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
ret
Andrew.
