Re^2: adding/subtracting calender dates

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xor = by hubb0r (Pilgrim) on Jun 02, 2005 at 20:50 UTC
FYI - xor is an "exclusive or". It has the following rules: `(1 xor 1) == 0 (1 xor 0) == 1 (0 xor 1) == 1 (0 xor 0) == 0 (0 xor 1 xor 1) == 0 (0 xor 1 xor 0) == 1 etc etc` [download] Meaning that it will return true if and only if exactly 1 of the xor'd variables returns true.	[reply] [d/l]
Re: xor = by fishbot_v2 (Chaplain) on Jun 02, 2005 at 22:28 UTC
Close, but not quite. Try this for example: `print "foo" if (1 xor 1 xor 1);` [download] prints foo, since `(1 xor 1 xor 1) == ((1 xor 1) xor 1) == ( ( 0 ) xor 1 ) == 1` [download] In general terms, `xor` actually returns true if an odd number of true values exist in the set. In the cases where the set is of size two, this means exactly one truth. That doesn't generalize, though. addendum: In the OPs case, though, it worked because his alternatives were all mutually exclusive. However, since `xor` can't shortcircuit, it took a lot longer to work than a normal `or`.	[reply] [d/l] [select]