ebevhan added inline comments.

================
Comment at: clang/lib/Basic/FixedPoint.cpp:242
+  } else
+    Overflowed = Result < Min || Result > Max;
+
----------------
rjmccall wrote:
> ebevhan wrote:
> > ebevhan wrote:
> > > rjmccall wrote:
> > > > If the maximum expressible value is *k*, and the fully-precise 
> > > > multiplication yields *k+e* for some epsilon *e* that isn't 
> > > > representable in the result semantics, is that considered an overflow?  
> > > > If so, I think you need to do the shift after these bound checks, since 
> > > > the shift destroys the difference between *k* and *k+e*.  That is, 
> > > > unless there's a compelling mathematical argument that it's not 
> > > > possible to overflow only in the fully-precision multiplication — but 
> > > > while I think that's possibly true of `_Fract` (since *k^2 < k*), it 
> > > > seems unlikely to be true of `_Accum`, although I haven't looked for a 
> > > > counter-example.  And if there is a compelling argument, it should 
> > > > probably be at least alluded to in a comment.
> > > > 
> > > > Would this algorithm be simpler if you took advantage of the fact that 
> > > > `APFixedPointSemantics` doesn't have to correspond to a real type?  You 
> > > > could probably just convert to a double-width common semantics, right?
> > > > If the maximum expressible value is *k*, and the fully-precise 
> > > > multiplication yields *k+e* for some epsilon *e* that isn't 
> > > > representable in the result semantics, is that considered an overflow? 
> > > > If so, I think you need to do the shift after these bound checks, since 
> > > > the shift destroys the difference between *k* and *k+e*.
> > > 
> > > I don't think I would consider that to be overflow; that's precision 
> > > loss. E-C considers these to be different:
> > > 
> > > > If the source value cannot be represented exactly by the fixed-point 
> > > > type, the source value is rounded to either the closest fixed-point 
> > > > value greater than the source value (rounded up) or to the closest 
> > > > fixed-point value less than the source value (rounded down).
> > > >
> > > > When the source value does not fit within the range of the fixed-point 
> > > > type, the conversion overflows. [...]
> > > >
> > > > [...]
> > > >
> > > > If the result type of an arithmetic operation is a fixed-point type, 
> > > > [...] the calculated result is the mathematically exact result with 
> > > > overflow handling and rounding performed to the full precision of the 
> > > > result type as explained in 4.1.3. 
> > > 
> > > There is also no value of `e` that would affect saturation. Any full 
> > > precision calculation that gives `k+e` must be `k` after downscaling, 
> > > since the bits that represent `e` must come from the extra precision 
> > > range. Even though `k+e` is technically larger than `k`, saturation would 
> > > still just give us `k` after truncating out `e`, so the end result is the 
> > > same.
> > > 
> > > > Would this algorithm be simpler if you took advantage of the fact that 
> > > > APFixedPointSemantics doesn't have to correspond to a real type? You 
> > > > could probably just convert to a double-width common semantics, right?
> > > 
> > > It's likely possible to use APFixedPoint in the calculations here, but I 
> > > used APInt to make the behavior explicit and not accidentally be 
> > > dependent on the behavior of APFixedPoint's conversions or operations.
> > Although.,. I guess I see your point in that an intermediate result of k+e 
> > technically "does not fit within the range of the fixed-point type"... but 
> > I wonder if treating such cases as overflow is particularly meaningful. I 
> > don't find there to be much of a distinction between such a case and the 
> > case where the exact result lands inbetween two representable values. We 
> > just end up with a less precise result.
> Right, I was wondering if there was an accepted answer here.  For saturating 
> arithmetic, it's equivalent to truncate this extra precision down to *k* or 
> to saturate to the maximum representable value, since by assumption that was 
> just *k*; but for non-saturating arithmetic, it affects whether the operation 
> has UB.  All else being the same, it's better to have fewer corner-case 
> sources of UB.
> 
> My read is that Embedded C is saying there's a sequence here: compute the 
> exact mathematical result; round that to the precision of the result type; 
> the operation overflows if the rounded result is not representable in the 
> result type.  Is the rounding direction completely unspecified, down to being 
> potentially operand-specific?  If so, we could just say that we always  round 
> to avoid overflow if possible.  The main consideration here is that we need 
> to give the operation the same semantics statically and dynamically, and I 
> don't know if there's any situation where those semantics would affect the 
> performance of the operation when done dynamically.
> For saturating arithmetic, it's equivalent to truncate this extra precision 
> down to *k* or to saturate to the maximum representable value, since by 
> assumption that was just *k*; but for non-saturating arithmetic, it affects 
> whether the operation has UB.

I'm fairly sure that the conclusions here about k and e only hold if k truly is 
the maximum representable value. If k is anything else (even 
epsilon-of-the-representable-range less), k+e can never be greater than the 
maximum.

And actually, crunching the numbers on this... If we have integers a and b of 
width N, sign extended to the double bitwidth A and B, there can be no values 
for a and b for which A*B is greater than N_Max<<N (`k`). Taking 8-bit as an 
example: Max is 127, and Max<<8 is 32512. The maximum possible value attainable 
is -128*-128, which is 16384. That isn't even close to the k+e case.

I'm unsure if this reasoning applies in the minimum case as well.

> My read is that Embedded C is saying there's a sequence here: compute the 
> exact mathematical result; round that to the precision of the result type; 
> the operation overflows if the rounded result is not representable in the 
> result type.

I wonder if it's intended to be a sequence. It's starting to feel like it can't 
actually be both cases at the same time.



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  https://reviews.llvm.org/D73186/new/

https://reviews.llvm.org/D73186



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