Double.parseDouble lost of precision!

Hello,
I have a small problem regarding convertion from string to double.
double dTotal;
dTotal = Double.parseDouble("10.01") + Double.parseDouble("0.12");
Can someone explain me why my result is dTotal = 10.12999999999999 ?
Thanks in advance.
Remi.

and read
http://docs.sun.com/source/806-3568/ncg_goldberg.html
This, too:http://java.sun.com/developer/JDCTechTips/2003/tt0204.html#2

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How do I get System.out.format to print out doubles with the same precision

Using the System.out.format, how do I print out the value of a double such that it looks exactly like it does if it were print from a System.out.println.
For example, take the following code:
double d = 12.48564734342343;
System.out.format("d as format: %f\n", d);
System.out.println("d as sout: " + d);Running the code, I get:

d as format: 12.485647
d as sout: 12.48564734342343

The precision of d has been lost.
I could bump up the precision as follows:
double d = 12.48564734342343;
System.out.format("d as format: %.14f\n", d);
System.out.println("d as sout: " + d);That appears to work, I get:

d as format: 12.48564734342343
d as sout: 12.48564734342343

However, that solution fails if d has a different precision, say 12.48. In that case I get:

d as format: 12.48000000000000
d as sout: 12.48

So how do I get System.out.format to print out doubles with the same precision as System.out.println?
Thanks..

YoungWinston wrote:
Schmoe wrote:
Interesting, but this is not what I am looking for...Your original question was "how do I print out the value of a double such that it looks exactly like it does if it were print from a System.out.println", and you've been told how to do that (although the pattern given by sabre may be a bit excessive - you should only need 15 '#'s).The initial phrase from my question was "Using the System.out.format, how do I..".
It's worth remembering that, unlike the Format hierarchy, 'format()' is NOT native to Java. It's a convenience implementation of the 'printf()' and 'sprintf()' methods provided in C, and first appeared in Java 1.5. Those methods were designed to produced fixed-format output; 'println()' was not.Perhaps it is the case that this can't be done.
Furthermore, Double.toString(), which is what is used by println() does not produce the same format in all cases; format("%.14f\n", d) does. TrySystem.out.println(1.8236473845783d);
System.out.println(1823647384.5783d);and you'll see what I mean.I am fine with that. It still displays all the precision.
I am simply looking for a way to quickly print out multiple variables on a sysout while debugging. I want it as syntactically sweet as possible. System.out.println can be a pain when outputting multiple variables like the following:
"System.out.println("a: " + a + "; b:" + b + "; c: " + c);"
For some reason, my fingers always typo the plus key.
I was hoping that System.out.format would be easier,along the lines of:
"System.out.format("a: %f, b: %f, c: %f\n", a, b, c);"
From a syntactical sweetness point of view, it is easier. However, the %f on doubles truncates the precision. I figured there must be a way to get the full precision.
DecimalFormat is syntactically sour for this purpose, as you need to instantiate the DecimalFormat.
fwiw I have enjoyed reading the suggestions in this thread...

Double.parseDouble(String) - problems when string is in scientific notation

Hello guys,
I'm doing some numerical calculations and I wonder whether it is possible for Double.parseDouble(String) to parse string in the scientific notation i.e. 1.0824234234E-10. Is it the notation itself causing the exception : NumberFormatException or the number is just too big/small and double can't hold it ?
If it's just the notation how can I fix it ?
Regards

i'm not quite sure whether double odoes not allow it.
perhaps consider the api Double.valueOf() and the testing code provided; reproduced below:To avoid calling this method on a invalid string and having a NumberFormatException be thrown, the regular expression below can be used to screen the input string:
        final String Digits     = "(\\p{Digit}+)";
final String HexDigits = "(\\p{XDigit}+)";
        // an exponent is 'e' or 'E' followed by an optionally
        // signed decimal integer.
        final String Exp        = "[eE][+-]?"+Digits;
        final String fpRegex    =
            ("[\\x00-\\x20]*"+ // Optional leading "whitespace"
             "[+-]?(" + // Optional sign character
             "NaN|" +           // "NaN" string
             "Infinity|" +      // "Infinity" string
             // A decimal floating-point string representing a finite positive
             // number without a leading sign has at most five basic pieces:
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             // Since this method allows integer-only strings as input
             // in addition to strings of floating-point literals, the
             // two sub-patterns below are simplifications of the grammar
             // productions from the Java Language Specification, 2nd
             // edition, section 3.10.2.
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             "(\\.("+Digits+")("+Exp+")?)|"+
       // Hexadecimal strings
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        "(0[xX]" + HexDigits + "(\\.)?)|" +
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if (Pattern.matches(fpRegex, myString))
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http://java.sun.com/j2se/1.5.0/docs/api/java/lang/Double.html

Double.parseDouble(String) : Problem to parse large String

Hello,
I need to convert A String to a Double in my application without rounding of digits.
I have used the Double.parseDouble(String) method.
The maximum string length can be 17 including dot(.)
So if my String is (of length 17)
Eg.
S= �9999999999.999999� then I am getting the double value as
d= 9999999999.999998(compare the last digit)
If s = �9999999999.111111� then d = 9999999999.111110
If s = �9999999999.555555� then d = 9999999999.555555 (result that I want)
If s = �9999999999.666666� then d = 9999999999.666666 (result that I want)
If s = �9999999999.777777� then d = 9999999999.777777 (result that I want)
If s = �9999999999.888888� then d = 9999999999.888887
If s = �9999999999.999999� then d = 9999999999.999998
If s = �9123456789.123456� then d = 9123456789.123456
But string length up to 16 is giving me the accurate result
So any body can explain me that why it is happening? And how can i get the accurate result.
Thanks in advanced.

Hi,
Thank You for your suggestion. By which i can store
the big double number in Data base , but at some
point i require to parse a double value from a Double
reference.In that case if the Decimal value length
=> 16 then it the last digit changes or rounds.
coverting 9999999999.999999 Decimal value to double
value. It gives me 9999999999.999998 and
9999999999.9999999 gives me the 10000000000.0000000.
but the Double value 999999999.999999 is ok
and 999999999.99999999 rounds.Err, is there a follow up question in there?

Double.parseDouble help?

I have this String Number 20.000. When I do a parseDouble I get 20.0 which is right.
public class qwerty {
     public static void main(String[] args) {
          String num = "20.000";
          System.out.println(Double.parseDouble(num));
But is there a way to preserve the zeroes and it prints 20.000 itself after parse Double?

But is there a way to preserve the zeroes and it prints 20.000 itself after parse Double?Not directly. Doubles (and doubles, and all other Number subclasses and number primitives) store only information about the value of the number, not how it is to be displayed.
As mentioned, you'll need to use formatting provided either by a DecimalFormat object or a printf() function. To do what you're asking, you'd need to somehow parse the input string to determine the required format, and store that information separately (e.g. in a String or a DecimalFormat object), then use that whenever you want to display the number. Check out the documentation for DecimalFormat and PrintStream.printf().
Regex would probably be helpful if you decide to go that route.

Double.parseDouble

the user enters a premium amount and then I want to use that amount in a calcuation; can't I parseDouble it like this? apparently not, because when I try to multiply, I get an error "the operator * is undefined for argument type(s) double, String. I have done this in the past, what is wrong?
String pAmount = JOptionPane.showInputDialog("Enter the premium amount:");
double num = EndCancelDiffCounter;
               double denom = counter;
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               Double.parseDouble(pAmount);
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parseDouble returns a double, it does not change the
String given froma String to a double. You need to
store that result (return value) in a new double
variable and perform your calculation with that.
Edit: Man I'm slow. (I had to check that
parseDouble returned a double and not a Double. I
wasn't 100% sure, as I'm a little daft this morning.)I see, thank you

Double.parseDouble() is not Locale specific, how do I get round this?

I am parsing Strings into doubles using
String myNumber = "12,34";
double d = Double.parseDouble( myNumber );The problem is that my number is of French style (it has commas instead of fullstops for the decimal points)
The default Locale is English, so I get NumberFormat problems when parsing. (I can understand this)
I even get NumberFormat problems when I do this:
Locale.setDefault(Locale.FRENCH);
double d = Double.parseDouble( myNumber );Which I think is really strange. I guess the Locale of Double is set at a certain time and not reset.
OR maybe Double.parseDouble( <STRING> ); is not Locale specific?
Where is my understanding wrong?

Do you know if DecimalFormat reflects the default Locale if the default Locale changes?
Or will I have to make a new DecimalFormat whenever it changes?As Locale is invariant, a DecimalFormat instance (as well a any object with a reference to the default Locale) cannot reflect the default Locale changes.
But each time you create a new instance of DecimalFormat without specifying the Locale, it will get up-to-date Locale.
However, you might prefer setting the locale explicitely (instead of changing the default one.)
Note that Double valueOf() or parseDouble() methods are not related to default Locale, they use a fixed format.

What if Textfield t1 is blank?? num1=Double.parseDouble(t1.getText());

Hi there
Im coding a simple calculator with 2 Textfield for entering two numbers. If one of Textfields is blank and an operation button (+,-,*,/) is clciked, I want the 3rd Textfield t3 to read "ERROR".
From the code below, what is num1 if t1 is blank. Or could you tell me how to go about this if statement.
if (ae.getSource()==multi){
               num1=Double.parseDouble(t1.getText());
               num2=Double.parseDouble(t2.getText());
               res=num1*num2;
               t3.setText(""+res);
cheers
SERGIO

I look forward to you figuring this out yourself.
1. What does getText() return when then JTextField is empty?
2. What happens when you pass that to parseDouble?
3. Should you even bother passing that to parseDouble?

How to roundoff double to 2 digit precision??

I want to roundoff double to 2 digit precision.
for e.g. a=15.536565656456
then a should become 15.54

That will format the result as two digits. If you want 2 digits resolution but still in floating point:
((int)(.5+a*100))/100d

Problem on Double.parseDouble

Double.parseDouble(InputString); was error on my NetBeans 6.8 how to fixed it? I've tried to reinstall but still the same

Make sure you are using CLDC 1.1. Double is not included in CLDC 1.0.
Erik Wetterberg

Does browser support Double.parseDouble()???

hi,
i have used Double.parseDouble() method in my program. After i run that program in browser, it shows error "Double.parseDouble() method not found". Doesnt browser support Double.parseDouble()??? Can anybody tell me how to solve this error?? (i have also tried Float.parseFloat(),but same error occurs) Please help me.
Thanks in advance,
Kalpana.

Newer non-Microsoft browsers such as Netscape 6 and Opera 5 support JDK 1.3. I think IE 5.1 for Mac OS X will use the JDK 1.3 installed on the computer, also. It's only older browsers and IE for Windows and Mac OS 9.1 and before that do not support JDK 1.3. Unfortunately, that's 90% of all browsers! :-(

Variants of Double.parseDouble

I have an applet that runs perfectly on appletviewer but screws up when viewed through my web browser. It doesn't parse the string( number ) found in my textfield( input ). I did it like this: "x = Double.parseDouble( input.getText() );". what is the syntax that i should follow to conform with those that my browser could support? Probably an older one. My browser is IE 4 or 5... Thanks!

Sun must have been struggling for a bit 'cos it wasn't until JDK1.2 that they introduced Double.parseDouble, even though Integer.parseInt has been there since the dawn of time itself.
And since Microsoft stopped supported Java around JDK1.1.6 you'll probably find that your Applet is complaining that there's no such method (have a look in the Java Console in IE to see what it's got to say for itself).
There are various things you could do:
1) install the Java Plug-in to bring your JVM into this century
2) see if DecimalFormat is supported by your runtime (can't find when it was introduced in the Javadocs - sorry)
3) write your own simple double-parsing routine
Personally I'd opt for 1. I'd be tempted to do option 2 as well but it's hardly necessary!
Hope this helps.

Issue with Double.parseDouble

ok i dono whats going on here im writing a console based bank management program and when converting a input srting from console.readline to Double.parseDouble it says the method parseDouble cannot be found, im using the latest jvm whats going on! help!

my bad i ment latest jdk not thinking tonight! heres the error:
D:\My Documents\Java\Bank Acount\AccountDriver.java:59: Method parseDouble(java.lang.String) not found in class java.lang.Double.
newAcc = Double.parseDouble (input);
^
1 error

Double.parseDouble returning 2.0E7

Hi,
I have a piece of code (snippet below)
double aggrAmount=0.0;
Tag t;
aggrAmount=Double.parseDouble(t.getValue().substring(t.getValue().lastIndexOf("/")+1, t.getValue().length()-1));
where t="AGGR//FAMT/20000000,"
when i check aggrAmount the value is 2.0E7. How can i get the value as it is?
PLs help...
THanks for your time and advice

when i check aggrAmount the value is 2.0E7. How can i get the value as it is?There's some confusion here between the value (a double value of twenty million), and the String you use to check it ("2.0E7"). They are different things.
If you want to see or use the value in a string without the exponent business, use String.format() or DecimalFormat. Note that the value remains what it is (twenty million); numbers don't have a format.
If you want the last part "as it is" ie as a String use String's lastIndexOf() and substring() methods and make value a String.

How to format a double to a certain precision?

Hi,
I have a variable of type 'double' and it's set to something like 2.3934583729. I was trying to print the double out to some standard system output with 2 decimal place, formatted precision, "2.39". It seems to me that there is no function like printf(".2f", var) in C/C++ that does the trick. Or am I overlooking something?
Any help would be greatly appreciated. Thanks.
Aaron

Yes you are over looking something.
Check this out...
import java.text.*;
public class TwoDigits
     public static final void main(String[] args) throws Exception
          double aDouble = 2.3934583729;
          DecimalFormat formatter = new DecimalFormat("#.00");
          System.out.println(" aDouble : " + formatter.format(aDouble));

Double.parseDouble lost of precision!

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