Hi!
On Mon, Jun 29, 2020 at 03:31:48PM -0700, Carl Love wrote:
> On Mon, 2020-06-29 at 16:58 -0500, Segher Boessenkool wrote:
> > On Mon, Jun 29, 2020 at 02:29:54PM -0700, Carl Love wrote:
> > > Segher:
> > >
> > > On Thu, 2020-06-25 at 17:39 -0500, Segher Boessenkool wrote:
> > > > > +;; Return 1 if op is a constant 32-bit floating point value
> > > > > +(define_predicate "f32bit_const_operand"
> > > > > + (match_code "const_double")
> > > > > +{
> > > > > + if (GET_MODE (op) == SFmode)
> > > > > + return 1;
> > > > > +
> > > > > + else if ((GET_MODE (op) == DFmode) && ((UINTVAL (op) >> 32)
> > > > > ==
> > > > > 0))
> > > > > + {
> > > > > + /* Value fits in 32-bits */
> > > > > + return 1;
> > > > > + }
> > > > > + else
> > > > > + /* Not the expected mode. */
> > > > > + return 0;
> > > > > +})
> > > >
> > > > I don't think this is the correct test. What you want to see is
> > > > if
> > > > the
> > > > number in "op" can be converted to an IEEE single-precision
> > > > number,
> > > > and
> > > > back again, losslessly. (And subnormal SP numbers aren't allowed
> > > > either, but NaNs and infinities are).
> > >
> > > The predicate is used with the xxsplitw_v4sf define_expand. The
> > > "user"
> > > claims the given immediate bit pattern is the bit pattern for a
> > > single
> > > precision floating point number. The immediate value is not
> > > converted
> > > to a float. Rather we are splatting a bit pattern that the "user"
> > > already claims represents a 32-bit floating point value. I just
> > > need
> > > to make sure the immediate value actually fits into 32-bits.
> > >
> > > I don't see that I need to check that the value can be converted to
> > > IEEE float and back.
> >
> > Ah, okay. Can you please put that in the function comment
> > then? Right
> > now it says
> > ;; Return 1 if op is a constant 32-bit floating point value
> > and that is quite misleading.
>
> Would the following be more clear
>
> ;; Return 1 if op is a constant bit pattern representing a floating
> ;; point value that fits in 32-bits.
Yes... But I still don't get it.
Say you have the number 785.066650390625 . As a single-precision IEEE
float, that is x4444_4444. But as a double-precision IEEE float, it is
0x4088_8888_8000_0000, which does not fit in 32 bits!
Seghr
