[ILUG] urg...

[John Tobin]

On Fri, Feb 01, 2002 at 06:47:31PM +0000, kevin lyda wrote:
? ok, this is driving me nuts.  i have a string in perl.  let's call it $s.
? now i want to take $s 6 bits at a time and put each 6 bits (plus 32) into
? a byte in $e.  in other words each 3 bytes of $s become 4 bytes in $e.
? 
? anyone else play with bit shifting in perl?  it doesn't seem to be
? working so far.
? 
? this doesn't work and i feel like shooting it at this point:
? 
?     #!/usr/bin/perl
? 
?     $s = shift;
?     $s .= "\0" x ((length($s) % 3)? 3 - (length($s) % 3): 0);
?     $e = "";
?     while (length($s) > 0) {
?         $c = substr($s, 0, 3);
? 	$s = substr($s, 3);
? 	foreach $x (1...4){
? 	    $e .= chr((ord(substr($c, 0, 1)) & 0xfc) + 32);
? 	    $c <<= 6;
? 	}
?     }
?     $d = "";
?     print("original: $s(".length($s).")\nencoded: $e\ndecoded: $d\n");

This is completely untested but might be a place to start.

@bits = unpack ( "b*", $s );
# make sure we have a multiple of 8 bits ( or whatever number )
# may not be necessary if unpack doesn't strip leading 0s
$num_needed = 8 - ( ( scalar @bits ) % 8 );
unshift @bits, 0 x $num_needed;

$e = 0;
while ( @bits ) {
	$e << 2;
	for ( 1 .. 6 ) {
		$e = ( $e << 1 ) + shift @bits;
	}
	$e = $e + 32;
}

Hope this helps,

? 
? kevin

-- 
John
"That would preempt a bunch of problems involved in trying to reconstruct
exactly how the Perl 5 parser thinks, which nobody entirely understands."
			Larry Wall, 2001/04/20, perl6-language at perl.org
"Finger to spiritual emptiness underlying everything." -- How a Japanese C
manual referred to a "pointer to void".