java的-0和0,Java中的0.0和-0.0（IEEE 754）

Java is totally compatible with IEEE 754 right? But I'm confused about how java decide the sign of float point addition and substraction.

Here is my test result：

double a = -1.5;

double b = 0.0;

double c = -0.0;

System.out.println(b * a); //-0.0

System.out.println(c * a); //0.0

System.out.println(b + b); //0.0

System.out.println(c + b); //0.0

System.out.println(b + c); //0.0

System.out.println(b - c); //0.0

System.out.println(c - b); //-0.0

System.out.println(c + c); //-0.0

I think in the multiplication and division, the sign is decided like: sign(a) xor sign(b),

but I wonder why 0.0 + -0.0 = 0.0, how does Java decide the sign in addition and substraction? Is it described in IEEE 754?

Also I found Java can somehow distinguish the similarities between 0.0 and -0.0, since

System.out.println(c == b); //true

System.out.println(b == c); //true

How does "==" in java works?

Is it treated as a special case?

解决方案

There's nothing here specific to Java, it's specified by IEEE754.

According to the IEEE 754 standard, negative zero and positive zero

should compare as equal with the usual (numerical) comparison

operators, like the == operators of C and Java.

So the following numbers compare equal:

(+0) - (-0) == +0

You'll get the same behavior in all modern languages when dealing with raw floating point numbers.