Warning: lots of 0 and 1 ahead. Typos are lurking...
On Thu, Jul 25, 2002 at 12:08:09AM +0800, Connie Chan wrote:
> 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
>
> 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 ?
XORing a variable twice with the same value restores its original value:
$a = 0x55
$b = 0xff
$a xor $b xor $b gives the old $a
0x55 xor 0xff xor 0xff is a no-op.
Now, letst do one xor and store this in $a. No information is lost and
$a now documents the differences between $a_old and $b.
Note that the order in an xor operation is interchangable
$a xor $b <=> $b xor $a
So, by doing the xor again, but not storing it in $a and recovering the
initial state, but storing it in $b, we've assigned the initial $a to
$b.
Now, $a still contains the initial differences between the two and $b
contains the initial $a.
Remember, the XORs are interchangeble. We have the initial $a and the
differences, which must result in the initial $b when XORed.
Assigning the result to $a finishes the swap.
0. $a = 0x55 => 01010101 binary
$b = 0xff => 11111111 binary
1. $a 01010101
XOR $b 11111111
Is 10101010 which goes into $a (These are the differences)
2. $a 10101010
XOR $b 11111111
Is 01010101 which goes into $b ($b now contains $a_initial)
3. $a 10101010
XOR $b 11111111
Is 01010101 which then goes into $a ($a now contains $b_initial)
<voice name="Kelly Bundy">Vioooola!</voice>
Try different numbers like 0x55 vs 0xaa or 0x33 vs 0xaa...
It's an old trick from the Assembler times. XOR was 1cycle op and the
whole thing only needed 2 registers. The alternative would have been a
3rd register or a memory/stack operation.
Drieux might provide us with some benchmarks about which one's faster
these days... ;-)
--
Well, then let's give that Java-Wussie a beating... (me)
Michael Lamertz | +49 221 445420 / +49 171 6900 310
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