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

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 double

Solution 1:^[1]

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.

Solution 2:^[2]

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

Solution 3:^[3]

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.

Solution 4:^[4]

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

Solution 5:^[5]

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";

Solution 6:^[6]

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.

Solution 7:^[7]

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() )
    }

Solution 8:^[8]

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.

Solution 9:^[9]

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);
}

Solution 10:^[10]

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!

Solution 11:^[11]

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

Solution 12:^[12]

use String.format ("%.0f", number)

%.0f for zero decimal

String numSring = String.format ("%.0f", firstNumber);
System.out.println(numString);

Solution 13:^[13]

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

Solution 14:^[14]

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);
}

Solution 15:^[15]

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());

Solution 16:^[16]

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

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution	Source
Solution 1	Master
Solution 2
Solution 3	Peter Mortensen
Solution 4	Obl Tobl
Solution 5
Solution 6	Eric Leschinski
Solution 7
Solution 8	Peter Mortensen
Solution 9	Theresa
Solution 10	Arsen Davtyan
Solution 11	frianH
Solution 12	Mahmoud Abu Alheja
Solution 13	Peter Mortensen
Solution 14
Solution 15	Rusty Sage
Solution 16	Nguy?n Th?ng

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

Solution 1:[1]

Solution 2:[2]

Java prevent E notation in a double:

Solution 3:[3]

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