On 20/04/11 09:21, charfi asma wrote:
>
>
>
>
> ----- Message d'origine ----
> De : Andrew Haley <aph@xxxxxxxxxx>
> À : gcc-help@xxxxxxxxxxx
> Envoyé le : Mar 19 avril 2011, 18h 53min 58s
> Objet : Re: OR in Generic or Gimple
>
> On 04/19/2011 05:48 PM, charfi asma wrote:
>> Hello,
>> In my front end I would like to translate to generic the OR operation (if ( y
>> =a) || (y=b)) ...
>> I tried to look at the gimple generated from this C++ program
>>
>> int main()
>> {
>> int x=6;
>> int y = x+10;
>> if ((y == 1) || (y==0))
>> return 10;
>> else
>> return 0;
>> }
>>
>>
>> I found this:
>>
>>
>> int main() ()
>> {
>> unsigned int y.0;
>> unsigned int D.2070;
>> int D.2073;
>>
>> {
>> int x;
>> int y;
>>
>> x = 6;
>> y = x + 10;
>> y.0 = (unsigned int) y;
>> D.2070 = y.0 + 4294967295;
>> if (D.2070 <= 1) goto <D.2071>; else goto <D.2072>;
>> <D.2071>:
>> D.2073 = 10;
>> return D.2073;
>> <D.2072>:
>> D.2073 = 0;
>> return D.2073;
>> }
>> D.2073 = 0;
>> return D.2073;
>> }
>>
>> I do not really understand how gimple is generated for the OR test.
>
> I don't understand it either. I think it's a bug. I get:
>
> int main() ()
> {
> unsigned int y.0;
> int D.2099;
>
> {
> int x;
> int y;
>
> x = 6;
> y = x + 10;
> y.0 = (unsigned int) y;
> if (y.0 <= 1) goto <D.2097>; else goto <D.2098>;
> <D.2097>:
> D.2099 = 10;
> return D.2099;
> <D.2098>:
> D.2099 = 0;
> return D.2099;
> }
> D.2099 = 0;
> return D.2099;
> }
>
> Andrew.
>
> Hello Andrew,
>
> I get this using both gcc4.4.1 and gcc4.6.
> I even tried with gcc (compile c file) instead of g++ (compile c++ file), but I
> get the same gimple.
>
> however when I change values:
> int x=6;
> int y =x+10;
> if ((y > 1) || (y < 9)) ....
>
> I get this:
>
> int main() ()
> {
> int D.2071;
>
> {
> int x;
> int y;
>
> x = 6;
> y = x + 10;
> if (1 != 0) goto <D.2069>; else goto <D.2070>;
> <D.2069>:
> D.2071 = 10;
> return D.2071;
> <D.2070>:
> D.2071 = 0;
> return D.2071;
> }
> D.2071 = 0;
> return D.2071;
> }
>
> Any way, did so know which generic tree code or gimple statement used to
> generate this ? why if ((y > 1) || (y < 9)) is translated to if (1!=0) ??
Well, we know that x == 6, so y == 6 + 10. Therefore we know that
(y == 1) || (y==0) is false.
So, the expression (y == 1) || (y==0) can be replaced with 1 != 0 .
Andrew.
