> -----Original Message----- > From: Connie Chan [mailto:[EMAIL PROTECTED]] > Sent: Wednesday, July 24, 2002 12:08 PM > To: [EMAIL PROTECTED] > Subject: That seems interesting ? but I don't know why ? > > > In normal case, when we want to swap 2 var, > , say $x and $y, we do in this way : > > $z = $x; $x = $y; $y = $z; # Swapped
Nah, that's the C way. See below... > > today, I suddenly found a code like this : > > $x ^= $y ; $y ^= $x ; $x ^= $y; # Swapped > > It works !! but how that works ? > Could anybody tell me ? Well, it only works for integers. It's because of the way ^ (bitwise xor) works. Given x=0 and y=1, for instance: x = x ^ y y = y ^ x x = x ^ y x = 0 ^ 1 y = 1 ^ 1 x = 1 ^ 0 x = 1 y = 0 x = 1 So, now x=1 and y=0. You can prove this for any combination of x and y in the range 0..1 (1 bit). The ^ performs the operation on each bit of a larger integer. But the "Perl way" to swap any given $x and $y is: ($x, $y) = ($y, $x); -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
