On 2011/3/18 12:18 AM, Richard Henderson wrote:
> On 03/16/2011 06:55 PM, Chung-Lin Tang wrote:
>> You have to use DeMorgan's Law to distribute the ~ operator:
>
> Duh. Not sure where my head was yesterday.
>
> Let's enhance the comment for someone else having an off day. Perhaps
>
> /*
On 03/16/2011 06:55 PM, Chung-Lin Tang wrote:
> You have to use DeMorgan's Law to distribute the ~ operator:
Duh. Not sure where my head was yesterday.
Let's enhance the comment for someone else having an off day. Perhaps
/* Given (xor (and A B) C), using P^Q == ~P&Q | ~Q&P and DeMorga
On 2011/3/17 03:21 AM, Richard Henderson wrote:
> 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 simpli
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 do
On 2011/3/11 10:14 PM, Chung-Lin Tang wrote:
> Hi,
> this patch adds a bit more sophistication to the handled xor RTX cases
> in foo().
>
> This may look a bit ad hoc, but I am seeing it useful for some cases
> where we combine zero_extend with (not (shift ...)). The supplied ARM
> testcase demon
