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Is there a way to sum up Decimal values in java

I've been trying to sum up decimal values using double in java and it doesn't work well, got a wrong answer.
I've already tried with Double, Float, BigDecimal.
{
double a = 2595.00;
double b = -1760.76;
double c = -834.00;
double d = -.24;
System.out.print(a+b+c+d);
}
The expected result should be "0" But Got 9.1038288019262836314737796783447265625E-15

You can use BigDecimal for this purpose and make sure you input the numbers as String in the BigDecimal constructor:
BigDecimal a = new BigDecimal("2595.00");
BigDecimal b = new BigDecimal("-1760.76");
BigDecimal c = new BigDecimal("-834.00");
BigDecimal d = new BigDecimal("-.24");
System.out.println(a.add(b).add(c).add(d));
Live Example
Output is:
0.00
From the Java docs for BigDecimal(String):
This is generally the preferred way to convert a float or double into
a BigDecimal, as it doesn't suffer from the unpredictability of the
BigDecimal(double) constructor.
Check this SO thread for why double results in a loss of precision.

As already pointed by the previous answers about double precision, the value here is very close to zero. You can see it with System.out.format as well.
System.out.format("%.14f%n",a+b+c+d);
System.out.format("%1.1f%n",a+b+c+d); //to print 0.0

Java Display 2 decimal with float variable [duplicate]

I need to format a float to "n"decimal places.
was trying to BigDecimal, but the return value is not correct...
public static float Redondear(float pNumero, int pCantidadDecimales) {
// the function is call with the values Redondear(625.3f, 2)
BigDecimal value = new BigDecimal(pNumero);
value = value.setScale(pCantidadDecimales, RoundingMode.HALF_EVEN); // here the value is correct (625.30)
return value.floatValue(); // but here the values is 625.3
}
I need to return a float value with the number of decimal places that I specify.
I need Float value return not Double
.

You may also pass the float value, and use:
String.format("%.2f", floatValue);
Documentation

Take a look at DecimalFormat. You can easily use it to take a number and give it a set number of decimal places.
Edit: Example

Try this this helped me a lot
BigDecimal roundfinalPrice = new BigDecimal(5652.25622f).setScale(2,BigDecimal.ROUND_HALF_UP);
Result will be
roundfinalPrice --> 5652.26

Of note, use of DecimalFormat constructor is discouraged. The javadoc for this class states:
In general, do not call the DecimalFormat constructors directly, since the NumberFormat factory methods may return subclasses other than DecimalFormat.
https://docs.oracle.com/javase/8/docs/api/java/text/DecimalFormat.html
So what you need to do is (for instance):
NumberFormat formatter = NumberFormat.getInstance(Locale.US);
formatter.setMaximumFractionDigits(2);
formatter.setMinimumFractionDigits(2);
formatter.setRoundingMode(RoundingMode.HALF_UP);
Float formatedFloat = new Float(formatter.format(floatValue));

Here's a quick sample using the DecimalFormat class mentioned by Nick.
float f = 12.345f;
DecimalFormat df = new DecimalFormat("#.00");
System.out.println(df.format(f));
The output of the print statement will be 12.35. Notice that it will round it for you.

Kinda surprised nobody's pointed out the direct way to do it, which is easy enough.
double roundToDecimalPlaces(double value, int decimalPlaces)
{
double shift = Math.pow(10,decimalPlaces);
return Math.round(value*shift)/shift;
}
Pretty sure this does not do half-even rounding though.
For what it's worth, half-even rounding is going to be chaotic and unpredictable any time you mix binary-based floating-point values with base-10 arithmetic. I'm pretty sure it cannot be done. The basic problem is that a value like 1.105 cannot be represented exactly in floating point. The floating point value is going to be something like 1.105000000000001, or 1.104999999999999. So any attempt to perform half-even rounding is going trip up on representational encoding errors.
IEEE floating point implementations will do half-rounding, but they do binary half-rounding, not decimal half-rounding. So you're probably ok

public static double roundToDouble(float d, int decimalPlace) {
BigDecimal bd = new BigDecimal(Float.toString(d));
bd = bd.setScale(decimalPlace, BigDecimal.ROUND_HALF_UP);
return bd.doubleValue();
}

This is a much less professional and much more expensive way but it should be easier to understand and more helpful for beginners.
public static float roundFloat(float F, int roundTo){
String num = "#########.";
for (int count = 0; count < roundTo; count++){
num += "0";
}
DecimalFormat df = new DecimalFormat(num);
df.setRoundingMode(RoundingMode.HALF_UP);
String S = df.format(F);
F = Float.parseFloat(S);
return F;
}

I was looking for an answer to this question and later I developed a method! :) A fair warning, it's rounding up the value.
private float limitDigits(float number) {
return Float.valueOf(String.format(Locale.getDefault(), "%.2f", number));
}

I think what you want ist
return value.toString();
and use the return value to display.
value.floatValue();
will always return 625.3 because its mainly used to calculate something.

You can use Decimal format if you want to format number into a string, for example:
String a = "123455";
System.out.println(new
DecimalFormat(".0").format(Float.valueOf(a)));
The output of this code will be:
123455.0
You can add more zeros to the decimal format, depends on the output that you want.

Fomat Double to 2 Decimal Places WITHOUT Using a String

I'm working on a project which requires me to have double values at exactly 2 decimal place. Using some math, I currently have most numbers rounded correctly, except if there are trailing zeros. I want to keep these 0's for results like 0.00 in particular This list is displayed in a TableView, so therefore must be sortable numerically. I'd like the double values to actually be doubles so that it may properly sort.
Is there any way I can keep 2 decimal places without String.format?
Thank you in advance.

You can use DecimalFormat
DecimalFormat df = new DecimalFormat("#.00");
00 = exactly two decimal places.
But this will give to you a String that you can convert again to a double, if you don't want to have this String-conversion check this question to see how it can be achieved using BigDecimal.
IMHO, if this is not for learning pourposes DecimalFormat will be enough for you. True, you will have to use String in some moment of the conversion, but you will only read and store double values and your sorting will be the correct...

This answer expands on the suggestion to use java.math.BigDecimal rather than double.
It is better for this application than double because every two decimal place number is exactly representable in BigDecimal. The rounding from an arbitrary double-representable value down to two decimal places can be easily done according to any of several rounding modes, including the one used for double arithmetic.
On the other hand, it is better than String because the natural sort order is numeric value.
Here is a short demo program illustrating these points:
import java.math.BigDecimal;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
public class Test {
public static void main(String[] args) throws Throwable {
List<BigDecimal> list = new ArrayList<BigDecimal>();
double[] in = {
3.5,
Math.PI,
-100.123456,
1e6
};
System.out.println("Original doubles: "+Arrays.toString(in));
for(double d : in){
list.add(doubleToBigDecimal(d,1));
}
System.out.println("Before sorting: " + list);
Collections.sort(list);
System.out.println("After sorting: " + list);
}
public static BigDecimal doubleToBigDecimal(double in, int places) {
BigDecimal result = new BigDecimal(in).setScale(2,
BigDecimal.ROUND_HALF_EVEN);
return result;
}
}
Output:
Original doubles: [3.5, 3.141592653589793, -100.123456, 1000000.0]
Before sorting: [3.50, 3.14, -100.12, 1000000.00]
After sorting: [-100.12, 3.14, 3.50, 1000000.00]

You can use DecimalFormat, Just set format in DecimalFormat's constructor as your required 2dp so:
double d = 1.234567;
DecimalFormat df = new DecimalFormat("#.##");
System.out.print(df.format(d));
output:
1.23
this code will round the decimal upto to decimal places However it's result will be not upto 2 decimal if value it self is not minimum 2 decimal places. e.g.
double d = 1.2;
DecimalFormat df = new DecimalFormat("#.##");
System.out.print(df.format(d));
output:
1.2
as Thisaru Guruge said and Jordi Castilla answer you can use like:
double d = 1.2;
DecimalFormat df = new DecimalFormat("#.00");
System.out.print(df.format(d));
output:
1.20

You can try BigDecimal, or you can simply create your own object that implements the Comparable interface for sorting and has a toString() method that outputs the value formatted to two decimal places.
class TableValue implements Comparable<TableValue> {
double value;
public TableValue(double value) {
this.value = value;
}
public int compareTo(TableValue o) {
return Double.compare(this.value, o.value);
}
public String toString() {
DecimalFormat df = new DecimalFormat("0.00");
return df.format(value);
}
}

Doubles are imprecise, BigDecimal has an extra attribute: its precision.
So new BigDecimal("5.20") would have a precision of 2, whereas double 5.20 * 10000 probably sufficiently deviates from 52000.0.
Mind the avoidable new BigDecimal(5.20) neither knows a precision, neither is precise.
BigDecimal unfortunately has a COBOListic usage.

You can use DecimalFormat class upto how much precision point you want you can set in class constructor suppose you want 2 precision you can set like
Double total= 12.15356789;
DecimalFormat df = new DecimalFormat("##.00");
Format the value using format method
String dx = df.format(total);
This will return String value of your Double value if you want Double with formatted 2 precision then use below line of code
Double totalVal = Double.valueOf(df.format(total));
This will return you Double value with 2 precision output would be like 12.15

convert double values in decimal (fraction) values java

I have below code:
Double a = new Double((123456798/1000000)); //123456798 this value comes from client side as a `int`
DecimalFormat df = new DecimalFormat("###.###");
log.info("a :"+a+" df "+df.format(a.doubleValue()));
output:
a :123.0 df 123
//i want output like this, a :123.xxx fd 123.xxx
please help
UPDATE:
123456798 this value comes from client side as a int so i cant do it as 123456798.0 (or something)

123456798 and 1000000 are int literals, so dividing them will use integer arithmetic, and yield 123.
Instead, you could use floating point literals in order to use floating point arithmetic:
Double a = new Double((123456798.0/1000000.0));
DecimalFormat df = new DecimalFormat("###.###");
log.info("a :"+a+" df "+df.format(a.doubleValue()));

Any one value in the division should be float or double.
Double a = new Double((123456798.0/1000000));
or
Double a = new Double((123456798/1000000.0));
if you are getting these values in variables, then multiply it with 1.0
like
Double a = new Double((variable*1.0/1000000));

Put it like that
Double a = new Double((123456798.0/1000000.0)); // <- note ".0"
the reason of the misbehavior is the integer division:
123456798/1000000
is the integer value, while
123456798.0/1000000.0
is the floating point one (double)

Double a = new Double((123456798/1000000));
You are doing integer division here. Make one of the constants a double, so that floating-point division is done. Also, why are you using Double? It's better to use the primitive type double.
double a = 123456798.0 / 1000000;
Or simply, since they are constants:
double a = 123.456789;

You perform an integer division, thats why a is incorrect:
Double a = new Double(123456798.0/1000000);

Format Float to n decimal places

I need to format a float to "n"decimal places.
was trying to BigDecimal, but the return value is not correct...
public static float Redondear(float pNumero, int pCantidadDecimales) {
// the function is call with the values Redondear(625.3f, 2)
BigDecimal value = new BigDecimal(pNumero);
value = value.setScale(pCantidadDecimales, RoundingMode.HALF_EVEN); // here the value is correct (625.30)
return value.floatValue(); // but here the values is 625.3
}
I need to return a float value with the number of decimal places that I specify.
I need Float value return not Double
.

You may also pass the float value, and use:
String.format("%.2f", floatValue);
Documentation

Take a look at DecimalFormat. You can easily use it to take a number and give it a set number of decimal places.
Edit: Example

Try this this helped me a lot
BigDecimal roundfinalPrice = new BigDecimal(5652.25622f).setScale(2,BigDecimal.ROUND_HALF_UP);
Result will be
roundfinalPrice --> 5652.26

Of note, use of DecimalFormat constructor is discouraged. The javadoc for this class states:
In general, do not call the DecimalFormat constructors directly, since the NumberFormat factory methods may return subclasses other than DecimalFormat.
https://docs.oracle.com/javase/8/docs/api/java/text/DecimalFormat.html
So what you need to do is (for instance):
NumberFormat formatter = NumberFormat.getInstance(Locale.US);
formatter.setMaximumFractionDigits(2);
formatter.setMinimumFractionDigits(2);
formatter.setRoundingMode(RoundingMode.HALF_UP);
Float formatedFloat = new Float(formatter.format(floatValue));

Here's a quick sample using the DecimalFormat class mentioned by Nick.
float f = 12.345f;
DecimalFormat df = new DecimalFormat("#.00");
System.out.println(df.format(f));
The output of the print statement will be 12.35. Notice that it will round it for you.

Kinda surprised nobody's pointed out the direct way to do it, which is easy enough.
double roundToDecimalPlaces(double value, int decimalPlaces)
{
double shift = Math.pow(10,decimalPlaces);
return Math.round(value*shift)/shift;
}
Pretty sure this does not do half-even rounding though.
For what it's worth, half-even rounding is going to be chaotic and unpredictable any time you mix binary-based floating-point values with base-10 arithmetic. I'm pretty sure it cannot be done. The basic problem is that a value like 1.105 cannot be represented exactly in floating point. The floating point value is going to be something like 1.105000000000001, or 1.104999999999999. So any attempt to perform half-even rounding is going trip up on representational encoding errors.
IEEE floating point implementations will do half-rounding, but they do binary half-rounding, not decimal half-rounding. So you're probably ok

public static double roundToDouble(float d, int decimalPlace) {
BigDecimal bd = new BigDecimal(Float.toString(d));
bd = bd.setScale(decimalPlace, BigDecimal.ROUND_HALF_UP);
return bd.doubleValue();
}

This is a much less professional and much more expensive way but it should be easier to understand and more helpful for beginners.
public static float roundFloat(float F, int roundTo){
String num = "#########.";
for (int count = 0; count < roundTo; count++){
num += "0";
}
DecimalFormat df = new DecimalFormat(num);
df.setRoundingMode(RoundingMode.HALF_UP);
String S = df.format(F);
F = Float.parseFloat(S);
return F;
}

I was looking for an answer to this question and later I developed a method! :) A fair warning, it's rounding up the value.
private float limitDigits(float number) {
return Float.valueOf(String.format(Locale.getDefault(), "%.2f", number));
}

I think what you want ist
return value.toString();
and use the return value to display.
value.floatValue();
will always return 625.3 because its mainly used to calculate something.

You can use Decimal format if you want to format number into a string, for example:
String a = "123455";
System.out.println(new
DecimalFormat(".0").format(Float.valueOf(a)));
The output of this code will be:
123455.0
You can add more zeros to the decimal format, depends on the output that you want.