Java DecimalFormat class and E-notation

Question

I'm attempting to learn Java(I have novice level programming experience in other languages) and am currently reading Absolute Java 5th Edition. Everything is smooth sailing so far except for a small bit related to the DecimalFormat class as it relates to E-notation. I understand the basics but some of the logic I just can't seem to "get".

For example, the number 12345 formatted with ##0.##E0 ends up as 12.3E3 according to the book. Why did it determine that there are two digits before the decimal instead of, say, one or three? I know the # is an optional digit, but it after playing with some formatting constraints on different numbers it almost seems like the formatting is somewhat arbitrary(although I know it can't be). I've searched for a good explanation outside of the book and have come up short. If someone could "dumb it down" for me I would be really appreciative.

Also, how often is this type of formatting used in real world application?

Thanks a bunch.

Answer 1

It is the number of # in the front part of the pattern, thus if it was #0.##E0 it will be 1.234E4

DecimalFormat df = new DecimalFormat("#0.##E0");
System.out.println(df.format(12345));

The best way to learn a new language if to write little test programs like this.

Also read more on the Oracle site