The bit that's about to be shifted off is the rightmost one, which I think you'll have to grab before you bitshift by masking:
for my $num (64, 65) {
my $popped = $num & 1;
my $left = $num >> 1;
print "$left leftover, $popped was popped\n";
}
__OUTPUT__
32 leftover, 0 was popped
32 leftover, 1 was popped
More generally, to get the stuff that goes away when you bitshift right by N bits, do a bitwise AND with 2^N-1.
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[d/l] |
| [reply] |
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");
}
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[d/l] |