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I want to simply have a function that converts a double with as many decimal places into 4 decimal places without rounding.
I have this code that has been working fine but found a random instance where it turned .0 into .99
Here are some sample outputs
4.12897456 ->4.1289
4.5 ->4.5
4.5231->4.5231
5.53->5.53
5.52->5.199 (Wrong conversion, I want it to be 5.52)
private static double get4Donly(double val){
double converted = ((long)(val * 1e4)) / 1e4;
return converted
}
EDIT: This conversion is called thousands of times, so please suggest a method where I dont have to create a new string all the time.
You can use DecimalFormat
import java.text.DecimalFormat;
import java.math.RoundingMode;
import java.util.Arrays;
public class MyClass {
public static void main(String args[]) {
DecimalFormat df = new DecimalFormat("#.####");
df.setRoundingMode(RoundingMode.DOWN);
for (Number n : Arrays.asList(4.12897456, 4.5, 4.5231, 5.53, 5.52)) {
Double d = n.doubleValue();
System.out.println(df.format(d));
}
}
}
RoundingMode.DOWN rounds towards zero, new DecimalFormat("#.####") creates a DecimalFormat instance that formats numbers to a maximum of 4 decimal places. Put those two together and the above code produces the following output, which I believe matches your expectations:
4.1289
4.5
4.5231
5.53
5.52
Doubles just don't work like you think they do.
They are stored in a binary form, not a decimal form. Just like '1 divided by 3' is not representable in a decimal double (0.3333333333 is not enough, it's infinite 3s, so not representable, so you get a rounding error), but '1 divided by 5' is representable just fine, there are numbers that are representable, and numbers that end up rounded when storing things in a double type, but crucially things that seem perfectly roundable in decimal may not be roundable in binary.
Given that they don't match up, your idea of 'eh, I will multiply by 4, turn it to a long, then convert back to a double, then divide by 1000' is not going to let those digits go through unmolested. This is not how you round things, as you're introducing additional loss in addition to the loss you already started out with due to using doubles.
You have 3 solutions available:
Just print it properly
A double cannot be considered to 'have 4 digits after the decimal separator' because a double isn't decimal.
Therefore, it doesn't even make sense to say: Please round this double to at most 4 fractional digits.
That is the crucial realisation. Once you understand that you'll be well along the way :)
What you CAN do is 'please take this double and print it by using no more than 4 digits after the decimal separator'.
String out = String.format("%.4f", 5.52);
or you can use System.printf(XXX) which is short for System.print(String.format(XXX)).
This is probably what you want
forget doubles entirely
For some domains its better to ditch doubles and switch to longs or ints. For example, if you're doing finances, it's better to store the atomic unit as per that currency in a long, and forego doubles instead. So, for dollars, store cents-in-a-long. For euros, the same. For bitcoin, store satoshis. Write custom rendering to render back in a form that is palatable for that currency:
long value = 450; // $4.50
String formatCurrency(long cents) {
return String.format("%s%s%d.%02d", cents < 0 ? "-" : " ", "$", Math.abs(cents) / 100, Math.abs(cents) % 100);
}
Use BigDecimal
This is generally more trouble than it is worth, but it stores every digit, decimally - it represent everything decimal notation can (and it also cannot represent anything else - 1 divided by 3 is impossible in BigDecimal).
I would recommend using the .substring() method by converting the double to a String. It is much easier to understand and achieve since you do not require the number to be rounded.
Moreover, it is the most simple out of all the other methods, such as using DecimalFormat
In that case, you could do it like so:
private static double get4Donly(double val){
String num = String.valueOf(val);
return Double.parseDouble(num.substring(0, 6));
}
However, if the length of the result is smaller than 6 characters, you can do:
private static double get4Donly(double val){
String num = String.valueOf(val);
if(num.length()>6) {
return Double.parseDouble(num.substring(0, 6));
}else {
return val;
}
}
public class doublePrecision {
public static void main(String[] args) {
double total = 0;
total += 5.6;
total += 5.8;
System.out.println(total);
}
}
The above code prints:
11.399999999999
How would I get this to just print (or be able to use it as) 11.4?
As others have mentioned, you'll probably want to use the BigDecimal class, if you want to have an exact representation of 11.4.
Now, a little explanation into why this is happening:
The float and double primitive types in Java are floating point numbers, where the number is stored as a binary representation of a fraction and a exponent.
More specifically, a double-precision floating point value such as the double type is a 64-bit value, where:
1 bit denotes the sign (positive or negative).
11 bits for the exponent.
52 bits for the significant digits (the fractional part as a binary).
These parts are combined to produce a double representation of a value.
(Source: Wikipedia: Double precision)
For a detailed description of how floating point values are handled in Java, see the Section 4.2.3: Floating-Point Types, Formats, and Values of the Java Language Specification.
The byte, char, int, long types are fixed-point numbers, which are exact representions of numbers. Unlike fixed point numbers, floating point numbers will some times (safe to assume "most of the time") not be able to return an exact representation of a number. This is the reason why you end up with 11.399999999999 as the result of 5.6 + 5.8.
When requiring a value that is exact, such as 1.5 or 150.1005, you'll want to use one of the fixed-point types, which will be able to represent the number exactly.
As has been mentioned several times already, Java has a BigDecimal class which will handle very large numbers and very small numbers.
From the Java API Reference for the BigDecimal class:
Immutable,
arbitrary-precision signed decimal
numbers. A BigDecimal consists of an
arbitrary precision integer unscaled
value and a 32-bit integer scale. If
zero or positive, the scale is the
number of digits to the right of the
decimal point. If negative, the
unscaled value of the number is
multiplied by ten to the power of the
negation of the scale. The value of
the number represented by the
BigDecimal is therefore (unscaledValue
× 10^-scale).
There has been many questions on Stack Overflow relating to the matter of floating point numbers and its precision. Here is a list of related questions that may be of interest:
Why do I see a double variable initialized to some value like 21.4 as 21.399999618530273?
How to print really big numbers in C++
How is floating point stored? When does it matter?
Use Float or Decimal for Accounting Application Dollar Amount?
If you really want to get down to the nitty gritty details of floating point numbers, take a look at What Every Computer Scientist Should Know About Floating-Point Arithmetic.
When you input a double number, for example, 33.33333333333333, the value you get is actually the closest representable double-precision value, which is exactly:
33.3333333333333285963817615993320941925048828125
Dividing that by 100 gives:
0.333333333333333285963817615993320941925048828125
which also isn't representable as a double-precision number, so again it is rounded to the nearest representable value, which is exactly:
0.3333333333333332593184650249895639717578887939453125
When you print this value out, it gets rounded yet again to 17 decimal digits, giving:
0.33333333333333326
If you just want to process values as fractions, you can create a Fraction class which holds a numerator and denominator field.
Write methods for add, subtract, multiply and divide as well as a toDouble method. This way you can avoid floats during calculations.
EDIT: Quick implementation,
public class Fraction {
private int numerator;
private int denominator;
public Fraction(int n, int d){
numerator = n;
denominator = d;
}
public double toDouble(){
return ((double)numerator)/((double)denominator);
}
public static Fraction add(Fraction a, Fraction b){
if(a.denominator != b.denominator){
double aTop = b.denominator * a.numerator;
double bTop = a.denominator * b.numerator;
return new Fraction(aTop + bTop, a.denominator * b.denominator);
}
else{
return new Fraction(a.numerator + b.numerator, a.denominator);
}
}
public static Fraction divide(Fraction a, Fraction b){
return new Fraction(a.numerator * b.denominator, a.denominator * b.numerator);
}
public static Fraction multiply(Fraction a, Fraction b){
return new Fraction(a.numerator * b.numerator, a.denominator * b.denominator);
}
public static Fraction subtract(Fraction a, Fraction b){
if(a.denominator != b.denominator){
double aTop = b.denominator * a.numerator;
double bTop = a.denominator * b.numerator;
return new Fraction(aTop-bTop, a.denominator*b.denominator);
}
else{
return new Fraction(a.numerator - b.numerator, a.denominator);
}
}
}
Observe that you'd have the same problem if you used limited-precision decimal arithmetic, and wanted to deal with 1/3: 0.333333333 * 3 is 0.999999999, not 1.00000000.
Unfortunately, 5.6, 5.8 and 11.4 just aren't round numbers in binary, because they involve fifths. So the float representation of them isn't exact, just as 0.3333 isn't exactly 1/3.
If all the numbers you use are non-recurring decimals, and you want exact results, use BigDecimal. Or as others have said, if your values are like money in the sense that they're all a multiple of 0.01, or 0.001, or something, then multiply everything by a fixed power of 10 and use int or long (addition and subtraction are trivial: watch out for multiplication).
However, if you are happy with binary for the calculation, but you just want to print things out in a slightly friendlier format, try java.util.Formatter or String.format. In the format string specify a precision less than the full precision of a double. To 10 significant figures, say, 11.399999999999 is 11.4, so the result will be almost as accurate and more human-readable in cases where the binary result is very close to a value requiring only a few decimal places.
The precision to specify depends a bit on how much maths you've done with your numbers - in general the more you do, the more error will accumulate, but some algorithms accumulate it much faster than others (they're called "unstable" as opposed to "stable" with respect to rounding errors). If all you're doing is adding a few values, then I'd guess that dropping just one decimal place of precision will sort things out. Experiment.
You may want to look into using java's java.math.BigDecimal class if you really need precision math. Here is a good article from Oracle/Sun on the case for BigDecimal. While you can never represent 1/3 as someone mentioned, you can have the power to decide exactly how precise you want the result to be. setScale() is your friend.. :)
Ok, because I have way too much time on my hands at the moment here is a code example that relates to your question:
import java.math.BigDecimal;
/**
* Created by a wonderful programmer known as:
* Vincent Stoessel
* xaymaca#gmail.com
* on Mar 17, 2010 at 11:05:16 PM
*/
public class BigUp {
public static void main(String[] args) {
BigDecimal first, second, result ;
first = new BigDecimal("33.33333333333333") ;
second = new BigDecimal("100") ;
result = first.divide(second);
System.out.println("result is " + result);
//will print : result is 0.3333333333333333
}
}
and to plug my new favorite language, Groovy, here is a neater example of the same thing:
import java.math.BigDecimal
def first = new BigDecimal("33.33333333333333")
def second = new BigDecimal("100")
println "result is " + first/second // will print: result is 0.33333333333333
Pretty sure you could've made that into a three line example. :)
If you want exact precision, use BigDecimal. Otherwise, you can use ints multiplied by 10 ^ whatever precision you want.
As others have noted, not all decimal values can be represented as binary since decimal is based on powers of 10 and binary is based on powers of two.
If precision matters, use BigDecimal, but if you just want friendly output:
System.out.printf("%.2f\n", total);
Will give you:
11.40
You're running up against the precision limitation of type double.
Java.Math has some arbitrary-precision arithmetic facilities.
You can't, because 7.3 doesn't have a finite representation in binary. The closest you can get is 2054767329987789/2**48 = 7.3+1/1407374883553280.
Take a look at http://docs.python.org/tutorial/floatingpoint.html for a further explanation. (It's on the Python website, but Java and C++ have the same "problem".)
The solution depends on what exactly your problem is:
If it's that you just don't like seeing all those noise digits, then fix your string formatting. Don't display more than 15 significant digits (or 7 for float).
If it's that the inexactness of your numbers is breaking things like "if" statements, then you should write if (abs(x - 7.3) < TOLERANCE) instead of if (x == 7.3).
If you're working with money, then what you probably really want is decimal fixed point. Store an integer number of cents or whatever the smallest unit of your currency is.
(VERY UNLIKELY) If you need more than 53 significant bits (15-16 significant digits) of precision, then use a high-precision floating-point type, like BigDecimal.
private void getRound() {
// this is very simple and interesting
double a = 5, b = 3, c;
c = a / b;
System.out.println(" round val is " + c);
// round val is : 1.6666666666666667
// if you want to only two precision point with double we
// can use formate option in String
// which takes 2 parameters one is formte specifier which
// shows dicimal places another double value
String s = String.format("%.2f", c);
double val = Double.parseDouble(s);
System.out.println(" val is :" + val);
// now out put will be : val is :1.67
}
Use java.math.BigDecimal
Doubles are binary fractions internally, so they sometimes cannot represent decimal fractions to the exact decimal.
/*
0.8 1.2
0.7 1.3
0.7000000000000002 2.3
0.7999999999999998 4.2
*/
double adjust = fToInt + 1.0 - orgV;
// The following two lines works for me.
String s = String.format("%.2f", adjust);
double val = Double.parseDouble(s);
System.out.println(val); // output: 0.8, 0.7, 0.7, 0.8
Doubles are approximations of the decimal numbers in your Java source. You're seeing the consequence of the mismatch between the double (which is a binary-coded value) and your source (which is decimal-coded).
Java's producing the closest binary approximation. You can use the java.text.DecimalFormat to display a better-looking decimal value.
Short answer: Always use BigDecimal and make sure you are using the constructor with String argument, not the double one.
Back to your example, the following code will print 11.4, as you wish.
public class doublePrecision {
public static void main(String[] args) {
BigDecimal total = new BigDecimal("0");
total = total.add(new BigDecimal("5.6"));
total = total.add(new BigDecimal("5.8"));
System.out.println(total);
}
}
Multiply everything by 100 and store it in a long as cents.
Computers store numbers in binary and can't actually represent numbers such as 33.333333333 or 100.0 exactly. This is one of the tricky things about using doubles. You will have to just round the answer before showing it to a user. Luckily in most applications, you don't need that many decimal places anyhow.
Floating point numbers differ from real numbers in that for any given floating point number there is a next higher floating point number. Same as integers. There's no integer between 1 and 2.
There's no way to represent 1/3 as a float. There's a float below it and there's a float above it, and there's a certain distance between them. And 1/3 is in that space.
Apfloat for Java claims to work with arbitrary precision floating point numbers, but I've never used it. Probably worth a look.
http://www.apfloat.org/apfloat_java/
A similar question was asked here before
Java floating point high precision library
Use a BigDecimal. It even lets you specify rounding rules (like ROUND_HALF_EVEN, which will minimize statistical error by rounding to the even neighbor if both are the same distance; i.e. both 1.5 and 2.5 round to 2).
Why not use the round() method from Math class?
// The number of 0s determines how many digits you want after the floating point
// (here one digit)
total = (double)Math.round(total * 10) / 10;
System.out.println(total); // prints 11.4
Check out BigDecimal, it handles problems dealing with floating point arithmetic like that.
The new call would look like this:
term[number].coefficient.add(co);
Use setScale() to set the number of decimal place precision to be used.
If you have no choice other than using double values, can use the below code.
public static double sumDouble(double value1, double value2) {
double sum = 0.0;
String value1Str = Double.toString(value1);
int decimalIndex = value1Str.indexOf(".");
int value1Precision = 0;
if (decimalIndex != -1) {
value1Precision = (value1Str.length() - 1) - decimalIndex;
}
String value2Str = Double.toString(value2);
decimalIndex = value2Str.indexOf(".");
int value2Precision = 0;
if (decimalIndex != -1) {
value2Precision = (value2Str.length() - 1) - decimalIndex;
}
int maxPrecision = value1Precision > value2Precision ? value1Precision : value2Precision;
sum = value1 + value2;
String s = String.format("%." + maxPrecision + "f", sum);
sum = Double.parseDouble(s);
return sum;
}
You can Do the Following!
System.out.println(String.format("%.12f", total));
if you change the decimal value here %.12f
So far I understand it as main goal to get correct double from wrong double.
Look for my solution how to get correct value from "approximate" wrong value - if it is real floating point it rounds last digit - counted from all digits - counting before dot and try to keep max possible digits after dot - hope that it is enough precision for most cases:
public static double roundError(double value) {
BigDecimal valueBigDecimal = new BigDecimal(Double.toString(value));
String valueString = valueBigDecimal.toPlainString();
if (!valueString.contains(".")) return value;
String[] valueArray = valueString.split("[.]");
int places = 16;
places -= valueArray[0].length();
if ("56789".contains("" + valueArray[0].charAt(valueArray[0].length() - 1))) places--;
//System.out.println("Rounding " + value + "(" + valueString + ") to " + places + " places");
return valueBigDecimal.setScale(places, RoundingMode.HALF_UP).doubleValue();
}
I know it is long code, sure not best, maybe someone can fix it to be more elegant. Anyway it is working, see examples:
roundError(5.6+5.8) = 11.399999999999999 = 11.4
roundError(0.4-0.3) = 0.10000000000000003 = 0.1
roundError(37235.137567000005) = 37235.137567
roundError(1/3) 0.3333333333333333 = 0.333333333333333
roundError(3723513756.7000005) = 3.7235137567E9 (3723513756.7)
roundError(3723513756123.7000005) = 3.7235137561237E12 (3723513756123.7)
roundError(372351375612.7000005) = 3.723513756127E11 (372351375612.7)
roundError(1.7976931348623157) = 1.797693134862316
Do not waste your efford using BigDecimal. In 99.99999% cases you don't need it. java double type is of cource approximate but in almost all cases, it is sufficiently precise. Mind that your have an error at 14th significant digit. This is really negligible!
To get nice output use:
System.out.printf("%.2f\n", total);
I want to return a double with 2 decimal places (i.e. 123.00). The following are part of the codes. The output is always 123.0. How to get a two decimal places?
public static double getPrice(){
double price = Double.valueOf(showInputDialog("Stock's price : "));
DecimalFormat rounded = new DecimalFormat("###.00");
double newPrice = Double.valueOf(rounded.format(price));
System.out.println(newPrice);
return newPrice;
}
As long as the return type of your function is double, the short answer is "you can't".
Many numbers that have no more than two decimal digits cannot be represented exactly as double. One common example is 0.1. The double that's nearest to it is 0.100000000000000005551115...
No matter how much rounding you do, you wouldn't be able to get exactly 0.1.
As far as your options go, you could:
accept the rounding issues associated with using double;
return an int (i.e. the rounded value multiplied by 100);
return a String;
return a BigDecimal.
In particular, if you're representing monetary amounts, using double is almost certainly a bad idea.
When formatting, you're making a string from a double. Don't convert your formatted string to a double again :
String formattedPrice = rounded.format(price);
System.out.println(formattedPrice); // prints 123.00 or 123,00, depending on your locale
A double keeps a numerical value following IEEE754, it doesn't keep any formatting information and it's only as precise as the IEEE754 double precision allows. If you need to keep a rendering information like the number of digits after the comma, you need something else, like BigDecimal.
In Java, I want to convert a double to an integer, I know if you do this:
double x = 1.5;
int y = (int)x;
you get y=1. If you do this:
int y = (int)Math.round(x);
You'll likely get 2. However, I am wondering: since double representations of integers sometimes look like 1.9999999998 or something, is there a possibility that casting a double created via Math.round() will still result in a truncated down number, rather than the rounded number we are looking for (i.e.: 1 instead of 2 in the code as represented) ?
(and yes, I do mean it as such: Is there any value for x, where y will show a result that is a truncated rather than a rounded representation of x?)
If so: Is there a better way to make a double into a rounded int without running the risk of truncation?
Figured something: Math.round(x) returns a long, not a double. Hence: it is impossible for Math.round() to return a number looking like 3.9999998. Therefore, int(Math.round()) will never need to truncate anything and will always work.
is there a possibility that casting a double created via Math.round() will still result in a truncated down number
No, round() will always round your double to the correct value, and then, it will be cast to an long which will truncate any decimal places. But after rounding, there will not be any fractional parts remaining.
Here are the docs from Math.round(double):
Returns the closest long to the argument. The result is rounded to an integer by adding 1/2, taking the floor of the result, and casting the result to type long. In other words, the result is equal to the value of the expression:
(long)Math.floor(a + 0.5d)
For the datatype Double to int, you can use the following:
Double double = 5.00;
int integer = double.intValue();
Double perValue = 96.57;
int roundVal= (int) Math.round(perValue);
Solved my purpose.
public class doublePrecision {
public static void main(String[] args) {
double total = 0;
total += 5.6;
total += 5.8;
System.out.println(total);
}
}
The above code prints:
11.399999999999
How would I get this to just print (or be able to use it as) 11.4?
As others have mentioned, you'll probably want to use the BigDecimal class, if you want to have an exact representation of 11.4.
Now, a little explanation into why this is happening:
The float and double primitive types in Java are floating point numbers, where the number is stored as a binary representation of a fraction and a exponent.
More specifically, a double-precision floating point value such as the double type is a 64-bit value, where:
1 bit denotes the sign (positive or negative).
11 bits for the exponent.
52 bits for the significant digits (the fractional part as a binary).
These parts are combined to produce a double representation of a value.
(Source: Wikipedia: Double precision)
For a detailed description of how floating point values are handled in Java, see the Section 4.2.3: Floating-Point Types, Formats, and Values of the Java Language Specification.
The byte, char, int, long types are fixed-point numbers, which are exact representions of numbers. Unlike fixed point numbers, floating point numbers will some times (safe to assume "most of the time") not be able to return an exact representation of a number. This is the reason why you end up with 11.399999999999 as the result of 5.6 + 5.8.
When requiring a value that is exact, such as 1.5 or 150.1005, you'll want to use one of the fixed-point types, which will be able to represent the number exactly.
As has been mentioned several times already, Java has a BigDecimal class which will handle very large numbers and very small numbers.
From the Java API Reference for the BigDecimal class:
Immutable,
arbitrary-precision signed decimal
numbers. A BigDecimal consists of an
arbitrary precision integer unscaled
value and a 32-bit integer scale. If
zero or positive, the scale is the
number of digits to the right of the
decimal point. If negative, the
unscaled value of the number is
multiplied by ten to the power of the
negation of the scale. The value of
the number represented by the
BigDecimal is therefore (unscaledValue
× 10^-scale).
There has been many questions on Stack Overflow relating to the matter of floating point numbers and its precision. Here is a list of related questions that may be of interest:
Why do I see a double variable initialized to some value like 21.4 as 21.399999618530273?
How to print really big numbers in C++
How is floating point stored? When does it matter?
Use Float or Decimal for Accounting Application Dollar Amount?
If you really want to get down to the nitty gritty details of floating point numbers, take a look at What Every Computer Scientist Should Know About Floating-Point Arithmetic.
When you input a double number, for example, 33.33333333333333, the value you get is actually the closest representable double-precision value, which is exactly:
33.3333333333333285963817615993320941925048828125
Dividing that by 100 gives:
0.333333333333333285963817615993320941925048828125
which also isn't representable as a double-precision number, so again it is rounded to the nearest representable value, which is exactly:
0.3333333333333332593184650249895639717578887939453125
When you print this value out, it gets rounded yet again to 17 decimal digits, giving:
0.33333333333333326
If you just want to process values as fractions, you can create a Fraction class which holds a numerator and denominator field.
Write methods for add, subtract, multiply and divide as well as a toDouble method. This way you can avoid floats during calculations.
EDIT: Quick implementation,
public class Fraction {
private int numerator;
private int denominator;
public Fraction(int n, int d){
numerator = n;
denominator = d;
}
public double toDouble(){
return ((double)numerator)/((double)denominator);
}
public static Fraction add(Fraction a, Fraction b){
if(a.denominator != b.denominator){
double aTop = b.denominator * a.numerator;
double bTop = a.denominator * b.numerator;
return new Fraction(aTop + bTop, a.denominator * b.denominator);
}
else{
return new Fraction(a.numerator + b.numerator, a.denominator);
}
}
public static Fraction divide(Fraction a, Fraction b){
return new Fraction(a.numerator * b.denominator, a.denominator * b.numerator);
}
public static Fraction multiply(Fraction a, Fraction b){
return new Fraction(a.numerator * b.numerator, a.denominator * b.denominator);
}
public static Fraction subtract(Fraction a, Fraction b){
if(a.denominator != b.denominator){
double aTop = b.denominator * a.numerator;
double bTop = a.denominator * b.numerator;
return new Fraction(aTop-bTop, a.denominator*b.denominator);
}
else{
return new Fraction(a.numerator - b.numerator, a.denominator);
}
}
}
Observe that you'd have the same problem if you used limited-precision decimal arithmetic, and wanted to deal with 1/3: 0.333333333 * 3 is 0.999999999, not 1.00000000.
Unfortunately, 5.6, 5.8 and 11.4 just aren't round numbers in binary, because they involve fifths. So the float representation of them isn't exact, just as 0.3333 isn't exactly 1/3.
If all the numbers you use are non-recurring decimals, and you want exact results, use BigDecimal. Or as others have said, if your values are like money in the sense that they're all a multiple of 0.01, or 0.001, or something, then multiply everything by a fixed power of 10 and use int or long (addition and subtraction are trivial: watch out for multiplication).
However, if you are happy with binary for the calculation, but you just want to print things out in a slightly friendlier format, try java.util.Formatter or String.format. In the format string specify a precision less than the full precision of a double. To 10 significant figures, say, 11.399999999999 is 11.4, so the result will be almost as accurate and more human-readable in cases where the binary result is very close to a value requiring only a few decimal places.
The precision to specify depends a bit on how much maths you've done with your numbers - in general the more you do, the more error will accumulate, but some algorithms accumulate it much faster than others (they're called "unstable" as opposed to "stable" with respect to rounding errors). If all you're doing is adding a few values, then I'd guess that dropping just one decimal place of precision will sort things out. Experiment.
You may want to look into using java's java.math.BigDecimal class if you really need precision math. Here is a good article from Oracle/Sun on the case for BigDecimal. While you can never represent 1/3 as someone mentioned, you can have the power to decide exactly how precise you want the result to be. setScale() is your friend.. :)
Ok, because I have way too much time on my hands at the moment here is a code example that relates to your question:
import java.math.BigDecimal;
/**
* Created by a wonderful programmer known as:
* Vincent Stoessel
* xaymaca#gmail.com
* on Mar 17, 2010 at 11:05:16 PM
*/
public class BigUp {
public static void main(String[] args) {
BigDecimal first, second, result ;
first = new BigDecimal("33.33333333333333") ;
second = new BigDecimal("100") ;
result = first.divide(second);
System.out.println("result is " + result);
//will print : result is 0.3333333333333333
}
}
and to plug my new favorite language, Groovy, here is a neater example of the same thing:
import java.math.BigDecimal
def first = new BigDecimal("33.33333333333333")
def second = new BigDecimal("100")
println "result is " + first/second // will print: result is 0.33333333333333
Pretty sure you could've made that into a three line example. :)
If you want exact precision, use BigDecimal. Otherwise, you can use ints multiplied by 10 ^ whatever precision you want.
As others have noted, not all decimal values can be represented as binary since decimal is based on powers of 10 and binary is based on powers of two.
If precision matters, use BigDecimal, but if you just want friendly output:
System.out.printf("%.2f\n", total);
Will give you:
11.40
You're running up against the precision limitation of type double.
Java.Math has some arbitrary-precision arithmetic facilities.
You can't, because 7.3 doesn't have a finite representation in binary. The closest you can get is 2054767329987789/2**48 = 7.3+1/1407374883553280.
Take a look at http://docs.python.org/tutorial/floatingpoint.html for a further explanation. (It's on the Python website, but Java and C++ have the same "problem".)
The solution depends on what exactly your problem is:
If it's that you just don't like seeing all those noise digits, then fix your string formatting. Don't display more than 15 significant digits (or 7 for float).
If it's that the inexactness of your numbers is breaking things like "if" statements, then you should write if (abs(x - 7.3) < TOLERANCE) instead of if (x == 7.3).
If you're working with money, then what you probably really want is decimal fixed point. Store an integer number of cents or whatever the smallest unit of your currency is.
(VERY UNLIKELY) If you need more than 53 significant bits (15-16 significant digits) of precision, then use a high-precision floating-point type, like BigDecimal.
private void getRound() {
// this is very simple and interesting
double a = 5, b = 3, c;
c = a / b;
System.out.println(" round val is " + c);
// round val is : 1.6666666666666667
// if you want to only two precision point with double we
// can use formate option in String
// which takes 2 parameters one is formte specifier which
// shows dicimal places another double value
String s = String.format("%.2f", c);
double val = Double.parseDouble(s);
System.out.println(" val is :" + val);
// now out put will be : val is :1.67
}
Use java.math.BigDecimal
Doubles are binary fractions internally, so they sometimes cannot represent decimal fractions to the exact decimal.
/*
0.8 1.2
0.7 1.3
0.7000000000000002 2.3
0.7999999999999998 4.2
*/
double adjust = fToInt + 1.0 - orgV;
// The following two lines works for me.
String s = String.format("%.2f", adjust);
double val = Double.parseDouble(s);
System.out.println(val); // output: 0.8, 0.7, 0.7, 0.8
Doubles are approximations of the decimal numbers in your Java source. You're seeing the consequence of the mismatch between the double (which is a binary-coded value) and your source (which is decimal-coded).
Java's producing the closest binary approximation. You can use the java.text.DecimalFormat to display a better-looking decimal value.
Short answer: Always use BigDecimal and make sure you are using the constructor with String argument, not the double one.
Back to your example, the following code will print 11.4, as you wish.
public class doublePrecision {
public static void main(String[] args) {
BigDecimal total = new BigDecimal("0");
total = total.add(new BigDecimal("5.6"));
total = total.add(new BigDecimal("5.8"));
System.out.println(total);
}
}
Multiply everything by 100 and store it in a long as cents.
Computers store numbers in binary and can't actually represent numbers such as 33.333333333 or 100.0 exactly. This is one of the tricky things about using doubles. You will have to just round the answer before showing it to a user. Luckily in most applications, you don't need that many decimal places anyhow.
Floating point numbers differ from real numbers in that for any given floating point number there is a next higher floating point number. Same as integers. There's no integer between 1 and 2.
There's no way to represent 1/3 as a float. There's a float below it and there's a float above it, and there's a certain distance between them. And 1/3 is in that space.
Apfloat for Java claims to work with arbitrary precision floating point numbers, but I've never used it. Probably worth a look.
http://www.apfloat.org/apfloat_java/
A similar question was asked here before
Java floating point high precision library
Use a BigDecimal. It even lets you specify rounding rules (like ROUND_HALF_EVEN, which will minimize statistical error by rounding to the even neighbor if both are the same distance; i.e. both 1.5 and 2.5 round to 2).
Why not use the round() method from Math class?
// The number of 0s determines how many digits you want after the floating point
// (here one digit)
total = (double)Math.round(total * 10) / 10;
System.out.println(total); // prints 11.4
Check out BigDecimal, it handles problems dealing with floating point arithmetic like that.
The new call would look like this:
term[number].coefficient.add(co);
Use setScale() to set the number of decimal place precision to be used.
If you have no choice other than using double values, can use the below code.
public static double sumDouble(double value1, double value2) {
double sum = 0.0;
String value1Str = Double.toString(value1);
int decimalIndex = value1Str.indexOf(".");
int value1Precision = 0;
if (decimalIndex != -1) {
value1Precision = (value1Str.length() - 1) - decimalIndex;
}
String value2Str = Double.toString(value2);
decimalIndex = value2Str.indexOf(".");
int value2Precision = 0;
if (decimalIndex != -1) {
value2Precision = (value2Str.length() - 1) - decimalIndex;
}
int maxPrecision = value1Precision > value2Precision ? value1Precision : value2Precision;
sum = value1 + value2;
String s = String.format("%." + maxPrecision + "f", sum);
sum = Double.parseDouble(s);
return sum;
}
You can Do the Following!
System.out.println(String.format("%.12f", total));
if you change the decimal value here %.12f
So far I understand it as main goal to get correct double from wrong double.
Look for my solution how to get correct value from "approximate" wrong value - if it is real floating point it rounds last digit - counted from all digits - counting before dot and try to keep max possible digits after dot - hope that it is enough precision for most cases:
public static double roundError(double value) {
BigDecimal valueBigDecimal = new BigDecimal(Double.toString(value));
String valueString = valueBigDecimal.toPlainString();
if (!valueString.contains(".")) return value;
String[] valueArray = valueString.split("[.]");
int places = 16;
places -= valueArray[0].length();
if ("56789".contains("" + valueArray[0].charAt(valueArray[0].length() - 1))) places--;
//System.out.println("Rounding " + value + "(" + valueString + ") to " + places + " places");
return valueBigDecimal.setScale(places, RoundingMode.HALF_UP).doubleValue();
}
I know it is long code, sure not best, maybe someone can fix it to be more elegant. Anyway it is working, see examples:
roundError(5.6+5.8) = 11.399999999999999 = 11.4
roundError(0.4-0.3) = 0.10000000000000003 = 0.1
roundError(37235.137567000005) = 37235.137567
roundError(1/3) 0.3333333333333333 = 0.333333333333333
roundError(3723513756.7000005) = 3.7235137567E9 (3723513756.7)
roundError(3723513756123.7000005) = 3.7235137561237E12 (3723513756123.7)
roundError(372351375612.7000005) = 3.723513756127E11 (372351375612.7)
roundError(1.7976931348623157) = 1.797693134862316
Do not waste your efford using BigDecimal. In 99.99999% cases you don't need it. java double type is of cource approximate but in almost all cases, it is sufficiently precise. Mind that your have an error at 14th significant digit. This is really negligible!
To get nice output use:
System.out.printf("%.2f\n", total);