Hmm. I didnt actually test it and had assumed that the result of binary xor was going to be signed, but of course its unsigned, hence the xor ideas doesnt work as written. My bad. Sorry. Id just use the $x<0 == $y<0 approach anyway. :-)
In reply to Re^3: Seeing if two numbers have the same sign
by demerphq
in thread Seeing if two numbers have the same sign
by grinder
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