how to fix double precision issue in java
Solution 1
You should really be using java.math.BigDecimal to avoid any precision issues, and always use a BigDecimal(String) constructor.
BigDecimal result = new BigDecimal("0.3").multiply( new BigDecimal("3.0") );
Solution 2
The problem isn't with multiplication. It starts with Double.valueOf("0.3"). That value can't be represented exactly in floating-point. You should use java.math.BigDecimal, and you should also Google for a page entitled "What every computer scientist should know about floating point".
prajul
Updated on July 22, 2022Comments
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prajul almost 2 years
Possible Duplicate:
Java floats and doubles, how to avoid that 0.0 + 0.1 + … + 0.1 == 0.9000001?How can I overcome the precision issue with double multiplication in java android??Please note that I am converting a string value into double value.
eg: when I multiply two double value:
double d1 = Double.valueOf("0.3").doubleValue() * Double.valueOf("3").doubleValue(); System.out.println("Result of multiplication : "+d1);
I am getting the following result : 0.8999999999999999
Some of the results that i am getting are.
0.6*3=1.7999999999999998;
0.2*0.2=0.04000000000000001;
etc.Instead of the above results I would like to get the following results.
0.3*3=0.9;
0.6*3=1.8;
0.2*0.2=0.04;Please remember that I am not trying to round it to the nearest integer.