On Sun, 26 May 1996, Eric Smith wrote:

> > numbers. To illustrate: what happens to -1 when you do this?
>
> It gets divided by two yielding -0.5.
> In integer representation there is no 2**-1 bit, so division by right shift
> truncates, effectively rounding down (toward negative infinity).
> -0.5 rounds to -1.

No.  That is not truncation you're describing: what Clyde posted results
in truncation (-1/2 truncated results in zero).

> Your proposed alternate code simply biases the result for negative numbers
> by 0.5, implementing round toward zero.  It is not obvious to me that
> round toward zero is more "correct".
>
> I would be inclined to prefer IEEE-style round to even.

De gustibus non est disputandum, but having -1 / 2 give a result of -1
will be surprising to anyone who's accustomed to integer division as
usually implented.

> If you really care about what happens to the 0.5, you should probably be
> using fixed-point representation.

Of couse, this argues equally against either result.  ;-)