On Mon, Feb 29, 2016 at 04:26:12PM +0100, Richard Biener wrote:
> *************** get_unary_op (tree name, enum tree_code
> *** 621,626 ****
> --- 641,680 ----
> return NULL_TREE;
> }
>
> + /* Return true if OP1 and OP2 have the same value if casted to either type.
> */
> +
> + static bool
> + ops_equal_values_p (tree op1, tree op2)
> + {
> + if (op1 == op2)
> + return true;
> +
> + if (TREE_CODE (op1) == SSA_NAME)
> + {
> + gimple *stmt = SSA_NAME_DEF_STMT (op1);
> + if (gimple_nop_conversion_p (stmt))
> + {
> + op1 = gimple_assign_rhs1 (stmt);
> + if (op1 == op2)
> + return true;
> + }
> + }
> +
> + if (TREE_CODE (op2) == SSA_NAME)
> + {
> + gimple *stmt = SSA_NAME_DEF_STMT (op2);
> + if (gimple_nop_conversion_p (stmt))
> + {
> + op2 = gimple_assign_rhs1 (stmt);
> + if (op1 == op2)
> + return true;
> + }
> + }
This will not work if you have:
x_1 = (nop) x_0;
x_2 = (nop) x_1;
and op1 is x_1 and op2 is x_2 (but will work
if op1 is x_2 and op1 is x_1).
Wouldn't it be better to also remember the original
tree orig_op1 = op1;
at the beginning and in the last comparison do
if (op1 == op2 || orig_op1 == op2)
?
Jakub
