How do I print a double value without scientific notation using Java?

Question

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";