On Tue, Mar 20, 2012 at 1:26 PM, Richard Sandiford
<rdsandif...@googlemail.com> wrote:
> Richard Guenther <richard.guent...@gmail.com> writes:
>>> I've no objection to moving the assert down to after the GEN_INT.
>>> But it sounds like I'm on my own with the whole CONST_DOUBLE sign thing.
>>> (That is, if we remove the assert altogether, we effectively treat the
>>> number as sign-extended if it happens to fit in a CONST_INT, and
>>> zero-extended otherwise.
>>
>> Why do we treat it zero-extended otherwise? Because we use
>> gen_int_mode for CONST_INTs, which sign-extends?
>
> Just to make sure we're not talking past each other, I meant
> moving the assert to:
>
> /* If this integer fits in one word, return a CONST_INT. */
> [A] if ((i1 == 0 && i0 >= 0) || (i1 == ~0 && i0 < 0))
> return GEN_INT (i0);
>
> <---HERE--->
>
> /* We use VOIDmode for integers. */
> value = rtx_alloc (CONST_DOUBLE);
> PUT_MODE (value, VOIDmode);
>
> CONST_DOUBLE_LOW (value) = i0;
> CONST_DOUBLE_HIGH (value) = i1;
>
> for (i = 2; i < (sizeof CONST_DOUBLE_FORMAT - 1); i++)
> XWINT (value, i) = 0;
>
> return lookup_const_double (value);
>
> [A] treats i0 and i1 as a sign-extended value. So if we
> removed the assert (or moved it to the suggested place):
>
> immed_double_const (-1, -1, 4_hwi_mode)
>
> would create -1 in 4_hwi_mode, represented as a CONST_INT.
> The three implicit high-order HWIs are -1. That's fine,
> because CONST_INT has long been defined as sign-extending
> rather than zero-extending.
>
> But if we fail the [A] test, we go on to create a CONST_DOUBLE.
> The problem is that AIUI we have never defined what happens for
> CONST_DOUBLE if the mode is wider than 2 HWIs. Again AIUI,
> that's why the assert is there.
>
> This matters because of things like the handling in simplify_immed_subreg
> (which, e.g., we use to generate CONST_DOUBLE pool constants, split
> constant moves in lower-subreg.c, etc.). CONST_INT is already
> well-defined to be a sign-extended constant, and we handle it correctly:
>
> switch (GET_CODE (el))
> {
> case CONST_INT:
> for (i = 0;
> i < HOST_BITS_PER_WIDE_INT && i < elem_bitsize;
> i += value_bit)
> *vp++ = INTVAL (el) >> i;
> /* CONST_INTs are always logically sign-extended. */
> for (; i < elem_bitsize; i += value_bit)
> *vp++ = INTVAL (el) < 0 ? -1 : 0;
> break;
>
> But because of this assert, the equivalent meaning for
> CONST_DOUBLE has never been defined, and the current code
> happens to zero-extend it:
>
> case CONST_DOUBLE:
> if (GET_MODE (el) == VOIDmode)
> {
> /* If this triggers, someone should have generated a
> CONST_INT instead. */
> gcc_assert (elem_bitsize > HOST_BITS_PER_WIDE_INT);
>
> for (i = 0; i < HOST_BITS_PER_WIDE_INT; i += value_bit)
> *vp++ = CONST_DOUBLE_LOW (el) >> i;
> while (i < HOST_BITS_PER_WIDE_INT * 2 && i < elem_bitsize)
> {
> *vp++
> = CONST_DOUBLE_HIGH (el) >> (i - HOST_BITS_PER_WIDE_INT);
> i += value_bit;
> }
> /* It shouldn't matter what's done here, so fill it with
> zero. */
> for (; i < elem_bitsize; i += value_bit)
> *vp++ = 0;
> }
>
> So the upshot is that:
>
> immed_double_const (-1, -1, 4_hwi_mode)
>
> sign-extends i1 (the second -1), creating (-1, -1, -1, -1). But:
>
> immed_double_const (0, -1, 4_hwi_mode)
>
> effectively (as the code falls out at the moment) zero-extends it,
> creating (0, -1, 0, 0). That kind of inconsistency seems wrong.
>
> So what I was trying to say was that if we remove the assert
> altogether, and allow CONST_DOUBLEs to be wider than 2 HWIs,
> we need to define what the "implicit" high-order HWIs of a
> CONST_DOUBLE are, just like we already do for CONST_INT.
> If we remove the assert altogether, it very much matters
> what is done by that last "*vp" line.
>
> If Mike or anyone is up to doing that, then great. But if instead
> it's just a case of handling zero correctly, moving rather than
> removing the assert seems safer.
>
> I'm obviously not explaining this well :-)
Ok, I see what you mean. Yes, moving the assert past the GEN_INT
case (though that is specifically meant to deal with the VOIDmode case
I think?) is ok.
Thanks,
Richard.
> Richard
