Logical XOR in C/C++?

[juncode]

your best way to do it would be

(!b1 ^ !b2)


where b1 and b2 are expressions.

This does not implement XOR. Try substituting b1<-0, b2<-0
You can also see it in the xor karnaugh map which cannot be simplified

simplest way is to use bitwise ^ or (!b1&&b2)||(b1&&!b2)

cheers!