问题
I was wondering, why my simple addition of some double values leads to the following results in Java:
double a = 1 + 1E-10; // 1.0000000001 (as expected)
double b = 1 + 1E-15; // 1.000000000000001 (as expected)
double c = 1 + 1E-20; // 1.0 (why?)
I thought I can at least add a value of Double.MIN_VALUE, which seems to be 4.9E-324.
What am I doing wrong here?
回答1:
As @Turing85 points out, a double has 11 bits of exponent and 53 bits of mantissa.
What we are calculating here is 1.0 + 1E-20. To represent that number (more precisely than 1.0) we need at least 21 decimal digits of precision or 71 bits. That is more precision than a double provides in the mantissa.
So, the nearest representable double number to 1.0 + 1E-20 is .... 1.0. And that's the result you get.
Welcome to mysterious world of floating point arithmetic.
回答2:
You are not doing anything wrong here. The concept of decimal precision is at hand. There is always a possibility of error in floating point numbers. And the common error value is represented as:
- 1/2E-n <= error <= 1/2E-n
where n is the number of decimal digits you have defined.
More about floating point errors can be found here.
来源：https://stackoverflow.com/questions/43542063/addition-of-very-small-double-values-java