Funny question on double value

Question

hi all,
I have a rather funny question ...
how to getrid of the E from a very huge double number without using the java.math.BigDecimal or java.math.BigInteger classes.
ex:-
Double d1 = new Double("1487194731984573985734095734905873059734059734957340958734957309457389457983475");
Double d2 = new Double(d1.doubleValue() * d1.doubleValue());
d2.doubleValue() will give 7.246073001982964E147
but the result should not have the E but the complete value.
the thing is not to use java.math package.
thx
chandra

Answer 1

Well you can use java.text.DecimalFormat to format the output of a double value. However you'll find that the huge value you think you stored in the Double is not in fact what java has kept - double does not have the precision you have requested and using the DecimalFormat this will be evident... EG doing something like
double d = 123456789012345678901234567890.0;
DecimalFormat df = new DecimalFormat(
System.out.println("123456789012345678901234567890.0");
System.out.println(d);
System.out.println(df.format(d));Should clearly show that the value stored has lost precision. The reason you do not get an overflow as with an int value is that the values scale can be represented in the double not its precision.
Hope that helps...
Talden