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View Full Version : Help converting JS to PHP



b_hole
03-19-2007, 12:21 PM
Hi,

I'm trying to convert some JS functions to PHP.
My main problem is tyhat I didn't wrote the JS, so I'm not fully understand it.

Anyway, here's a given JS function for example:

a -= b; a -= c; a ^= (zeroFill(c,13));
b -= c; b -= a; b ^= (a<<8);
c -= a; c -= b; c ^= (zeroFill(b,13));
a -= b; a -= c; a ^= (zeroFill(c,12));
b -= c; b -= a; b ^= (a<<16);
c -= a; c -= b; c ^= (zeroFill(b,5));
a -= b; a -= c; a ^= (zeroFill(c,3));
b -= c; b -= a; b ^= (a<<10);
c -= a; c -= b; c ^= (zeroFill(b,15));
var ret = new Array(a,b,c);

My PHP is this:

$a=$a - $b;
$a=$a - $c;
$a=$a ^ (zeroFill($c, 13));
$b=$b - $c;
$b=$b - $a;
$b=$b ^ ($a<<8);
$c=$c - $a;
$c=$c - $b;
$c=$c ^ (zeroFill($b, 13));
$a=$a - $b;
$a=$a - $c;
$a=$a ^ (zeroFill($c, 12));
$b=$b - $c;
$b=$b - $a;
$b=$b ^ ($a<<16);
$c=$c - $a;
$c=$c - $b;
$c=$c ^ (zeroFill($b, 5));
$a=$a - $b;
$a=$a - $c;
$a=$a ^ (zeroFill($c, 3));
$b=$b - $c;
$b=$b - $a;
$b=$b ^ ($a<<10);
$c=$c - $a;
$c=$c - $b;
$c=$c ^ (zeroFill($b, 15));
$ret=Array($a, $b, $c);
I can't see a reason for, but ret and $ret are not the same (checked using a loop to print all values).
Help please?

Inigoesdr
03-19-2007, 11:31 PM
What is that supposed to do exactly?



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