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Hi there,
I think what you're probably missing is that a signed 16-bit integer has
one more possible value: -32768 (0x8000). Apparently you want to clamp
the result to +/-32767 (not including -32768), so you'll need to check
for this result and correct accordingly.

For adding positive integers you've already got it. The smallest
possible overflow is something like 16384+16384=32768=0x8000=-32768 and
so the sign bit flips. The largest possible overflow is something like
32767+32767=65534=0xFFFE=-2 and so the sign bit also flips.

For adding a negative and a nonnegative integer you're also right (and
be aware that I'm saying *nonnegative* and not *positive* for a reason -
zero has MSB=0 here, but it's not strictly positive). The general case
could potentially end with -32768 (if you added 0 + -32768), but you say
neither of your input numbers can ever be -32768 so this isn't a problem
here.

For adding two negative numbers, the largest addition that doesn't
overflow is something like -16384-16383=-32767=0x8001. You then have the
special overflow case, -16384-16384=-32768=0x8000, which should be
checked for and translated to -32767. You then have the general overflow
cases, which range from -16384-16385=-32769=0x7FFF=32767 and so on up to
- -32767-32767=-65534=2. So you can see here that the overflow cases all
lead to either positive numbers *or* the number -32768.

Note that it's fairly easy to test for both the special and general
overflow cases in the negative-negative addition, because if you
consider the result as though it were unsigned, you just need to check
"is result <= 0x8000".

There may be fancier ways to combine these cases using the STATUS
register bits.

Chris

Matthew Rhys-Roberts wrote:
[snip]
| If input and output words are both negative... well this is where I go
| snow-blind.
|
| I first thought that if the sign bit went from 1 to 0, then a rollover
| had occurred. But this didn't account for all cases.
[snip]
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