What are the 22 possible values of x where x== -x (true)?

Question

Recently this question asked in Java interview. Tried and searched the solution but can't find. kindly, comment the solution if anybody knows. It would be helpful.

Answer 1

Zeroes in the numeric basic types. Float and double have two zeroes each. That's nine values. Then there's MIN_VALUE for int and long. That's eleven.

So:

int x = 0;
int x = Integer.MIN_VALUE;
long x = 0;
long x = Long.MIN_VALUE;
byte x = 0;
short x = 0;
char x = 0;
double x = 0.0;
double x = -0.0;
float x = 0f;
float x = -0f;

Then each of those values wrapped as an object:

Integer x = 0;
Integer x = Integer.MIN_VALUE;
Long x = 0L;
Long x = Long.MIN_VALUE;
Byte x = 0;
Short x = 0;
Character x = 0;
Double x = 0.0;
Double x = -0.0;
Float x = 0f;
Float x = -0f;

That's 22 total.

(I wouldn't have called the objects more values. They're the same 11 values again but wrapped in objects. But if you're supposed to find 22 total, I think this must be it.)

Note that for smaller integral types, like short, performing -x would widen them to an int, so x==-x does not work for Short.MIN_VALUE.

x==-x evaluates as true for floating-point zeroes, because even though positive zero and negative zero are different values, they are regarded as equal to each other.