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Is Java assigning incorrect value to double variable? [duplicate]

I want to print a double value in Java without exponential form.
double dexp = 12345678;
System.out.println("dexp: "+dexp);
It shows this E notation: 1.2345678E7.
I want it to print it like this: 12345678
What is the best way to prevent this?

Java prevent E notation in a double:
Five different ways to convert a double to a normal number:
import java.math.BigDecimal;
import java.text.DecimalFormat;
public class Runner {
public static void main(String[] args) {
double myvalue = 0.00000021d;
//Option 1 Print bare double.
System.out.println(myvalue);
//Option2, use decimalFormat.
DecimalFormat df = new DecimalFormat("#");
df.setMaximumFractionDigits(8);
System.out.println(df.format(myvalue));
//Option 3, use printf.
System.out.printf("%.9f", myvalue);
System.out.println();
//Option 4, convert toBigDecimal and ask for toPlainString().
System.out.print(new BigDecimal(myvalue).toPlainString());
System.out.println();
//Option 5, String.format
System.out.println(String.format("%.12f", myvalue));
}
}
This program prints:
2.1E-7
.00000021
0.000000210
0.000000210000000000000001085015324114868562332958390470594167709350585
0.000000210000
Which are all the same value.
Protip: If you are confused as to why those random digits appear beyond a certain threshold in the double value, this video explains: computerphile why does 0.1+0.2 equal 0.30000000000001?
http://youtube.com/watch?v=PZRI1IfStY0

You could use printf() with %f:
double dexp = 12345678;
System.out.printf("dexp: %f\n", dexp);
This will print dexp: 12345678.000000. If you don't want the fractional part, use
System.out.printf("dexp: %.0f\n", dexp);
0 in %.0f means 0 places in fractional part i.e no fractional part. If you want to print fractional part with desired number of decimal places then instead of 0 just provide the number like this %.8f. By default fractional part is printed up to 6 decimal places.
This uses the format specifier language explained in the documentation.
The default toString() format used in your original code is spelled out here.

In short:
If you want to get rid of trailing zeros and Locale problems, then you should use:
double myValue = 0.00000021d;
DecimalFormat df = new DecimalFormat("0", DecimalFormatSymbols.getInstance(Locale.ENGLISH));
df.setMaximumFractionDigits(340); // 340 = DecimalFormat.DOUBLE_FRACTION_DIGITS
System.out.println(df.format(myValue)); // Output: 0.00000021
Explanation:
Why other answers did not suit me:
Double.toString() or System.out.println or FloatingDecimal.toJavaFormatString uses scientific notations if double is less than 10^-3 or greater than or equal to 10^7
By using %f, the default decimal precision is 6, otherwise you can hardcode it, but it results in extra zeros added if you have fewer decimals. Example:
double myValue = 0.00000021d;
String.format("%.12f", myvalue); // Output: 0.000000210000
By using setMaximumFractionDigits(0); or %.0f you remove any decimal precision, which is fine for integers/longs, but not for double:
double myValue = 0.00000021d;
System.out.println(String.format("%.0f", myvalue)); // Output: 0
DecimalFormat df = new DecimalFormat("0");
System.out.println(df.format(myValue)); // Output: 0
By using DecimalFormat, you are local dependent. In French locale, the decimal separator is a comma, not a point:
double myValue = 0.00000021d;
DecimalFormat df = new DecimalFormat("0");
df.setMaximumFractionDigits(340);
System.out.println(df.format(myvalue)); // Output: 0,00000021
Using the ENGLISH locale makes sure you get a point for decimal separator, wherever your program will run.
Why using 340 then for setMaximumFractionDigits?
Two reasons:
setMaximumFractionDigits accepts an integer, but its implementation has a maximum digits allowed of DecimalFormat.DOUBLE_FRACTION_DIGITS which equals 340
Double.MIN_VALUE = 4.9E-324 so with 340 digits you are sure not to round your double and lose precision.

You can try it with DecimalFormat. With this class you are very flexible in parsing your numbers.
You can exactly set the pattern you want to use.
In your case for example:
double test = 12345678;
DecimalFormat df = new DecimalFormat("#");
df.setMaximumFractionDigits(0);
System.out.println(df.format(test)); //12345678

I've got another solution involving BigDecimal's toPlainString(), but this time using the String-constructor, which is recommended in the javadoc:
this constructor is compatible with the values returned by Float.toString and Double.toString. 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.
It looks like this in its shortest form:
return new BigDecimal(myDouble.toString()).stripTrailingZeros().toPlainString();
NaN and infinite values have to be checked extra, so looks like this in its complete form:
public static String doubleToString(Double d) {
if (d == null)
return null;
if (d.isNaN() || d.isInfinite())
return d.toString();
return new BigDecimal(d.toString()).stripTrailingZeros().toPlainString();
}
This can also be copied/pasted to work nicely with Float.
For Java 7 and below, this results in "0.0" for any zero-valued Doubles, so you would need to add:
if (d.doubleValue() == 0)
return "0";

Java/Kotlin compiler converts any value greater than 9999999 (greater than or equal to 10 million) to scientific notation ie. Epsilion notation.
Ex: 12345678 is converted to 1.2345678E7
Use this code to avoid automatic conversion to scientific notation:
fun setTotalSalesValue(String total) {
var valueWithoutEpsilon = total.toBigDecimal()
/* Set the converted value to your android text view using setText() function */
salesTextView.setText( valueWithoutEpsilon.toPlainString() )
}

This will work as long as your number is a whole number:
double dnexp = 12345678;
System.out.println("dexp: " + (long)dexp);
If the double variable has precision after the decimal point it will truncate it.

I needed to convert some double to currency values and found that most of the solutions were OK, but not for me.
The DecimalFormat was eventually the way for me, so here is what I've done:
public String foo(double value) //Got here 6.743240136E7 or something..
{
DecimalFormat formatter;
if(value - (int)value > 0.0)
formatter = new DecimalFormat("0.00"); // Here you can also deal with rounding if you wish..
else
formatter = new DecimalFormat("0");
return formatter.format(value);
}
As you can see, if the number is natural I get - say - 20000000 instead of 2E7 (etc.) - without any decimal point.
And if it's decimal, I get only two decimal digits.

I think everyone had the right idea, but all answers were not straightforward.
I can see this being a very useful piece of code. Here is a snippet of what will work:
System.out.println(String.format("%.8f", EnterYourDoubleVariableHere));
the ".8" is where you set the number of decimal places you would like to show.
I am using Eclipse and it worked no problem.
Hope this was helpful. I would appreciate any feedback!

The following code detects if the provided number is presented in scientific notation. If so it is represented in normal presentation with a maximum of '25' digits.
static String convertFromScientificNotation(double number) {
// Check if in scientific notation
if (String.valueOf(number).toLowerCase().contains("e")) {
System.out.println("The scientific notation number'"
+ number
+ "' detected, it will be converted to normal representation with 25 maximum fraction digits.");
NumberFormat formatter = new DecimalFormat();
formatter.setMaximumFractionDigits(25);
return formatter.format(number);
} else
return String.valueOf(number);
}

This may be a tangent.... but if you need to put a numerical value as an integer (that is too big to be an integer) into a serializer (JSON, etc.) then you probably want "BigInterger"
Example:
value is a string - 7515904334
We need to represent it as a numerical in a Json message:
{
"contact_phone":"800220-3333",
"servicer_id":7515904334,
"servicer_name":"SOME CORPORATION"
}
We can't print it or we'll get this:
{
"contact_phone":"800220-3333",
"servicer_id":"7515904334",
"servicer_name":"SOME CORPORATION"
}
Adding the value to the node like this produces the desired outcome:
BigInteger.valueOf(Long.parseLong(value, 10))
I'm not sure this is really on-topic, but since this question was my top hit when I searched for my solution, I thought I would share here for the benefit of others, lie me, who search poorly. :D

use String.format ("%.0f", number)
%.0f for zero decimal
String numSring = String.format ("%.0f", firstNumber);
System.out.println(numString);

I had this same problem in my production code when I was using it as a string input to a math.Eval() function which takes a string like "x + 20 / 50"
I looked at hundreds of articles... In the end I went with this because of the speed. And because the Eval function was going to convert it back into its own number format eventually and math.Eval() didn't support the trailing E-07 that other methods returned, and anything over 5 dp was too much detail for my application anyway.
This is now used in production code for an application that has 1,000+ users...
double value = 0.0002111d;
String s = Double.toString(((int)(value * 100000.0d))/100000.0d); // Round to 5 dp
s display as: 0.00021

This will work not only for a whole numbers:
double dexp = 12345678.12345678;
BigDecimal bigDecimal = new BigDecimal(Double.toString(dexp));
System.out.println("dexp: "+ bigDecimal.toPlainString());

My solution:
String str = String.format ("%.0f", yourDouble);

For integer values represented by a double, you can use this code, which is much faster than the other solutions.
public static String doubleToString(final double d) {
// check for integer, also see https://stackoverflow.com/a/9898613/868941 and
// https://github.com/google/guava/blob/master/guava/src/com/google/common/math/DoubleMath.java
if (isMathematicalInteger(d)) {
return Long.toString((long)d);
} else {
// or use any of the solutions provided by others, this is the best
DecimalFormat df =
new DecimalFormat("0", DecimalFormatSymbols.getInstance(Locale.ENGLISH));
df.setMaximumFractionDigits(340); // 340 = DecimalFormat.DOUBLE_FRACTION_DIGITS
return df.format(d);
}
}
// Java 8+
public static boolean isMathematicalInteger(final double d) {
return StrictMath.rint(d) == d && Double.isFinite(d);
}

This works for me. The output will be a String.
String.format("%.12f", myvalue);

Good way to convert scientific e notation
String.valueOf(YourDoubleValue.longValue())

tinylog formatting for double values [duplicate]

I want to print a double value in Java without exponential form.
double dexp = 12345678;
System.out.println("dexp: "+dexp);
It shows this E notation: 1.2345678E7.
I want it to print it like this: 12345678
What is the best way to prevent this?

Java prevent E notation in a double:
Five different ways to convert a double to a normal number:
import java.math.BigDecimal;
import java.text.DecimalFormat;
public class Runner {
public static void main(String[] args) {
double myvalue = 0.00000021d;
//Option 1 Print bare double.
System.out.println(myvalue);
//Option2, use decimalFormat.
DecimalFormat df = new DecimalFormat("#");
df.setMaximumFractionDigits(8);
System.out.println(df.format(myvalue));
//Option 3, use printf.
System.out.printf("%.9f", myvalue);
System.out.println();
//Option 4, convert toBigDecimal and ask for toPlainString().
System.out.print(new BigDecimal(myvalue).toPlainString());
System.out.println();
//Option 5, String.format
System.out.println(String.format("%.12f", myvalue));
}
}
This program prints:
2.1E-7
.00000021
0.000000210
0.000000210000000000000001085015324114868562332958390470594167709350585
0.000000210000
Which are all the same value.
Protip: If you are confused as to why those random digits appear beyond a certain threshold in the double value, this video explains: computerphile why does 0.1+0.2 equal 0.30000000000001?
http://youtube.com/watch?v=PZRI1IfStY0

You could use printf() with %f:
double dexp = 12345678;
System.out.printf("dexp: %f\n", dexp);
This will print dexp: 12345678.000000. If you don't want the fractional part, use
System.out.printf("dexp: %.0f\n", dexp);
0 in %.0f means 0 places in fractional part i.e no fractional part. If you want to print fractional part with desired number of decimal places then instead of 0 just provide the number like this %.8f. By default fractional part is printed up to 6 decimal places.
This uses the format specifier language explained in the documentation.
The default toString() format used in your original code is spelled out here.

In short:
If you want to get rid of trailing zeros and Locale problems, then you should use:
double myValue = 0.00000021d;
DecimalFormat df = new DecimalFormat("0", DecimalFormatSymbols.getInstance(Locale.ENGLISH));
df.setMaximumFractionDigits(340); // 340 = DecimalFormat.DOUBLE_FRACTION_DIGITS
System.out.println(df.format(myValue)); // Output: 0.00000021
Explanation:
Why other answers did not suit me:
Double.toString() or System.out.println or FloatingDecimal.toJavaFormatString uses scientific notations if double is less than 10^-3 or greater than or equal to 10^7
By using %f, the default decimal precision is 6, otherwise you can hardcode it, but it results in extra zeros added if you have fewer decimals. Example:
double myValue = 0.00000021d;
String.format("%.12f", myvalue); // Output: 0.000000210000
By using setMaximumFractionDigits(0); or %.0f you remove any decimal precision, which is fine for integers/longs, but not for double:
double myValue = 0.00000021d;
System.out.println(String.format("%.0f", myvalue)); // Output: 0
DecimalFormat df = new DecimalFormat("0");
System.out.println(df.format(myValue)); // Output: 0
By using DecimalFormat, you are local dependent. In French locale, the decimal separator is a comma, not a point:
double myValue = 0.00000021d;
DecimalFormat df = new DecimalFormat("0");
df.setMaximumFractionDigits(340);
System.out.println(df.format(myvalue)); // Output: 0,00000021
Using the ENGLISH locale makes sure you get a point for decimal separator, wherever your program will run.
Why using 340 then for setMaximumFractionDigits?
Two reasons:
setMaximumFractionDigits accepts an integer, but its implementation has a maximum digits allowed of DecimalFormat.DOUBLE_FRACTION_DIGITS which equals 340
Double.MIN_VALUE = 4.9E-324 so with 340 digits you are sure not to round your double and lose precision.

You can try it with DecimalFormat. With this class you are very flexible in parsing your numbers.
You can exactly set the pattern you want to use.
In your case for example:
double test = 12345678;
DecimalFormat df = new DecimalFormat("#");
df.setMaximumFractionDigits(0);
System.out.println(df.format(test)); //12345678

I've got another solution involving BigDecimal's toPlainString(), but this time using the String-constructor, which is recommended in the javadoc:
this constructor is compatible with the values returned by Float.toString and Double.toString. 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.
It looks like this in its shortest form:
return new BigDecimal(myDouble.toString()).stripTrailingZeros().toPlainString();
NaN and infinite values have to be checked extra, so looks like this in its complete form:
public static String doubleToString(Double d) {
if (d == null)
return null;
if (d.isNaN() || d.isInfinite())
return d.toString();
return new BigDecimal(d.toString()).stripTrailingZeros().toPlainString();
}
This can also be copied/pasted to work nicely with Float.
For Java 7 and below, this results in "0.0" for any zero-valued Doubles, so you would need to add:
if (d.doubleValue() == 0)
return "0";

Java/Kotlin compiler converts any value greater than 9999999 (greater than or equal to 10 million) to scientific notation ie. Epsilion notation.
Ex: 12345678 is converted to 1.2345678E7
Use this code to avoid automatic conversion to scientific notation:
fun setTotalSalesValue(String total) {
var valueWithoutEpsilon = total.toBigDecimal()
/* Set the converted value to your android text view using setText() function */
salesTextView.setText( valueWithoutEpsilon.toPlainString() )
}

This will work as long as your number is a whole number:
double dnexp = 12345678;
System.out.println("dexp: " + (long)dexp);
If the double variable has precision after the decimal point it will truncate it.

I needed to convert some double to currency values and found that most of the solutions were OK, but not for me.
The DecimalFormat was eventually the way for me, so here is what I've done:
public String foo(double value) //Got here 6.743240136E7 or something..
{
DecimalFormat formatter;
if(value - (int)value > 0.0)
formatter = new DecimalFormat("0.00"); // Here you can also deal with rounding if you wish..
else
formatter = new DecimalFormat("0");
return formatter.format(value);
}
As you can see, if the number is natural I get - say - 20000000 instead of 2E7 (etc.) - without any decimal point.
And if it's decimal, I get only two decimal digits.

I think everyone had the right idea, but all answers were not straightforward.
I can see this being a very useful piece of code. Here is a snippet of what will work:
System.out.println(String.format("%.8f", EnterYourDoubleVariableHere));
the ".8" is where you set the number of decimal places you would like to show.
I am using Eclipse and it worked no problem.
Hope this was helpful. I would appreciate any feedback!

The following code detects if the provided number is presented in scientific notation. If so it is represented in normal presentation with a maximum of '25' digits.
static String convertFromScientificNotation(double number) {
// Check if in scientific notation
if (String.valueOf(number).toLowerCase().contains("e")) {
System.out.println("The scientific notation number'"
+ number
+ "' detected, it will be converted to normal representation with 25 maximum fraction digits.");
NumberFormat formatter = new DecimalFormat();
formatter.setMaximumFractionDigits(25);
return formatter.format(number);
} else
return String.valueOf(number);
}

This may be a tangent.... but if you need to put a numerical value as an integer (that is too big to be an integer) into a serializer (JSON, etc.) then you probably want "BigInterger"
Example:
value is a string - 7515904334
We need to represent it as a numerical in a Json message:
{
"contact_phone":"800220-3333",
"servicer_id":7515904334,
"servicer_name":"SOME CORPORATION"
}
We can't print it or we'll get this:
{
"contact_phone":"800220-3333",
"servicer_id":"7515904334",
"servicer_name":"SOME CORPORATION"
}
Adding the value to the node like this produces the desired outcome:
BigInteger.valueOf(Long.parseLong(value, 10))
I'm not sure this is really on-topic, but since this question was my top hit when I searched for my solution, I thought I would share here for the benefit of others, lie me, who search poorly. :D

use String.format ("%.0f", number)
%.0f for zero decimal
String numSring = String.format ("%.0f", firstNumber);
System.out.println(numString);

I had this same problem in my production code when I was using it as a string input to a math.Eval() function which takes a string like "x + 20 / 50"
I looked at hundreds of articles... In the end I went with this because of the speed. And because the Eval function was going to convert it back into its own number format eventually and math.Eval() didn't support the trailing E-07 that other methods returned, and anything over 5 dp was too much detail for my application anyway.
This is now used in production code for an application that has 1,000+ users...
double value = 0.0002111d;
String s = Double.toString(((int)(value * 100000.0d))/100000.0d); // Round to 5 dp
s display as: 0.00021

This will work not only for a whole numbers:
double dexp = 12345678.12345678;
BigDecimal bigDecimal = new BigDecimal(Double.toString(dexp));
System.out.println("dexp: "+ bigDecimal.toPlainString());

My solution:
String str = String.format ("%.0f", yourDouble);

For integer values represented by a double, you can use this code, which is much faster than the other solutions.
public static String doubleToString(final double d) {
// check for integer, also see https://stackoverflow.com/a/9898613/868941 and
// https://github.com/google/guava/blob/master/guava/src/com/google/common/math/DoubleMath.java
if (isMathematicalInteger(d)) {
return Long.toString((long)d);
} else {
// or use any of the solutions provided by others, this is the best
DecimalFormat df =
new DecimalFormat("0", DecimalFormatSymbols.getInstance(Locale.ENGLISH));
df.setMaximumFractionDigits(340); // 340 = DecimalFormat.DOUBLE_FRACTION_DIGITS
return df.format(d);
}
}
// Java 8+
public static boolean isMathematicalInteger(final double d) {
return StrictMath.rint(d) == d && Double.isFinite(d);
}

This works for me. The output will be a String.
String.format("%.12f", myvalue);

Good way to convert scientific e notation
String.valueOf(YourDoubleValue.longValue())

Double number formatting in java

I am working on a calculator program in Java and I need to display a number with no more than 15 digits in length (decimal point included).
I've tried to format this number using String.format("%g", myDouble), also with %f, but can't get the result I'm looking for.
The thing is that I need the number:
to be displayed as scientific notation if it has more than 15 digits,
or display it without rounding it if it has less than 15 digits.
Also, I do not want any fixed number of digits before or after the decimal point.
Any suggestions?
This is the code snipet where my number gets displayed:
if (("" + myDouble).length() > 15) {
screen.setText(String.format("%g", myDouble));
} else {
screen.setText("" + myDouble);
}
One problem is that when the else clause is executed, sometimes the numbers get displayed as 9.0000E7, which is not favorable and is different from the String.format("%g", myDouble) scientific notation which looks more like 9.000e+7.

This seems to be close to what you're asking for:
static String format(double n) {
if(Double.isInfinite(n) || Double.isNaN(n))
return Double.toString(n);
String result = BigDecimal.valueOf(n).toPlainString();
if(result.length() > 15)
result = String.format("%.15e", n);
return result;
}
Since your requirement is that the decimal places aren't fixed, I don't really see a way to do it with a formatter exclusively.
If you want a particular number of digits for the scientific notation, then you can use a format like "%.15e" where 15 is the number of decimal digits you want.
Or possibly you want something like:
if(result.length() > 15)
result = new DecimalFormat("0.###############E0").format(n);
But if you're trying to "block align" to some maximum, then things get complicated because the number of fractional digits depends on the number of digits in the exponent.

Just use the NumberFormat object. See the javadoc for details.
NumberFormat formatter = new DecimalFormat();
if (("" + myDouble).length() > 15) {
formatter = new DecimalFormat("0.######E0");
screen.setText(formatter.format(myDouble));
} else {
formatter = new DecimalFormat("#");
screen.setText(formatter.format(myDouble));
}

How to use Java's DecimalFormat for "smart" currency formatting?

I'd like to use Java's DecimalFormat to format doubles like so:
#1 - 100 -> $100
#2 - 100.5 -> $100.50
#3 - 100.41 -> $100.41
The best I can come up with so far is:
new DecimalFormat("'$'0.##");
But this doesn't work for case #2, and instead outputs "$100.5"
Edit:
A lot of these answers are only considering cases #2 and #3 and not realizing that their solution will cause #1 to format 100 as "$100.00" instead of just "$100".

Does it have to use DecimalFormat?
If not, it looks like the following should work:
String currencyString = NumberFormat.getCurrencyInstance().format(currencyNumber);
//Handle the weird exception of formatting whole dollar amounts with no decimal
currencyString = currencyString.replaceAll("\\.00", "");

Use NumberFormat:
NumberFormat n = NumberFormat.getCurrencyInstance(Locale.US);
double doublePayment = 100.13;
String s = n.format(doublePayment);
System.out.println(s);
Also, don't use doubles to represent exact values. If you're using currency values in something like a Monte Carlo method (where the values aren't exact anyways), double is preferred.
See also: Write Java programs to calculate and format currency

Try
new DecimalFormat("'$'0.00");
Edit:
I Tried
DecimalFormat d = new DecimalFormat("'$'0.00");
System.out.println(d.format(100));
System.out.println(d.format(100.5));
System.out.println(d.format(100.41));
and got
$100.00
$100.50
$100.41

Try using
DecimalFormat.setMinimumFractionDigits(2);
DecimalFormat.setMaximumFractionDigits(2);

You can check "is number whole or not" and choose needed number format.
public class test {
public static void main(String[] args){
System.out.println(function(100d));
System.out.println(function(100.5d));
System.out.println(function(100.42d));
}
public static String function(Double doubleValue){
boolean isWholeNumber=(doubleValue == Math.round(doubleValue));
DecimalFormatSymbols formatSymbols = new DecimalFormatSymbols(Locale.GERMAN);
formatSymbols.setDecimalSeparator('.');
String pattern= isWholeNumber ? "#.##" : "#.00";
DecimalFormat df = new DecimalFormat(pattern, formatSymbols);
return df.format(doubleValue);
}
}
will give exactly what you want:
100
100.50
100.42

You can use the following format:
DecimalFormat dformat = new DecimalFormat("$#.##");

I know its too late. However following worked for me :
DecimalFormatSymbols otherSymbols = new DecimalFormatSymbols(Locale.UK);
new DecimalFormat("\u00A4#######0.00",otherSymbols).format(totalSale);
\u00A4 : acts as a placeholder for currency symbol
#######0.00 : acts as a placeholder pattern for actual number with 2 decimal
places precision.
Hope this helps whoever reads this in future :)

You can try by using two different DecimalFormat objects based on the condition as follows:
double d=100;
double d2=100.5;
double d3=100.41;
DecimalFormat df=new DecimalFormat("'$'0.00");
if(d%1==0){ // this is to check a whole number
DecimalFormat df2=new DecimalFormat("'$'");
System.out.println(df2.format(d));
}
System.out.println(df.format(d2));
System.out.println(df.format(d3));
Output:-
$100
$100.50
$100.41

You could use the Java Money API to achieve this. (although this is not using DecialFormat)
long amountInCents = ...;
double amountInEuro = amountInCents / 100.00;
String customPattern;
if (minimumOrderValueInCents % 100 == 0) {
customPattern = "# ¤";
} else {
customPattern = "#.## ¤";
}
Money minDeliveryAmount = Money.of(amountInEuro, "EUR");
MonetaryAmountFormat formatter = MonetaryFormats.getAmountFormat(AmountFormatQueryBuilder.of(Locale.GERMANY)
.set(CurrencyStyle.SYMBOL)
.set("pattern", customPattern)
.build());
System.out.println(minDeliveryAmount);

printf also works.
Example:
double anyNumber = 100;
printf("The value is %4.2f ", anyNumber);
Output:
The value is 100.00
4.2 means force the number to have two digits after the decimal. The 4 controls how many digits to the right of the decimal.

How can I format a String number to have commas and round?

What is the best way to format the following number that is given to me as a String?
String number = "1000500000.574" //assume my value will always be a String
I want this to be a String with the value: 1,000,500,000.57
How can I format it as such?

You might want to look at the DecimalFormat class; it supports different locales (eg: in some countries that would get formatted as 1.000.500.000,57 instead).
You also need to convert that string into a number, this can be done with:
double amount = Double.parseDouble(number);
Code sample:
String number = "1000500000.574";
double amount = Double.parseDouble(number);
DecimalFormat formatter = new DecimalFormat("#,###.00");
System.out.println(formatter.format(amount));

This can also be accomplished using String.format(), which may be easier and/or more flexible if you are formatting multiple numbers in one string.
String number = "1000500000.574";
Double numParsed = Double.parseDouble(number);
System.out.println(String.format("The input number is: %,.2f", numParsed));
// Or
String numString = String.format("%,.2f", numParsed);
For the format string "%,.2f" - "," means separate digit groups with commas, and ".2" means round to two places after the decimal.
For reference on other formatting options, see https://docs.oracle.com/javase/tutorial/java/data/numberformat.html

Given this is the number one Google result for format number commas java, here's an answer that works for people who are working with whole numbers and don't care about decimals.
String.format("%,d", 2000000)
outputs:
2,000,000

Once you've converted your String to a number, you can use
// format the number for the default locale
NumberFormat.getInstance().format(num)
or
// format the number for a particular locale
NumberFormat.getInstance(locale).format(num)

I've created my own formatting utility. Which is extremely fast at processing the formatting along with giving you many features :)
It supports:
Comma Formatting E.g. 1234567 becomes 1,234,567.
Prefixing with "Thousand(K),Million(M),Billion(B),Trillion(T)".
Precision of 0 through 15.
Precision re-sizing (Means if you want 6 digit precision, but only have 3 available digits it forces it to 3).
Prefix lowering (Means if the prefix you choose is too large it lowers it to a more suitable prefix).
The code can be found here. You call it like this:
public static void main(String[])
{
int settings = ValueFormat.COMMAS | ValueFormat.PRECISION(2) | ValueFormat.MILLIONS;
String formatted = ValueFormat.format(1234567, settings);
}
I should also point out this doesn't handle decimal support, but is very useful for integer values. The above example would show "1.23M" as the output. I could probably add decimal support maybe, but didn't see too much use for it since then I might as well merge this into a BigInteger type of class that handles compressed char[] arrays for math computations.

you can also use the below solution
public static String getRoundOffValue(double value){
DecimalFormat df = new DecimalFormat("##,##,##,##,##,##,##0.00");
return df.format(value);
}

public void convert(int s)
{
System.out.println(NumberFormat.getNumberInstance(Locale.US).format(s));
}
public static void main(String args[])
{
LocalEx n=new LocalEx();
n.convert(10000);
}

You can do the entire conversion in one line, using the following code:
String number = "1000500000.574";
String convertedString = new DecimalFormat("#,###.##").format(Double.parseDouble(number));
The last two # signs in the DecimalFormat constructor can also be 0s. Either way works.

Here is the simplest way to get there:
String number = "10987655.876";
double result = Double.parseDouble(number);
System.out.println(String.format("%,.2f",result));
output:
10,987,655.88

The first answer works very well, but for ZERO / 0 it will format as .00
Hence the format #,##0.00 is working well for me.
Always test different numbers such as 0 / 100 / 2334.30 and negative numbers before deploying to production system.

According to chartGPT
Using DecimalFormat:
DecimalFormat df = new DecimalFormat("#,###.00");
String formattedNumber = df.format(yourNumber);
Using NumberFormat:
NumberFormat nf = NumberFormat.getNumberInstance();
nf.setGroupingUsed(true);
String formattedNumber = nf.format(yourNumber);
Using String.format():
String formattedNumber = String.format("%,.2f", yourNumber);
Note: In all the above examples, "yourNumber" is the double value that you want to format with a comma. The ".2f" in the format string indicates that the decimal places should be rounded to 2 decimal places. You can adjust this value as needed.