Comments on: Swap two variables using XOR

By: Anonymous

Anonymous — Sat, 27 Feb 2010 06:52:05 +0000

void main()
{
int a=25,b=15;
a=a+b;
b=a-b;
a=a-b;
b=a+b;
printf(”%d%d”,a,b);
}

By: Sriram

Sriram — Sun, 15 Nov 2009 04:30:17 +0000

Hey guys, what about swaping variables contains string values? All of your solutions will suck… Try this and swap integer or string without third variable… Happy Sensible Coding..

$v = ’sriram’;
$u = ‘lakshmi’;

$v .= $u;
$u = substr($v,0,(strlen($v) – strlen($u)));
$v = substr($v,(strlen($v) – strlen($u)-1), strlen($v));

echo ‘u = ‘ . $u .”;
echo ‘v = ‘ . $v;

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.