lkcl luke.leighton at gmail.com
Sun Apr 17 20:30:51 BST 2022

```
On April 17, 2022 7:17:41 PM UTC, Richard Wilbur <richard.wilbur at gmail.com> wrote:
>
>
>> On Apr 17, 2022, at 11:27, lkcl <luke.leighton at gmail.com> wrote:
>[…]
>>
>>    cat[0:127] = (RB) || (RS)
>>    sum[0:127] = cat + EXTZ(RA) + *128
>>    rhi[0:63] = sum[0:63]
>>    if (RA) <= 1 then rhi = rhi + (*63 || 1)
>>    RA = rhi
>>    RT = sum[64:127]
>
>A question of notation:
>
>I see ‘cat[0:127]’ and think that here we have a case of <signal
>name>[bit high:bit low]

yes.  MSB0 numbering.

>(or vice versa, depending on endianness).  In essence, all 128 bits of
>‘cat’.
>
>I see ‘(*63 || 1)’ and think 64-bit constant value of 0x0000 0000
>0000 0001.

yep.

>So when I see ‘*128’ I think 128-bit constant value of 0xFFFF FFFF
>FFFF FFFF FFFF FFFF FFFF FFFF.

yep

>  If that is the case, I can guarantee
>"sum[0:127] = cat + EXTZ(RA) + *128”
>will overflow!

nnnope.  truncation occurs (wrap) because the destination (sum) is also only 128 bit.

adding 0xffffffff is basically "subtract 1"

l.

```