In Java, I am trying to parse a string of format "###.##" to a float. The string should always have 2 decimal places.
Even if the String has value 123.00, the float should also be 123.00, not 123.0.
This is what I have so far:
System.out.println("string liters of petrol putting in preferences is " + stringLitersOfPetrol);
Float litersOfPetrol = Float.parseFloat(stringLitersOfPetrol);
DecimalFormat df = new DecimalFormat("0.00");
df.setMaximumFractionDigits(2);
litersOfPetrol = Float.parseFloat(df.format(litersOfPetrol));
System.out.println("liters of petrol before putting in editor: " + litersOfPetrol);
It prints:
string liters of petrol putting in preferences is 010.00
liters of petrol before putting in editor: 10.0
This line is your problem:
litersOfPetrol = Float.parseFloat(df.format(litersOfPetrol));
There you formatted your float to string as you wanted, but but then that string got transformed again to a float, and then what you printed in stdout was your float that got a standard formatting. Take a look at this code
import java.text.DecimalFormat;
String stringLitersOfPetrol = "123.00";
System.out.println("string liters of petrol putting in preferences is "+stringLitersOfPetrol);
Float litersOfPetrol=Float.parseFloat(stringLitersOfPetrol);
DecimalFormat df = new DecimalFormat("0.00");
df.setMaximumFractionDigits(2);
stringLitersOfPetrol = df.format(litersOfPetrol);
System.out.println("liters of petrol before putting in editor : "+stringLitersOfPetrol);
And by the way, when you want to use decimals, forget the existence of double and float as others suggested and just use BigDecimal object, it will save you a lot of headache.
Java convert a String to decimal:
String dennis = "0.00000008880000";
double f = Double.parseDouble(dennis);
System.out.println(f);
System.out.println(String.format("%.7f", f));
System.out.println(String.format("%.9f", new BigDecimal(f)));
System.out.println(String.format("%.35f", new BigDecimal(f)));
System.out.println(String.format("%.2f", new BigDecimal(f)));
This prints:
8.88E-8
0.0000001
0.000000089
0.00000008880000000000000106383001366
0.00
Use BigDecimal:
new BigDecimal(theInputString);
It retains all decimal digits. And you are sure of the exact representation since it uses decimal base, not binary base, to store the precision/scale/etc.
And it is not subject to precision loss like float or double are, unless you explicitly ask it to.
I just want to be sure that the float number will also have 2 decimal places after converting that string.
You can't, because floating point numbers don't have decimal places. They have binary places, which aren't commensurate with decimal places.
If you want decimal places, use a decimal radix.
litersOfPetrol = Float.parseFloat(df.format(litersOfPetrol));
System.out.println("liters of petrol before putting in editor : "+litersOfPetrol);
You print Float here, that has no format at all.
To print formatted float, just use
String formatted = df.format(litersOfPetrol);
System.out.println("liters of petrol before putting in editor : " + formatted);
Float.parseFloat() is the problem as it returns a new float.
Returns a new float initialized to the value represented by the specified String, as performed by the valueOf method of class Float.
You are formatting just for the purpose of display . It doesn't mean the float will be represented by the same format internally .
You can use java.lang.BigDecimal.
I am not sure why are you using parseFloat() twice. If you want to display the float in a certain format then just format it and display it.
Float litersOfPetrol=Float.parseFloat(stringLitersOfPetrol);
DecimalFormat df = new DecimalFormat("0.00");
df.setMaximumFractionDigits(2);
System.out.println("liters of petrol before putting in editor"+df.format(litersOfPetrol));
Related
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())
I want to parse Double value to a specific format. I don't want it as a string.
I am trying
DecimalFormat format3DigitsAfterDecimal = new DecimalFormat("#0.000");
double d = 57;
Double.parseDouble(format3DigitsAfterDecimal.format(d));
This returns 57.0, I want 57.000 returned by format method in double.
Am I possibly asking wrong question ?
A double doesn't have a specific format.
Getting a Double value as 57.0 or 57.000 is the same for the computer but not for the user.
As a human, you don't read Double, but String representation of Double.
If you want a Double to be displayed with 3 decimal, you have to turn it to String using DecimalFormat
double d = 57.0;
NumberFormat numFormat = NumberFormat.getInstance();
numFormat.setMaximumFractionDigits(3);
numFormat.setMinimumFractionDigits(3);
System.out.println(numFormat.format(d));
You can't change the value of a decimal, but you can use NumberFormat to display it with three decimal places by setting minimum and maximum fraction digits like in the example above.
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())
I'm doing a project where I need to save some user input as a double, formatted to 4 decimal places. I can manage to get the string to have the right format, but parsing it to a Double breaks it. Here is an example of what i'm doing..
DecimalFormat format = new DecimalFormat("#.0000");
String example = "1.23";
Double num = Double.parseDouble(example);
String str = format.format(num);
Double num1 = Double.parseDouble(str);
System.out.println(num1);
this prints
1.23
when i'd like it to print
1.2300
You're confusing the number with the String representation of the number. These are two very different things. A floating point number has an accuracy, but it doesn't understand what display significant digits mean. It's like asking the abstract number 0 to know that 00 is.
To display the number correctly, use the formatter that you have already created:
System.out.println(format.format(num1));
I'm programming a simple java program. I need to get a string from input and divide it into two parts: 1-double 2-string.
Then I need to do a simple calculation on the double and send the result to the output with specific precision(4). It works fine, but there is a problem when the input is 0, then it doesn't work properly.
For example for these input, output will be:
1 kg
output:2.2046
3.1 kg
output:6.8343
But when the input is 0, the output should be 0.0000, but it shows 0.0 .
What should I do to force it to show 0.0000?
I read similar post about double precision, they suggest something like BigDecimal class, but I can't use them in this case,
my code for doing this is:
line=input.nextLine();
array=line.split(" ");
value=Double.parseDouble(array[0]);
type=array[1];
value =value*2.2046;
String s = String.format("%.4f", value);
value = Double.parseDouble(s);
System.out.print(value+" kg\n");
DecimalFormat will allow you to define how many digits you want to display. A '0' will force an output of digits even if the value is zero, whereas a '#' will omit zeros.
System.out.print(new DecimalFormat("#0.0000").format(value)+" kg\n"); should to the trick.
See the documentation
Note: if used frequently, for performance reasons you should instantiate the formatter only once and store the reference: final DecimalFormat df = new DecimalFormat("#0.0000");. Then use df.format(value).
add this instance of DecimalFormat to the top of your method:
DecimalFormat four = new DecimalFormat("#0.0000"); // will round and display the number to four decimal places. No more, no less.
// the four zeros after the decimal point above specify how many decimal places to be accurate to.
// the zero to the left of the decimal place above makes it so that numbers that start with "0." will display "0.____" vs just ".____" If you don't want the "0.", replace that 0 to the left of the decimal point with "#"
then, call the instance "four" and pass your double value when displaying:
double value = 0;
System.out.print(four.format(value) + " kg/n"); // displays 0.0000
System.out.format("%.4f kg\n", 0.0d) prints '0.0000 kg'
I suggest you to use the BigDecimal class for calculating with floating point values. You will be able to control the precision of the floating point arithmetic. But back to the topic :)
You could use the following:
static void test(String stringVal) {
final BigDecimal value = new BigDecimal(stringVal).multiply(new BigDecimal("2.2046"));
DecimalFormat df = new DecimalFormat();
df.setMaximumFractionDigits(4);
df.setMinimumFractionDigits(4);
System.out.println(df.format(value) + " kg\n");
}
public static void main(String[] args) {
test("0");
test("1");
test("3.1");
}
will give you the following output:
0,0000 kg
2,2046 kg
6,8343 kg
String.format is just makign a String representation of the floating point value. If it doesnt provide a flag for a minimum precision, then just pad the end of the string with zeros.
Use DecimalFormat to format your double value to fixed precision string output.
DecimalFormat is a concrete subclass of NumberFormat that formats
decimal numbers. It has a variety of features designed to make it
possible to parse and format numbers in any locale, including support
for Western, Arabic, and Indic digits. It also supports different
kinds of numbers, including integers (123), fixed-point numbers
(123.4), scientific notation (1.23E4), percentages (12%), and currency
amounts ($123). All of these can be localized.
Example -
System.out.print(new DecimalFormat("##.##").format(value)+" kg\n");