atan2(0, -55) is different from atan2(-1*0, -55)

Just found that

    >> atan2(0, -55) 

    ans =

        3.1416
  >> atan2(-1*0, -55)

ans =

     -3.1416

I don't understand how is this possible? -1 *0 doesn't equal to 0?? How can I convert such behaviour to C++?

Answer 1

[Steven Lord]

Numerically they are equal.

z = 0;
mz = -0;
z == mz

However, they are not identical.

format hex
z
mz

Note that the sign bits are different. You can see this:

format
plusInf = 1/z
minusInf = 1/mz

As for how you can translate this to C++ ... why not just call std::atan2?

 If the implementation supports IEEE floating-point arithmetic (IEC 60559),
 ...
    If y is ±0 and x is negative or -0, ±π is returned

This is specifically called out in section 9.2.1 of IEEE 754-2008.

[Kyle]

Your answer explains well, thank you. std::atan2(-1*0, -55) will be the same as std::atan2(1*0, -55), although I tried all kinds of data type in the two inputs. My current solutions is std::atan2(-1*(0 + 0.000000001), -55).