> On Jul 14, 2014, at 8:57 AM, "Bin.Cheng" <amker.ch...@gmail.com> wrote:
>
> Hi,
> For a simple example like below.
>
> int
> f1 (int p, short i, short n)
> {
> int sum = 0;
>
> do
> {
> sum += i;
> i++;
This here is the same as i = i + 1; which is really
i = (short)(((int)i) + 1);
So it gets convert to using unsigned short to avoid undefined overflow ness of
signed types (which is part of gcc's ir).
Thanks,
Andrew
> }
> while (i < n);
>
> return sum;
> }
>
> When compiling with -O2 -fdump-tree-all options, GCC introduces
> unsigned type at the very beginning of gimple pass, for example, the
> dump for gimple pass is like below.
> f1 (int p, short int i, short int n)
> {
> int D.4116;
> short int i.0;
> unsigned short i.1;
> unsigned short D.4119;
> int D.4120;
> int sum;
>
> sum = 0;
> <D.4111>:
> D.4116 = (int) i;
> sum = D.4116 + sum;
> i.0 = i;
> i.1 = (unsigned short) i.0;
> D.4119 = i.1 + 1;
> i = (short int) D.4119;
> if (i < n) goto <D.4111>; else goto <D.4112>;
> <D.4112>:
> D.4120 = sum;
> return D.4120;
> }
>
> It uses i.1 to increase the induction variable and converts it back to
> signed type for comparison. Is it designed behavior? &why?
>
> Thanks,
> bin
