Comments on: Swap two variables using XOR

By: Anonymous

Anonymous — Mon, 02 Nov 2009 00:22:25 +0000

Well I’m not getting one single exchange instruction… at least in debugging mode. This is what I see in my VC++ 6 dissassembly window:

45: a.i ^= b.i;
00401450 mov edx,dword ptr [a]
00401453 xor edx,dword ptr [b]
00401456 mov dword ptr [a],edx
46: b.i = a.i^b.i;
00401459 mov eax,dword ptr [a]
0040145C xor eax,dword ptr [b]
0040145F mov dword ptr [b],eax
47: a.i ^= b.i;
00401462 mov ecx,dword ptr [a]
00401465 xor ecx,dword ptr [b]
00401468 mov dword ptr [a],ecx
48:
49: t = y;
0040146B mov edx,dword ptr [y]
0040146E mov dword ptr [t],edx
50: y = x;
00401471 mov eax,dword ptr [x]
00401474 mov dword ptr [y],eax
51: x = t;
00401477 mov ecx,dword ptr [t]
0040147A mov dword ptr [x],ecx

‘a’ and ‘b’ are what I call “flints,” a union containing a float and int.

By: Anonymous

Anonymous — Tue, 23 Jun 2009 13:41:53 +0000

Hi Kalid,
In response to Rich you said the following:

///////////////////
Hi Rich, good question.

When using pointers to the same location, when you do the first step (x = x xor y) => 0 you also set y to 0 as well since it’s at the same location. In this case, you’ve lost the value 1234 — x and y were sharing the same location and it was overwritten with 0.

Hope this helps.

Kalid — July 17, 2008 @ 12:55 pm
//////

I’m not able to understand why would pointer y change when pointer x changes. I think you meant
(*y) and (*x). Please Clarify.

Thanks.

By: Borsuc

Borsuc — Sun, 31 May 2009 20:13:38 +0000

There’s no reason why it wouldn’t work with floating point numbers, because they are just 1s and 0s as well.

This technique will swap the BITS, and everything made with BITS will work.

Sure you’ll first have to have a way to access float’s bits directly with XOR.

There’s nothing magical about float. Just bits that, together in a “special” format, represent some value. The XOR trick does not need to know the format, it just swaps the bits. (you could make your own format if you want, encryption for example, or compression, etc… and it would still work)

Also for the pointer-to-the-same-location case, this trick will also work. Operating directly on the pointer data is not a good concept by itself when doing computations — because the compiler will not optimize it in registers. If you access the same memory subsequently it’s better to use a temporary variable in C, so the compiler can make register optimizations (put it in registers, calculate on it, then put it in memory).

In this case it would work, since x and y would be on two registers or temporary variables.