how to fix double precision issue in java [duplica

问题:

This question already has answers here:

Closed 8 years ago.

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.

回答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") );

回答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".

回答3:

Unfortunately, I am not aware of a simple way of doing exactly what you ask for.

Like Strelok says, you should not be using a floating-point type if you need exact results. However, for most purposes, it is enough to just specify a rounding precision for output. The following code is close to, but not quite, what you want:

System.out.printf("Result of multiplication : %.1g\n", d1);

For more info on the syntax of printf, see the java.util.Formatter documentation.