to type double is by the usual round-to-nearest rule of IEEE 754 floating-point arithmetic.

You have to imagine the conversion being done by first calculating the nearest floating point number ignoring the exponent limitation, and then checking whether the exponent fits. Double.MAX_VALUE is the closest number under that rule to some numbers that are strictly greater than it.

To confirm this is normal rounding behavior, consider the following program:

    public class Test {
      public static void main(String[] args) {
        double ulp = Math.ulp(Double.MAX_VALUE);
        System.out.println(ulp);
        System.out.println(Double.MAX_VALUE);
        System.out.println(Double.MAX_VALUE+ulp/2.0000000001);
        System.out.println(Double.MAX_VALUE+ulp/2);
      }
    }

It outputs:

1.9958403095347198E292
1.7976931348623157E308
1.7976931348623157E308
Infinity

Adding something even slightly less than half a ulp to Double.MAX_VALUE does not change it. Adding half a ulp overflows to infinity.