On Tuesday 12 April 2005 23:46, Patrick R. Michaud wrote: > On Tue, Apr 12, 2005 at 11:13:18PM -0400, John Macdonald wrote: > > On Tuesday 12 April 2005 22:36, Patrick R. Michaud wrote: > > > It's entirely possible that I have my mathematics messed up here, > > > but C<xor> doesn't seem to me to be entirely associative, at least not > > > as I commonly think of associativity. Consider > > > > > > $x = (((0 xor 2) xor 4) xor 6); # ((a xor b) xor c) == 6 > > > > > > $y = ((0 xor 2) xor (4 xor 6)); # (a xor (b xor c)) == 2 > > > > You're both messed up your math and you've mixed the boolean > > xor function up with the extension of it to an array of bits that is used > > for integer xor. When you xor integers, you are really xor'ing each > > of the bit positions in the integers in parallel, using the boolean xor > > function for each bit independently. > > No, I'm afraid I haven't messed up my math. Integer-xor is &infix:<+^> . > Array-of-bits-xor is &infix:<~^> . I'm specifically talking about > &infix:<xor>, which "Perl 6 and Parrot Essentials" says (p. 36):
Time for me to see the optometrist. I could have sworn that your examples were returning 0 and 2, but that's definitely a 6 up there. We're violently in agreement. The "boolean" form of xor is associative about the true/false meaning of what it returns, but non-associative about what actual value is returned for true. I apologize. You were not mixed up or messed up. > The XOR relation ... returns the value of the true operand if any > one operand is true, and a false value if both are true or neither > is true. > > It does *not* say that XOR returns "true", it says it returns > "the value of the true operand". (Note that this is different from p5, > which at least on my system does return true/false for C<xor>.) > > > > You can see the same boolean/integer difference in perl5 with: > > > > (4&2) is 0 > > (4&&2) is 4 > > Yes, I know this difference. And in perl 6, the difference is > > (4 +& 2) is 0 > (4 && 2) is 4 > > (4 +^ 2) is 6 > (4 ^^ 2) is false > > > Meanwhile, with your example (using integer xor): > > (((0 xor 2) xor 4) xor 6) > > is ((2 xor 4) xor 6) > > is (6 xor 6) > > is 0 > > Integer xor is (((0 +^ 2) +^ 4) +^ 6) == 0. > > > The same example, using boolean xor: > > (((0 xor 2) xor 4) xor 6) > > --> (((false xor true) xor true) xor true) > > --> ((true xor true) xor true) > > --> (false xor true) > > --> true > > Boolean xor is (((0 ?^ 2) ?^ 4) ?^ 6) == 1. > > > While boolean xor and integer xor give different results (just as boolean > > and integer AND gave different results in my example above) they are > > both perfectly associative. > > Yes, boolean xor and integer xor are associative. But that's not what > I was talking about. In particular: > > (((0 xor 2) xor "four") xor "six") > --> ((2 xor "four") xor "six") > --> (false xor "six") > --> "six" > > > The issue for perl, in the boolean xor case, is which particular true value > > will be returned > > The &infix:<xor> operator in perl 6 is not a "boolean" operator, it's > a low-level "logical" operator, same as &infix:<and> and &infix:<or>. > And C<xor> returns one of its arguments (or "false"), again the same that > C<and> and C<or> do. And within these constraints, I don't believe > that C<xor> is mathematically associative according to any of the > definitions offerred thus far. > > Pm > >
