On 06/18/2010 10:11 AM, Georg Lay wrote:
> Andrew Haley schrieb:
>> On 06/18/2010 08:56 AM, Georg Lay wrote:
>>> Hi, I have a question on gcc's signed overflow optimisation in the
>>> following C function:
>>>
>>> int abssat2 (int x)
>>> {
>>> unsigned int y = x;
>>>
>>> if (x < 0)
>>> y = -y;
>>>
>>> if (y >= 0x80000000)
>>> y--;
>>>
>>> return y;
>>> }
>>>
>>> gcc optimises the second comparison and throws it away, and that's the
>>> part I do not understand because all computations are performed on
>>> unsigned int which has no undefined behaviour on overflow.
>>>
>>> For the unary - the standard says in 6.5.3.3.3:
>>> The result of the unary - operator is the negative of its (promoted)
>>> operand. The integer promotions are performed on the operand, and
>>> the result has the promoted type.
>>>
>>> And the promotion rules in 6.3.1.1.2:
>>> If an int can represent all values of the original type, the value
>>> is converted to an int; otherwise, it is converted to an unsigned int.
>>> These are called the integer promotions. All other types are unchanged
>>> by the integer promotions.
>>>
>>> As an int cannot represent all values that can be represented by an
>>> unsigned int, there is no signed int in the line y = -y.
>>>
>>> Could anyone explain this? I see this on gcc 4.4.3.
>>
>> Works for me on gcc 4.4.3:
>>
>> movl %edi, %eax
>> sarl $31, %eax
>> xorl %eax, %edi
>> subl %eax, %edi
>> movl %edi, %eax
>> shrl $31, %eax
>> subl %eax, %edi
>> movl %edi, %eax
>
> Ok. Is your code as it should be on any machine or is it just a "missed
> optimisation"?
Your problem sounds to me like incorrect code, not missed optimization.
> I observe it on a non-standard embedded target that has an abs
> instruction, i.e. there is an abssi2 insn that leads to abs:SI rtx.
> Maybe there is some standard target that also has native support of abs
> to see what happens there?
Perhaps. I think the problem may be that abs is being generated incorrectly.
What does -fdump-tree-optimized look like?
> Or is my confusion based on some misunderstandings of the language
> standard?
Your code is correct as far as I can see.
I don't get any warnings with -Wstrict-overflow.
Andrew.
;; Function abssat2 (abssat2)
Analyzing Edge Insertions.
abssat2 (int x)
{
int prephitmp.15;
unsigned int y;
<bb 2>:
y = (unsigned int) x;
if (x < 0)
goto <bb 3>;
else
goto <bb 4>;
<bb 3>:
y = -y;
<bb 4>:
prephitmp.15 = (int) y;
if (prephitmp.15 < 0)
goto <bb 5>;
else
goto <bb 6>;
<bb 5>:
prephitmp.15 = (int) (y + 4294967295);
<bb 6>:
return prephitmp.15;
}
