This question already has answers here:
Is floating point math broken?
(31 answers)
Closed 9 years ago.
Could you please explain, why I got next result:
when I run this:
System.out.println((0.2 - 0.1));
I got: 0.1
when I run this:
System.out.println((0.3 - 0.2));
I got: 0.09999999999999998
I know that number "0.1" doesn't have finite representation in binary, but it doesn't explain the results above. Most likely this is not about particular language but about how digits are stored in computer.
Java uses IEEE floating point to represent double values. It is not a precise representation, and some calculations result in tiny errors that manifest themselves in this way.
I agree with Bohemian above (float and double is not precise) so you will get oddities like this
but there is a solution for your problem:
NumberFormat nf = NumberFormat.getInstance();
nf.setMaximumFractionDigits(1);
nf.format(0.3f - 0.2f);
This will produce 0.1.
Related
This question already has answers here:
Is floating point math broken?
(31 answers)
Closed 2 months ago.
According to official java doc,
RoundingMode HALF_EVEN:
Rounding mode to round towards the nearest neighbor unless both neighbors are equidistant, in which case, round towards the even neighbor.
So as this is a case of equidistant neighbours, why is it still rounding off to 6.33 instead of 6.32
public static void main(String[] args) {
DecimalFormat df = new DecimalFormat("0.00");
df.setRoundingMode(RoundingMode.HALF_EVEN);
System.out.println((df.format(6.325)));
}
I am expecting the output of 6.32 for Half_EVEN roundoff but instead it yeilds 6.33.
Any help would be appreciated
Thanks
Floating point numbers can't represent some numbers exactly (there's lots of questions here on SO dealing with that).
The closest double to 6.325 is 6.32500000000000017763568394002504646778106689453125, and that's why it's rounded up to 6.33.
You can search for things like "double precision floating converter online" for web services that can show you the exact binary representations of floating point numbers.
This question already has answers here:
Is floating point math broken?
(31 answers)
Closed 8 years ago.
In Java, for the following code
Double d = 2.0-1.1;
System.out.println(d);
The result is
0.8999999999999999
If the program is dealing with sensitive information such as precentile/money or cents how do I solve this problem?
I also tried the following piece of code:
new BigDecimal(d)
which outputs
0.899999999999999911182158029987476766109466552734375
What should I do to get 0.90 for the above case?
Since double cannot accurately represent the result of 2 - 1.11, the precision has already been lost by the time the constructor is used.Therefore you need to chain BigDecimal using the String based constructor
BigDecimal result = new BigDecimal("2").subtract(new BigDecimal("1.1"));
The Standard Reference is What Every Computer Scientist Should Know About Floating-Point Arithmetic
A more digestable read: Floating Point Numbers
This question already has answers here:
Why can't decimal numbers be represented exactly in binary?
(22 answers)
Floating point arithmetic not producing exact results [duplicate]
(7 answers)
Closed 9 years ago.
I'm simply trying to calculate percentage_imp, but instead of 0.22 (exactly 0.22, no rounding error), I get 0.22000000000000003!!
I used to get similar odd results, and I've been told to move from float to double, but this one is still odd!
All the variables below are double!
double percentage_imp= budget - (sum_minlessi)/ (sum_i + sum_lessi);
Thats because of the floating point precision values.
You must read:- What Every Computer Scientist Should Know About Floating-Point Arithmetic
You must also read how floating point arithmetic and its internal representation works.
0.22 is not representable as a double.
As an example, 1/3 cannot be represented in base-10, so we approximate with 0.3333333333333333.
Its a rounding error inherit in binary calculations. Not all rational decimal numbers can be represented as a single decimal number in binary. As such, when you perform operations such as multiply and divide, you'll get some nasty error on the last few digits.
Moving from double to float doesn't change this as double is simply a double precision floating point number.
I suggest you look at this link
As a extra bonus in java, simple operations such as binary multiplications are optimized as much as possible utilizing any sort of hardware optimizations when possible yielding different answers on different machines. If you require consistent behavior across all machines use the strictfp keyword in your class declaration.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
how to fix double precision issue in java
I have a small piece of code like this:
double number1 = 6;
double number2 = 5.99;
double result = number1 - number2;
However, the result == 0.009999999999999787 instead of 0.01
I know it is the issue of IEEE 754 standard, but I don't understand why. Could you please explain it for me?
This is because float point numbers cannot be exactly represented with in binary system with limited bits (not without precision loss)
See: http://en.wikipedia.org/wiki/Loss_of_significance
Because there is no .01 in floating point numbers. The fractional bits are expressed as 1/root 2 so you can get something like .0125 or what you have there but there is not .01 in floating point numbers. If you need exact precision use integers instead.
This question already has answers here:
Why does floating-point arithmetic not give exact results when adding decimal fractions?
(31 answers)
Closed 6 years ago.
I have 2 numbers stored as Double, 1.4300 and 1.4350. When I subtract 1.4350 - 1.4300, it gives me the result: 0.0050000000000001155. Why does it add 1155 to the end and how can I solve this so that it returns 0.005 or 0.0050? I'm not sure rounding will work as I'm working with 2 and 4 decimal numbers.
Oh, I love these... these are caused by inaccuracy in the double representation and floating-point arithmetic is full of these. It is often caused by recurring numbers in binary (i.e. base-2 floating-point representation). For example, in decimal 1/3 = 0.3333' In binary 1/10 is a recurring number, which means it cannot be perfectly represented. Try this: 1 - 0.1 - 0.1 - 0.1 - 0.1. You wont get 0.6 :-)
To solve this, use BigDecimal (preferred) or manipulating the double by first multiplying it something like 10000, then rounding it and then dividing it again (less clean).
Good question... it has caused huge problems in the past. Missiles overshooting targets, satellites crashing after launch, etc. Search the web for some, you'll be amazed!
This is a common pitfall with some computer representations of fractional numbers, see this question or google for floating point precision.
Double is not the right type for very precision floating point calculations, if you want exact results you have to use BigDecimal.