Right shift vs. divide

[Paul Schauble]

I suspect that most of the people who have been recently flaming in 
the right-shift debate don't understand the problem. So...doning
asbestos suit...

The shift instruction that is normally taken as equivalent to divide is
right shift with sign extension. That is, the sign bit fills in the bits
vacated by the shift. Assuming a 16 bit binary machine and starting with
-15(decimal) or FFF1(hex) and repeatedly shifting right 1 bit gives:

Shifts     Hex      Decimal
-----------------------------------
   0      FFF1        -15
   1      FFF8        -8
   2      FFFC        -4
   3      FFFE        -2
   4      FFFF        -1
   5      FFFF        -1
  >5      FFFF        -1

Starting with -16 and dividing by 2, using the standard divide, gives

Divides   Decimal
-----------------------------------
   0       -15
   1       -7
   2       -3
   3       -1
   4        0
  >4        0

Anyone still think they're the same thing?  This optimization only works
if the numbers are known to be non-negative. A Pascal compiler may make
this optimization, becausethe programmer may specify the range for
variables, but not C.

BTW, is >> defined to be a sign-extended shift or a zero extended shift
and under what circumstances.
  
                                    Paul
                                    Schauble @ MIT-Multics