Re: popped bit in right shift operation

blokhead is right, you really want to do a bitwise and (&) BEFORE you shift.

You may also wish to check out Bit operations for beginners or Re: Re: What A Wonderful World: Bitmasks! for more information on bitwise operations.

--
"To err is human, but to really foul things up you need a computer." --Paul Ehrlich

Comment on Re: popped bit in right shift operation

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Re: Re: popped bit in right shift operation by vasu (Novice) on Oct 03, 2003 at 18:26 UTC
Thank you for the great pointers. It just seemed to me, since the right shift operator is going to be popping the bit out, that there would be some variable that would hold the popped value. I used the following logic to get the least significant bit that has a value of 1 in a given byte. I'm sure there is a more elegant solution available :) `$i=0; $j=1; if ($num != 0) # $num is the last octet of an IP address { while ($i<8 && (($j&$num) == 0)) { $i++; $j <<= 1; } # Only lists the possible class C subnets printf("Bit mask range = %d to 31\nPossible bitmasks:\n", 31-$i); for ($k = 0; $k <= $i; $k++) { printf("\t255.255.255.%d\n", 256-(1<<$k)); } } else { printf("Possible bitmask:\n\t255.255.255.0\n"); }` [download]	[reply] [d/l]

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Re: Re: popped bit in right shift operation
by vasu (Novice) on Oct 03, 2003 at 18:26 UTC

$i=0;
$j=1;
if ($num != 0) # $num is the last octet of an IP address
{
   while ($i<8 && (($j&$num) == 0))
   {
      $i++;
      $j <<= 1;
   }

   # Only lists the possible class C subnets
   printf("Bit mask range = %d to 31\nPossible bitmasks:\n",
          31-$i);
   for ($k = 0; $k <= $i; $k++)
   {
      printf("\t255.255.255.%d\n", 256-(1<<$k));
   }
}
else
{
   printf("Possible bitmask:\n\t255.255.255.0\n");
}
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