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 DeMorgan's
we can transform
A&B ^ C == ~(A&B)&C | ~C&(A&B)
== (~A|~B)&C | A&B&~C
== ~A&C | ~B&C | A&B&~C
Attempt a few simplifications when B and C are both constants. */
> + if (GET_CODE (op0) == AND
> + && CONST_INT_P (op1) && CONST_INT_P (XEXP (op0, 1)))
Should have a newline before the second && here.
Ok with those changes.
r~
