Java DecimalFormat省略小数点前的前导零

Question

Let's take below code snippet

double f = .123;        
System.out.println(f);
System.out.println(new DecimalFormat(".###").format(f));

So we can use .### to omit the leading zero 0, though we have to define the trailing digits after decimal point/period.

I want to define the DecimalFormat formatter that can serve unlimitted number of after-decimal-point digits.

How can I do that?

ps

I did try "." but that we print no decimal digits. May we have some formation syntax like ".#*" ?

Answer 1

Double is a finite approximation (64 bits including sign and exponent) so unlimited and exactness cannot be done with that.

Use a BigDecimal. A String constructor will take its precision by the string.

BigDecimal n = new BigDecimal("0.1234567890123456789012345678901234567890");
n = n.multiply(n); // n now has double precision.
String s = n.toPlainString().replaceFirst("^0\\.", "."); // Without exponent.

As you want unlimited representation simply use toPlainString . About dropping the zero before the period can be done a bit ugly with a regex replace.

Answer 2

You can turn the number to string and skip the first character:

double f = .123;        
System.out.println(Double.toString(f).substring(1));

After a bit of researching, I don't believe that this can be achieved by using the DecimalFormat . However, ".#*" , that you suggest could be a useful addition.

Answer 3

Since float has no unlimitted number of after-dicimal-points just use as many # as you expect at maximum.

You can also convert the number to a String and select a substring to display.

float f = 0.123;
String fs = Float.toString(f);
System.out.println(fs.substring(fs.indexOf('.'), fs.lenght());

This also works, if you don't know in advance, if you number has only one leading zero.

Unfortunately I can not tell you if this is also possible with DecimalFormatter.