That's useful to know - thanks Florian. So it's possible to forego the
-1 check if no downsizing occurs.
I suppose it makes sense... if we go by the signed 64-bit equivalents,
$FFFFFFFF80000000 div $FFFFFFFFFFFFFFFF = $0000000080000000, which when
typecast to a LongInt does result in a signed overflow that isn't
checked without -Co, so the answer is $80000000. On the flip side,
$0000000080000000 div $FFFFFFFFFFFFFFFF = $FFFFFFFF80000000, which when
typecast to a LongInt is also equal to $80000000, this time without an
overflow. Even though the original operand of $0000000080000000 is out
of range for a LongInt, it's perfectly fine as an Int64.
If my logic is correct, the -1 check is not needed in the following
conditions:
* The division is being upsized.
* There is no change in size and neither source is unsigned (e.g.
Cardinal --> LongInt would require the check).
Currently, as given in my original code example, LongInt(LongInt div
LongInt) is given the -1 check (assuming Integer = LongInt), and this
just seems like a waste. Would you approve this change Florian? (I'll
also add a comment to explain about the downsizing situation)
Kit
----
On 19/05/2023 21:55, Florian Klämpfl via fpc-devel wrote:
Am 19.05.23 um 21:14 schrieb J. Gareth Moreton via fpc-devel:
So I need to ask... should the check for a divisor of -1 still be
performed?
Yes. This is the result of "down sizing" a division. In case of
longint(int64 div int64) can be converted only into longint(int64) div
longint(int64) if this check is carried out. longint($80000000 div
$ffffffff) must silently result in $8000000 in this case.
The case of doing "min_int div -1", even with unsigned-to-signed
typecasting, seems very contrived and the programmer should expect
problems if "min_int" and "-1" appear as the operands. Is there a
specific example where this implicit check is absolutely necessary?
As others have pointed out, silently returning "min_int" as the
answer seems more unexpected (granted this is just the behaviour of
an optimisation that converts the nodes equating to "x div -1" to
"-x", and Intel's NEG instruction doesn't return an error if min_int
is its input operand, but I can't be sure if the same applies to
non-Intel processors and their equivalent instructions).
Kit
On 17/05/2023 09:51, J. Gareth Moreton via fpc-devel wrote:
Logically yes, but using 16-bit as an example, min_int is -32,768,
and signed 16-bit integers range from -32,768 to 32,767. So -32,768
÷ -1 = 32,768, which is out of range. This is where the problem lies.
Internally, negation involves inverting all of the bits and then
adding 1 (essentially how you subtract a number using two's
complement), so min_int, which is 1000000000000000, becomes
0111111111111111 and then, after incrementing, 1000000000000000,
which is min_int again.
Kit
On 16/05/2023 13:13, Jean SUZINEAU via fpc-devel wrote:
Le 16/05/2023 à 01:47, Stefan Glienke via fpc-devel a écrit :
min_int div -1
"min_int div -1" should give "- min_int" ?
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