Re: [patch][4.7] Enhance XOR handling in simplify-rtx.c

[Richard Henderson]

On 03/11/2011 06:14 AM, Chung-Lin Tang wrote:
> +      /* Given (xor (and A B) C), using P^Q == ~PQ | ~QP (concat as AND),
> +      we can transform (AB)^C into:
> +                              A(~CB) | ~AC | ~BC    
> +      Attempt a few simplifications when B and C are both constants.  */

I don't quite follow the step that gets you to that final form.

        P^Q == ~PQ | ~QP

        AB ^ C
        == ~(AB)C | (~C)AB
        == (~A)(~B)C | (~C)AB
        == ?

I.e. how does

        ~AC | ~BC == (~A)(~B)C

?


r~