Re: Bit shift oddity?

[Stefan Monnier]

>                       (ash shift (lognot 0))))))

Could there be a simple misunderstanding: the shift amount is the
*second* argument.  Here you always pass (lognot 0), which is -1, so
it's a right shift by 1, IOW a kind of "divide by 2".

>   shift=-1 res=-1
>   shift=-2 res=-1
>   shift=-3 res=-2
>   shift=-4 res=-2
>   shift=-5 res=-3
>   shift=-6 res=-3
>   shift=-7 res=-4
>   shift=-8 res=-4
>   shift=-9 res=-5
>   shift=-10 res=-5

That does indeed look like divide by 2 to me (rounding toward -∞).


        Stefan