Re: Help comparing data in byte by shifting (>>) bits

Guys: Sorry about that. I missed when he said "something like that", which means not exactly it. I am supposed to make it work... Ok so, it is now working. Here is what I have (not exactly...). ;-)

sub isbitset {

   my $bytetocheck = $_[0];
   my $bittocheck = $_[1];

   if (($bytetocheck >> $bittocheck) & 1) {
      return 1;
   } else {
      return 0;
   }

} # End of isbitset function.

##################################################
# I am testing all 8 bits to see which one is set.
##################################################
printf ("\tByte %08b bit 0 => %d\n\n", 
      ord $uh[0], &isbitset((ord $uh[0]), 0));

printf ("\tByte %08b bit 1 => %d\n\n", 
      ord $uh[0], &isbitset((ord $uh[0]), 1));

printf ("\tByte %08b bit 2 => %d\n\n", 
      ord $uh[0], &isbitset((ord $uh[0]), 2));

printf ("\tByte %08b bit 3 => %d\n\n", 
      ord $uh[0], &isbitset((ord $uh[0]), 3));

printf ("\tByte %08b bit 4 => %d\n\n", 
      ord $uh[0], &isbitset((ord $uh[0]), 4));

printf ("\tByte %08b bit 5 => %d\n\n", 
      ord $uh[0], &isbitset((ord $uh[0]), 5));

printf ("\tByte %08b bit 6 => %d\n\n", 
      ord $uh[0], &isbitset((ord $uh[0]), 6));

printf ("\tByte %08b bit 7 => %d\n\n", 
      ord $uh[0], &isbitset((ord $uh[0]), 7));
[download]

Is it looking any better? Thanks, RV

Comment on Re: Help comparing data in byte by shifting (>>) bits Download Code

Replies are listed 'Best First'.
Re^2: Help comparing data in byte by shifting (>>) bits by ikegami (Patriarch) on Jun 15, 2006 at 21:42 UTC
FYI, the following all accomplish the same: `if (($bytetocheck >> $bittocheck) & 1)` `if ($bytetocheck & (1 << $bittocheck))` `if ($bytetocheck & $bit[$bittocheck])` `if (vec($bytetocheck, $bittocheck, 1))` And in the interest of eliminating redundancy, `if (($bytetocheck >> $bittocheck) & 1) { return 1; } else { return 0; }` [download] simplifies to any of the following: `return (($bytetocheck >> $bittocheck) & 1);` `return ($bytetocheck & (1 << $bittocheck)) ? 1 : 0;` `return ($bytetocheck & $bit[$bittocheck]) ? 1 : 0;` `return vec($bytetocheck, $bittocheck, 1);` and the following, if you're only interested in returning a boolean value: `return ($bytetocheck & (1 << $bittocheck));` `return ($bytetocheck & $bit[$bittocheck]);`	[reply] [d/l] [select]
Re^2: Help comparing data in byte by shifting (>>) bits by japhy (Canon) on Jun 15, 2006 at 19:59 UTC
You could simply return the value of `($bytetocheck >> $bittocheck) & 1`, since it will return 0 or 1. And it's unnecessary to use the '&' when calling your function. Jeff `japhy` Pinyan, P.L., P.M., P.O.D, X.S.: Perl, regex, and `perl` hacker How can we ever be the sold short or the cheated, we who for every service have long ago been overpaid? ~~ Meister Eckhart	[reply] [d/l]
Re^3: Help comparing data in byte by shifting (>>) bits by unixhome (Novice) on Jun 15, 2006 at 20:06 UTC
Jeff: Done! Thanks, RV	[reply]
Re^2: Help comparing data in byte by shifting (>>) bits by unixhome (Novice) on Jun 15, 2006 at 19:49 UTC
Oh yeah, other than: `for ($tmp = 0 ; $tmp < 8 ; $tmp++) { printf ("\tByte %08b bit $tmp => %d\n\n", ord $uh[0], &isbitset((ord $uh[0]), $tmp)); }` [download]	[reply] [d/l]