Richard Guenther <> 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);


    /* 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;

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)
                    = 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 :-)


Reply via email to