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Java, Round double to two decimal (round to 0.05)

I would like to round a double in java. I would like it rounded like this:
24.04 should give 24.05
24.07 should give 24.05
24.02 = 24.00
24.08 = 24.10
I try to use round like this:
amount = amount *10;
amount = Math.round(amount)/10;
But, like you can see, it's not work.
Thank you for the answers!

Math.round(amount * 20) / 20f is one way.

Please not that rounding using float can generate errors due to the internal binary representation of float (same for double).
So an operation like Math.round(amount * 20) / 20f for an amount 23.01 can give as result 23.000000003 (is an example I didn't checked it, but is simple to find a problem using few tests).
Use BigDecimal insteads.

I would do it like this:
double[] amounts = new double[]{24.04, 24.07, 24.02, 24.08};
for(double d : amounts){
double d2 = Math.round(d * 20) / 20.0;
System.out.println(String.format("%.2f -> %.2f",d,d2));
}
Result is this:
24,04 -> 24,05
24,07 -> 24,05
24,02 -> 24,00
24,08 -> 24,10
[EDIT]
As other guys suggested there might be some precision error. It most likely will never be visible if you print it as suggested in my example. But it depends what do you need the number for....

How to round to the nearest 0.05? [duplicate]

This question already has an answer here:
Java rounding to nearest 0.05
(1 answer)
Closed 8 years ago.
I'm trying to find a way on how to round to the nearest 0.05 in java. Let's say that I have the following numbers:
0.33
0.02
0.874
0.876
This should become:
0.35
0.00
0.85
0.90
I tried many things and I can only get it to round to n places behind the comma by using BigDecimal, but I can't seem to find a way for this one.
Can someone help me?
EDIT: Thank you for all your help, I am amazed at how easy this could be done. And how do I get the double converted into a string properly? I can't use Double.toString(double d) because for example the string will be "0.9" instead of "0.90"?

0.05 == 1/20, right? Therefore, what you need is just the nearest number with dividing by 1/20, so, you may multiply this number by 20, get the nearest number with dividing by 1, then get the initial things.
TL;DR: you just may just multiply it by 20, round and divide by 20 again:
public double customRound(double num) {
return Math.round(num * 20) / 20.0;
}

A simple way would be:
double d = 0.33;
double roundedTimes20 = Math.round(d * 20);
double rounded = roundedTimes20 / 20; //0.35
but note that the resulting double is not necessarily the exact representation of the rounded number (usual floating point caveat) and that the method assumes that your original double times 20 can fit in a long.

Try a function:
public static double round05(double num) {
return Math.round(num * 20) / 20.0;
}

You can use String.format to format value to String
String s = String.format("%.2f", 0.9);

- Rounding a double up & removing the numbers after a decimal point -

Could someone tell me how I can round a double & then remove the numbers after the decimal places - & then convert this number to a String please?
e.g. start with a double value 55.6666666666667 - round it up to a double value of 56.0000000000 -
or start with 55.333333333333 - round down to a double value of 55.0000000000 -
& then remove the decimal point and trailing zeros - convert to String.
Thanks.

The best way to round a value to the nearest integer is:
int x = (int)Math.round(55.6666666666667);
x will be 56. You can also use Math.ceil() and Math.floor() to round up or down respectively. Finally to make x a string use String.valueOf(). Like this:
String xString = String.valueOf(x);
If you wanted to do it all on one line:
String xString = String.valueOf(Math.round(55.6666666666667));
You can read more about the Math class here and the String class here.

Use the Math.round() function. You can cast it to an int to eliminate the decimal places, or you can use Math.floor() or Math.ceil() to round it down or up before casting.

Assuming that you don't actually want to save the rounded value, but just want to see it as a string, this should do everything you want, where x is the double:
String s = String.format("%.0f", x);

Converting a percent, to a decimal?...using for volume control

This seems BEYOND easy but I cant see to find a clear answer...
Say I have a string (volume) that has a percentage...(80, 50, etc), I want to get the number in DECIMAL format, and then use that to figure out the volume to set the phone at. I already get the maxVolume...so it's just getting the correct value that's killing me...
Integer theVolume=Integer.parseInt(volume);
double decimalNumber = theVolume/100;
int calc=(int) (maxVolume*decimalNumber);
mgr.setStreamVolume(AudioManager.STREAM_MUSIC,calc, 0);
The decimalNumber is only coming out as 0.0...no matter what number I give it...?

Try doing it like this:
double decimalNumber = theVolume / 100.0;
The .0 at the end converts it to a decimal number.
edit: And for reference, it behaves like this because Java is trying to divide two integers (when using 100) so it tries to output the result as an integer. The result is a fraction so Java rounds it down by default to the closest integer to that fraction (0 in this case) and then assigns that integer value to your double. By instead using 100.0, you are telling Java to divide the integer by a double, which is then knows it can keep the fraction result and not round down.

Parse the volume as a double
double theVolume=Double.parseDouble(volume);
double decimalNumber = theVolume/100;
int calc=(int) (maxVolume*decimalNumber);
mgr.setStreamVolume(AudioManager.STREAM_MUSIC,calc, 0);

unexpected results with decimal arithmetic expression

I have some business logic with arithmetic expression and their results are as follows
(10.0 * 20.58)/1.0=205.7999..98
(6.0 * 37.9)/1.0=227.3999..98
(5.0 * 50.0)/1.0=250.0
(10.0 * 37.9)/1.0=379.0
But expected results are
(10.0 * 20.58)/1.0=205.8
(6.0 * 37.9)/1.0=227.4
(5.0 * 50.0)/1.0=250.0
(10.0 * 37.9)/1.0=379.0
I am not clear why we are getting that .999..98 fraction part? Due to that my equals comparison is failing and so business logic. For few cases we are using
amt = (double)Math.round(orderAmt*100000)/100000;
But that is not possible to do the same in each and every place where we have double arithmetic expression.
I want to know why we get such results randomly and is there any possibility to round the results to 5 decimal places instead of rounding every where?

With radix 10 there are some fractions who can't be expressed exactly with a finite number of digits, like for example 1/3 = 0.33333333....
It's the same with radix 2, except that the dividers that produce this kind of results are not the one we are accustomed to, and for example, for 20.58, which is 2058 / 100, it is the case.
Internally, doubles and floats are stored with bits (an not digit), so the exact value of the double or float just can't be stored in the computer's memory. Each time you perform an operation with this value, you get a small shift, because of the approximation, which becomes visible when converting back to decimal format for printing.
It's something you have to pay attention while perfoming computations where precision is important.
So you have two solutions:
Store all your numbers in decimal type and perform all your calculation with it. This will achieve accuracy but for the price of performance.
You can also keep all the calculation with double or float, and format with a fixed number of digits only for printing results.

You could use BigDecimal for roundoff
BigDecimal bd = new BigDecimal((10.0 * 20.58)/1.0) ;
bd = bd.setScale(4, RoundingMode.UP);
use with a static method
public static double round(double value, int digits) {
BigDecimal bd = new BigDecimal(value);
return bd.setScale(digits, RoundingMode.UP).doubleValue();
}
RoundingMode has :
RoundingMode.UP;
RoundingMode.DOWN;
RoundingMode.HALF_DOWN;
RoundingMode.HALF_EVEN;

The basic problem is that
System.out.println(10.0 * 20.58);
prints
205.79999999999998
has a small rounding error due to a representation error in 20.58
You either need to
round the result before comparing.
use a comparision which allows for some error
use BigDecimal (which is over kill in most cases)
use cents instead of dollars i.e. use int or long with fixed precision.
In the last case, the same operation would read
System.out.println(10 * 2058);
prints
20580
where this is 100x the value you need as its fixed precision e.g. cents instead of dollars.

You may want to use double with rounding as below:
double roundingPlaces = 10.0;//use 10.0 for single decimal digit rounding
double a1 = 10.0;
double b1 = 20.58;
double c1 = 1.0;
System.out.println(Math.round((a1*b1*roundingPlaces)/c1)/roundingPlaces);
This prints 205.8.
float fa1 = (float) 10.0;
float fb1 = (float)20.58;
float fc1 = (float)1.0;
System.out.println(fa1*fb1/fc1);
This also prints 205.8

Use float instead of double
http://ideone.com/L9vwR8
System.out.println((float)((10.0f * 20.58f)/1.0f));
output
205.8