On Tue, Sep 05, 2017 at 11:21:48PM -0600, Jeff Law wrote:
> --- a/gcc/tree-ssa-reassoc.c
> +++ b/gcc/tree-ssa-reassoc.c
> @@ -5763,14 +5763,15 @@ reassociate_bb (basic_block bb)
> "Width = %d was chosen for reassociation\n",
> width);
>
>
> - /* For binary bit operations, if the last operand in
> - OPS is a constant, move it to the front. This
> - helps ensure that we generate (X & Y) & C rather
> - than (X & C) & Y. The former will often match
> - a canonical bit test when we get to RTL. */
> - if ((rhs_code == BIT_AND_EXPR
> - || rhs_code == BIT_IOR_EXPR
> - || rhs_code == BIT_XOR_EXPR)
> + /* For binary bit operations, if there are at least 3
> + operands and the last last operand in OPS is a constant,
> + move it to the front. This helps ensure that we generate
> + (X & Y) & C rather than (X & C) & Y. The former will
> + often match a canonical bit test when we get to RTL. */
> + if (ops.length () != 2
So wouldn't it be clearer to write ops.length () > 2 ?
if (ops.length () == 0)
else if (ops.length () == 1)
come earlier, so it is the same thing, but might help the reader.
> + && (rhs_code == BIT_AND_EXPR
> + || rhs_code == BIT_IOR_EXPR
> + || rhs_code == BIT_XOR_EXPR)
> && TREE_CODE (ops.last ()->op) == INTEGER_CST)
> std::swap (*ops[0], *ops[ops_num - 1]);
Don't you then want to put the constant as second operand rather than first,
i.e. swap with *ops[1]?
And doesn't swap_ops_for_binary_stmt undo it again?
Jakub
