Issue adding double value

[Mary Joe]

public class TestDobleTwo {

	  public static void main(String[] args) {
		    double d1 = 2095475.03;
		    double d2 = 3225.0;		    
		      
			  System.out.println(" d1 + d2  : " + (d1 + d2)); 
	 }

    }

}

Printing out as
d1 + d2 : 2098700.0300000003
while I expect 2098700.03, can anyone explain the reason?

[Matthew Brown]

Floating point variables aren't very good at storing exact values. Remember that the number is stored internally in binary, and 2095475.03 won't be represented exactly in binary.

So you've got two options.

- If you need the calculations to be exact, then look at using the BigDecimal class to represent numbers and do calculations.

- If you don't care about these tiny inaccuracies, but all you care about is getting sensible output, do the calculations as you are, and try the DecimalFormat class to produce output to the accuracy you want.

[Wouter Oet]

That is how floating point calculations work. It basically is because not every number can be stored in a double so it will take the closed value. If you want it to have a certain format then you can use NumberFormat.

[Jesper de Jong]

See also question #20 in our FAQ.

[Mary Joe]

Thanks what I was trying to do was importing a file.

The header had a total which was giving the value 2098700.03
which will be compared against the total of the relevant filed(of Numeric(5,2)) which is a double in the body
which can be from upto a total of 50000 rows.
This total also represented as a double was returning as 2098700.0300000003 and hence it was throwing an error beacuse the values didn't match.
I will try using BigDecimal

Any example you have on addition on BigDecimal

[Vinay Chaudhary]

BigDecimal bd1= new BigDecimal("2095475.03");
BigDecimal bd2 = new BigDecimal("3225.0");
bd1.add(bd2);
System.out.println("BIG DECI " + bd1);


This will print the result you want

[Mary Joe]

What I didn't understand though was the same logic was used to calculate the header total as well, which didn't present with the same issue.

total += row.getFee().doubleValue();

[Winston Gutkowski]

Mary Joe wrote:Printing out as
d1 + d2 : 2098700.0300000003
while I expect 2098700.03, can anyone explain the reason?

Unfortunately, the original Goldberg page seems to have disappeared, but this one looks similar. If you really need fixed decimal values, you're usually better off using BigDecimal than double.

[fred rosenberger]

Comparing floating point numbers for equality is ALWAYS risky in any computer language. if you have to do so, you generally figure out some delta that the difference should be less than...i.e.:

if (float1 - float2 < 0.000001)

And even this is dangerous if the magnitude of your numbers can vary.

[Mary Joe]

I have overcome the issue firt mentioned in my post using BigDecimal.

Now I have a different one.


Bigdecimal one = 2948.0
Bigdecimal one = 2948 (this is being returned from a jdbc query of Double value new BigDecimal(resultRow.getFee().toString());)
Again failing on comparison, anway to impose the decimal ?

[Joshua Barrett]

I do believe your BigDecimal object has its current scale set to Zero.

The scale defines the number of decimal places the object will hold.

Take a look at the various setScale() Methods in the API, you can also define the scale in several constructors.

Java BigDecimal

Also I would suggest when setting scale you also define the rounding mode. BigDecimal.ROUND_HALF_UP is the standard rounding we all should know from school.

[Mary Joe]

Thanks Joshua that helped a lot.

All issues resolved