expr bit-wise "and" operator
Arguments must be integers, result is an integer.
Bit n of the result is 1 if bit n of each argument is 1. Otherwise, bit n of the result is 0.
For negative arguments, we use the extended definition of ~ that ~$a==-1-$a. There are then the following cases:
Case Result
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$a>=0, $b>=0 Ordinary bitwise &
$a>=0, $b<0 $a&$b == $a & ~(~$b) Contrapositive law
== $a & ~ (-1-$b) Extended definition of [~]
Since -1-$b is positive, $a & ~(-1-$b) can be
evaluated in bitwise fashion.
$a<0, $b>=0 Commute to ($b & $a) and use the calculation above
$a<0, $b<0 $a&$b == ~ (~$a | ~$b) De Morgan's Law
== ~ ((-1-$a) | (-1-$b)) Extended definition of [~]
== -1-((-1-$a) | (-1-$b)) Extended definition of [~]
Since -1-$a and -1-$b are both positive, the expression
((-1-$a) | (-1-$b)) can be evaluated in the ordinary
bitwise fashion.