On 20/04/11 09:40, charfi asma wrote:
>
>> 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.
>
> Andrew, 1!=0 is true ;) and the OR expression is if ((y > 1) || (y < 9)) that is
> also true. I did not say that gcc make a mistake in evaluating this but I would
> like to know how it is translated in gimple which generic tree code are used,
> where it does this transfo from if ((y > 1) || (y < 9)) to (1!=0) ? in the
> gimplify function or when it builds generic if it builds it ....
It looks like constant folding.
Hard to say, because no gcc version I have does it.
With 4.6 I get:
x = 6;
y = x + 10;
if (y == 1) goto <D.2069>; else goto <D.2071>;
<D.2071>:
if (y == 0) goto <D.2069>; else goto <D.2070>;
<D.2069>:
D.2072 = 10;
return D.2072;
<D.2070>:
D.2072 = 0;
return D.2072;
Are you using optimization?
Andrew.
