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

[download]
More generally, to get the stuff that goes away when you bitshift right by N bits, do a bitwise AND with 2^N-1.

blokhead

In reply to Re: popped bit in right shift operation by blokhead
in thread popped bit in right shift operation by vasu

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