On 04/20/2011 08:22 AM, Kai Tietz wrote:
> + if (TREE_CODE (arg0) == BIT_AND_EXPR
> + && TREE_CODE (arg1) == BIT_AND_EXPR)
> + {
> + tree a0, a1, l0, l1, n0, n1;
> +
> + a0 = fold_convert_loc (loc, type, TREE_OPERAND (arg1, 0));
> + a1 = fold_convert_loc (loc, type, TREE_OPERAND (arg1, 1));
> +
> + l0 = fold_convert_loc (loc, type, TREE_OPERAND (arg0, 0));
> + l1 = fold_convert_loc (loc, type, TREE_OPERAND (arg0, 1));
> +
> + n0 = fold_build1_loc (loc, BIT_NOT_EXPR, type, l0);
> + n1 = fold_build1_loc (loc, BIT_NOT_EXPR, type, l1);
> +
> + if ((operand_equal_p (n0, a0, 0)
> + && operand_equal_p (n1, a1, 0))
> + || (operand_equal_p (n0, a1, 0)
> + && operand_equal_p (n1, a0, 0)))
> + return fold_build2_loc (loc, TRUTH_XOR_EXPR, type, l0, n1);
First, you typoed BIT_XOR_EXPR in this first block.
Second, I don't see how you're arbitrarily choosing L0 and N1 in the
expansion. If you write the expression the other way around,
(~x & y) | (x & ~y)
don't you wind up with
(~x ^ ~y)
? Or do the extra NOT expressions get folded away anyway?
> + if (TREE_CODE (arg0) == TREE_CODE (arg1)
> + && (TREE_CODE (arg1) == TRUTH_AND_EXPR
> + || TREE_CODE (arg1) == TRUTH_ANDIF_EXPR))
I don't believe you want to apply this transformation with ANDIF.
r~
