expr bit-wise "xor" operator
Arguments must be integers, result is an integer.
Bit n of the result is 1 if bit n of the two arguments differ. Otherwise, bit n of the result is 0.
To evaluate $a^$b when either $a or $b is negative, we make use of the following reasoning:
Case Result
$a>=0, $b>=0 Bitwise operation
$a>=0, $b<0 $a^$b == ~($a ^ ~$b) Contrapositive law
== ~($a ^ (-1-$b)) Extended definition of [~]
== -1-~($a ^ (-1-$b)) Extended definition of [~]
Since $a and (-1-$b) are both positive, the [^] in the
last expression can be evaluated in bitwise fashion
$a<0, $b>=0 Commute to ($b^$a) and evaluate as above.
$a<0, $b<0 $a^$b == (~$a) ^ (~$b) Contrapositive law
== (-1-$a) ^ (-1-$b) Extended definition of [~]
Since (-1-$a) and (-1-$b) are both positive, the [^]
in the last expression can be evaluated in bitwise
fashion.% expr 0b010 | 0b000 2
[So, what do I need to add to this example so the result is binary as well? Some sort of format - but I don't see a binary conversion sequence in the docs...]