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How do You Find the Number of Places After the Decimal Point with BigDecimal?

I'm trying to use BigDecimals to calculate something and then round to 5 decimal places if it has more than that. How can I do this?
Would scale() work?

Did you already crawl through JavaDoc, especially the function precision()?
And this here might be a direct solution which makes it needless to check the precision first:
yournumber.round(new MathContext(5, HALF_UP));
And after that use stripTrailingZeros() (thx to #GregKopff)

How can I stop Java from shortening large double?

Let's suppose we have the following code:
System.out.println(String.valueOf(100000000000.0));
Now the output to that is 1.0E11. But that is not what I want. (Looks bad on a highscore)
I want it to output exactly 100000000000.0. Is there a way to do that?

Format it appropriately. For example:
System.out.printf("%.1f", 1654621658874684.0);
Be aware that double is not infinitely precise. It has a precision of about 15 to 17 decimal digits. If you need floating-point numbers with arbitrary precision, use BigDecimal instead of double.
Or you could use String.format():
System.out.println(String.format("%.0f", 1654621658874684.0d));

System.out.printf("Score: %.0f\n", 1e5); will print 100000.

Refer to this ...
Quest
You can use DecimalFormat to format your value for displaying

For those kind of big numbers, I think you should use BigDecimal.

Java - unexpected result when rounding a double/float to 2 decimals

I encounter a problem when round a double to 2 decimals. I know this questions have been asked in many places. But my question is slightly different and I cannot find it in other places.
So far as I know, there are 2 ways to do this.
Math.round(double*100.0)/100.0
DecimalFormat(“###.##”)
I am trying to use the first way: a custom way to round double.
When the second decimal is 0, the result will only print the first decimal and ignore the second one.
For example,
Math.round(1.23333*100.0)/100.0 The result is 1.23. This works good.
Math.round(3.90*100.0)/100.0. The result is 3.9. Problem occurs. I want to show 3.90 instead of 3.9
Math.round(3*100.0)/100.0. The result is 4.0. I want 4.00 instead of 4.0
So, my question is that how I can have a double value with 2 decimals no matter if the last decimal is 0 or not. I know I can use the second way- DecimalFormat(“###.##”) to achieve what I want! But is it possible to do it by using the first way?
Edit: Thanks for the answers. It looks like it is NOT possible to use the round() method to achieve it. However, some people suggest to use the combination of 2 ways to achieve it. But I think using only DecimalFormat(“###.##”) can get what I want. Can anyone confirm it?

I would suggest using String.format("%1$.2f",x). It rounds the value to the specified precision (2 digits in our example) and leaves the trailing zeros on the right.
System.out.println(String.format("%1$.2f",3.121)) gives 3.12
System.out.println(String.format("%1$.2f",3.129)) gives 3.13
System.out.println(String.format("%1$.2f",3.12)) gives 3.12
System.out.println(String.format("%1$.2f",3.10)) gives 3.10

Have you tried the following?
DecimalFormat(“###.00”)
If not mistaken, trailing zeros are left blank when using the #-sign.

I believe you need to use a combination of both to achieve your needs.
The rounding to obtain a number with two decimals, and the DecimalFormat to display it with two decimals.

You should be formatting your result with DecimalFormat
DecimalFormat format = new DecimalFormat("0.00");
System.out.println(Math.round(3.90*100.0)/100.0); // 3.9
System.out.println(format.format(Math.round(3.90*100.0)/100.0)); // after using format 3.90
System.out.println(format.format(Math.round(3*100.0)/100.0));
And the output is
3.9
3.90
3.00

Rounding off in java

I have a list that contains many decimal values.
If i get a value of .032 i have to
round it off to 3.2%
If i get a value
of 32.33 i have to round it off to
32%
If i get a value of 32.66 i have to round it off to 33%
if decimal digit is 5,6,7,8 or 9
round up to the next
how should i go bt dng this?

Use Math.round(). This correctly rounds a number upwards if the fractional part is 0.5 or greater.

If I understood you question correctly:
For the first point, you can use DecimalFormat class, see API
I'm not sure why you'd want to "round" 32.33 -> 32% as this is not rounding. If you are parsing user input data which could be given in this way, you can divide by 100 (giving 0.3233) and use DecimalFormat.
For the others you can use Math.round() as adviced below/above

Math.round() and convert it to percent form.
I doubt about .032 case. rest are looking good.

Use MathUtils (Apache Commons Math)
MathUtils.round(double, scale, roundingMethod);
scale - The number of digits to the right of the decimal point. (+/-)
roundingMethod : BigDecimal.ROUND_DOWN, #.ROUND_CEILING, #.ROUND_HALF_UP, ...

Displaying doubles to a certain precision in java

I am currently writing a calculator application. I know that a double is not the best choice for good math. Most of the functions in the application have great precision but the ones that don't get super ugly results. My solution is to show users only 12 decimals of precision. I chose 12 because my lowest precision comes from my numerical derive function.
The issue I am having is that if I multiply it by a scaler then round then divide by the scaler the precision will most likely be thrown out of whack. If I use DecimalFormat there is no way to show only 12 and have the E for scientific notation show up correctly, but not be there if it doesn’t need to be.
for example I want
1.23456789111213 to be 1.234567891112
but never
1.234567891112E0
but I also want
1.23456789111213E23 to be 1.234567891112E23
So basically I want to format the string of a number to 12 decimals places, preserving scientific notation, but not being scientific when it shouldn't

Use String.format("%.12G", doubleVariable);
That is how you use format() to display values in scientific notation, but without the scientific notation if not needed. The one caveat is that you end up with a '+' after the 'E', so yours would end up like 1.234567891112E+23

String.format("%.12d", doubleVariable);
Should give you what you are looking for in your first matter. I'm sorry but I don't know how to define when your E-notification is showed.

You'll be interested in BigDecimal, for example:
BigDecimal number = new BigDecimal("1.23456789111213");
number = number.setScale(12, RoundingMode.HALF_UP);
System.out.println(number);
Choose appropriate to you RoundingMode.