String Huge Decimal Numbers In Java

Question

double a=816992306.6297043221;
System.out.println("a="+a);

The problem with this is that the number is too large for double, and it gives the following output:

a=8.169923062970433E7

When I decrease the number, the result gets displayed correctly. What is the efficient solution for this problem?

The number is not too large for double. It's just shown in the scientific notation. There is nothing wrong with that, it's intended behavior. — Ben, May 30 '18 at 12:28
Possible duplicate of [Double vs. BigDecimal?](https://stackoverflow.com/questions/3413448/double-vs-bigdecimal) — Patrick Parker, May 30 '18 at 12:28
With the answer by Davide Lorenzo MARINO in mind it is indeed correct though that you are loosing precision on the last 3 decimal places. — Ben, May 30 '18 at 12:32

Davide Lorenzo MARINO · Accepted Answer · 2018-05-30T12:30:22.637

2

You can use a BigDecimal number instead of double:

Immutable, arbitrary-precision signed decimal numbers.

BigDecimal can represent any number with the choosen precision.

Double has limits on maximum and minimum values and on precision.

edited May 30 '18 at 12:30

answered May 30 '18 at 12:28

Davide Lorenzo MARINO

26,420
4
39
56

score 0 · Answer 2 · answered May 30 '18 at 12:34

I think it is a precision problem.

https://docs.oracle.com/javase/1.5.0/docs/api/java/util/Formatter.html#syntax

You can check this link out for learning about format.

You can use DecimalFormat Class

double d1 = 3.14159;
double d2 = 1.235;

DecimalFormat format = new DecimalFormat("#.##");

double roundedD1 = format.format(d1); // 3.14
double roundedD2 = format.format(d2); // 1.24

If you want to set the precision during runtime.

format.setMaximumFractionDigits(precision);

String Huge Decimal Numbers In Java

2 Answers2