I came across the following code snippet
if( 0 != ( x ^ 0x1 ) )
encode( x, m );
x ^ 0x1
The XOR operation (
x ^ 0x1) inverts bit 0. So the expression effectively means: if bit 0 of x is 0, or any other bit of x is 1, then the expression is true.
Conversely the expression is false if x == 1.
So the test is the same as:
if (x != 1)
and is therefore (arguably) unnecessarily obfuscated.