Jakub Jelinek <[email protected]> writes:
> On Sat, Nov 02, 2013 at 03:15:44PM +0000, Richard Sandiford wrote:
>> What should integer_onep mean if we have a signed 1-bit bitfield in
>> which the bit is set? Seen as a 1-bit value it's "obviously" 1,
>> but seen as a value extended to infinite precision it's -1.
>>
>> Current mainline returns false while wide-int returns true.
>
> Then current mainline is correct. signed 1-bit bitfield has values
> 0 and -1, not 0 and 1. And, signed 1-bit -1 should be just
> integer_minus_onep and integer_all_onesp.
OK, thanks. I should have realised this earlier, but we have:
/* Return 1 if EXPR is the integer constant one or the corresponding
complex constant. */
int
integer_onep (const_tree expr)
...
/* Return 1 if EXPR is the integer constant minus one. */
int
integer_minus_onep (const_tree expr)
which makes them sound like a pair. But integer_minus_onep returns
true for any all-ones INTEGER_CST (regardless of sign):
if (TREE_CODE (expr) == COMPLEX_CST)
return (integer_all_onesp (TREE_REALPART (expr))
&& integer_zerop (TREE_IMAGPART (expr)));
else
return integer_all_onesp (expr);
So a nonzero 1-bit unsigned bitfield is both integer_onep and
integer_minus_onep, but a 1-bit signed bitfield is only
integer_minus_onep. Should integer_minus_onep be changed so
that it always returns false for unsigned types?
Thanks,
Richard
