问题
I am wondering why I am losing precision when using this code :
double x = 12.0456; // or float : same result
System.out.println(x); // outputs 12.0456 obviously
x %= 1; // should now be equal to 0.0456 right?
System.out.println(x); // outputs 0.04560000000000031 or 0.045599937 when using float
12.0456 modulo 1 should equal 0.0456 right? But it shows a slightly different value, why do I keep losing precision? I mean the code should substract exactly 1 until the value is less than 1.
However, I found out a way to get the correct value :
double x = 12.0456;
System.out.println(x);
x %= 1;
System.out.println((float)x); //outputs 0.0456 exactly
This way works perfectly, but do you guys have a better solution?
I don't care which floating point type I should use, I just want to find a clean way to get the correct value! I don't like having to convert the value to a double and then to a float.
回答1:
Float and double are imprecise - they have a finite amount of bits to represent a value.
Because humans use base 10, and computers use base 2, numbers that appear "simple" to us can be impossible to represent accurately as a float/double, especially the results of computations due to way CPUs execute them.
来源：https://stackoverflow.com/questions/14792253/why-does-using-modulo-on-non-integer-values-lose-floating-point-precision