浮点算法和浮点值比较

Question

For these lines of code, I get back 0 as an output that is they are all equal to each other. Now if I understand correctly, ab and c may store slightly different versions of the true value .3 hence when do a Float.compare(...) against these values, I expect to get back as an output a value other than 0. Why do I get them as 0?

float a = 0.15f + 0.15f;
float b = 0.1f + 0.2f;
float c = 0.3f;
System.out.println(Float.compare(a,  b));  //<--- outputs 0
System.out.println(Float.compare(a,  c));  //<--- outputs 0
System.out.println(Float.compare(b,  c));  //<--- outputs 0

Answer 1

Because, as you say, they may store slightly different versions. But with these simple expressions, there is no loss of precision, so a, b and c all contain exactly the same version of .3f .

For fun, try this. Here you will lose precision, and the result of the comparison will not be 0 :

public static void main(String[] args) {
    float a = .3f;
    float b = .3f;
    a = (float) Math.cos(a);
    a = (float) Math.acos(a);
    System.out.println(Float.compare(a, b));
}