0.0 and -0.0 in Java (IEEE 754)

Question

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?

Answer 1

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

From the wikipedia article on the negative zero :

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.

Answer 2

IEEE754 specifies a signed zero. That is, -0.0 and +0.0 are represented individually.

They are defined to compare true on equality.

Java is implementing this correctly.