What does (x ^ 0x1) != 0 mean?
I came across the following code snippet
if( 0 != ( x ^ 0x1 ) )
encode( x, m );
What does x ^ 0x1 mean? Is this some standard technique?
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.
^ is the bitwise XOR operation 0x1 is 1 in hex notation x ^ 0x1 will invert the last bit of x (refer to the XOR truth table in the link above if that's not clear to you). So, the condition (0 != ( x ^ 0x1 )) will be true if x is greater than 1 or if the last bit of x is 0. Which only leaves x==1 as a value at which the condition will be false. So it is equivalent to
if (x != 1)
PS Hell of a way to implement such a simple condition, I might add. Don't do that. And if you must write complicated code, leave a comment . I beg of you.
This may seem as oversimplified explanation, but if someone would like to go through it slowly it is below:
^ is a bitwise XOR operator in c, c++ and c#.
A bitwise XOR takes two bit patterns of equal length and performs the logical exclusive OR operation on each pair of corresponding bits.
Exclusive OR is a logical operation that outputs true whenever both inputs differ (one is true, the other is false).
The truth table of a xor b :
a b a xor b
----------------------------
1 1 0
1 0 1
0 1 1
0 0 0
So let's illustrate the 0 == ( x ^ 0x1 ) expression on binary level:
what? xxxxxxxx (8 bits)
xor 00000001 (hex 0x1 or 0x01, decimal 1)
gives 00000000
---------------------------
the only answer is 00000001
so:
0 == ( x ^ 0x1 ) => x == 1
0 != ( x ^ 0x1 ) => x != 1
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