Thanks for pointing out. I have updated the post. I use the opposite shift since I store the binary in u64 with a different endian from the C++ version. Sorry for the confusion, my bad.
Endian indicates the byte order (MSB->LSB, LSB->MSB), but does not change the representation of a bit. The conditional logic says that an even result is 32 and an odd is 0 after a modulus two. In binary that (x & 1) is 0 for even and 1 for odd. When you shift by 5 that is 0 for even and 32 for odd, which is the opposite of the conditional logic. This is why I suggested using a binary NOT to flip the bits so that you get the same result as the original.