如何处理 java 中的下溢？

Question

Here's the issue

double a = 1.0665901062490447E-18;
double b = 1.0;
double c = a+b; // this is 1.0

I need that summed value. What do I do about it?

Answer 1

As I mentioned in my comment , 1.0665901062490447E-18 is a very small value. If you need that much precision, you should be using BigDecimal . Something like,

double a = 1.0665901062490447E-18, b = 1.0;
System.out.println(BigDecimal.valueOf(a).add(BigDecimal.valueOf(b)).toPlainString());

Which outputs

1.0000000000000000010665901062490447

Note that there is a predefined constant BigDecimal.ONE - so the above could also be written

System.out.println(BigDecimal.valueOf(a).add(BigDecimal.ONE).toPlainString());

Answer 2

There are limits to what primitive types can do in almost any language. In this case, the double type I believe is bound by IEEE-754 . In order to handle these class of problems Arbitrary Precision Libraries have been created for many languages.

The classic "What Every Computer Scientist Should Know About Floating Point Arithmetic" is required reading for all uses of floating point data types.