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"?
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?
Or is my confusion based on some misunderstandings of the language
standard?
