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Re: Option to make unsigned->signed conversion always well-defined?
On Thu, Oct 6, 2011 at 11:04 AM, Miles Bader <miles@gnu.org> wrote:
> How about:
>
> bool overflowbit2(unsigned int a, unsigned int b)
> {
> const unsigned int sum = a + b;
> return ~(a ^ b) & sum & 0x80;
> }
>
> ?
>
> I thinnnnk it has the same results as your function...
> [I just made a table of all 8 possibilities, and checked!]
Miles, it is not the same. Take for example (0xff, 0xff). In 8-bit 2's
complement, this is (-1, -1) and does not overflow. Your function says
it does.
Em 06-10-2011 12:23, Jeremy Hall escreveu:
> bool overflow(int16_t a, int16_t b)
> {
> const int16_t sum = a + b;
> return sum > INT8_MAX || sum < INT8_MIN;
> }
Jeremy, here you are ignoring the problem of converting from the
unsigned int (in the range 0 to 0xff) to the signed integer that it
represents in 8-bit two's complement. Example: 0xff -> -1.
In practice, casting the unsigned int to int8_t works in most cases, but
it is compiler-defined. We are trying to find a always well-defined
approach that is efficient as well.
> Ops, should have been
>
> return ~(a ^ b) & (a ^ sum) & 0x80
>
> ~(a ^ b) gives 1 in the sign bit position if the signs are the same,
> and (a ^ sum) gives 1 if it's different in the sum.
This is good. Do you think this is suboptimal? How are you evaluating
efficiency? In x86 this generates pretty small code.
<overflow2>:
400524: 8d 04 3e lea (%rsi,%rdi,1),%eax
400527: 31 f8 xor %edi,%eax
400529: 31 f7 xor %esi,%edi
40052b: f7 d7 not %edi
40052d: 21 f8 and %edi,%eax
40052f: 25 80 00 00 00 and $0x80,%eax
400534: c3 retq
--
Pedro Pedruzzi