--- Comment #4 from pinskia at gcc dot gnu dot org 2006-04-10 18:12 ---
Well really more like:
int f(int *a)
{
int t = *a;
unsigned *b = (unsigned *)a;
int *c = (int*)b;
return *c + t;
}
Which should be the same as:
int f(int *a)
{
return *a * 2;
}
--
--- Comment #3 from pinskia at gcc dot gnu dot org 2006-04-09 17:40 ---
Actually this testcase comes down to:
int f(int *a)
{
*a = 1;
unsigned *b = (unsigned *)a;
int *c = (int*)b;
return *c;
}
And this is caught correctly at the RTL level.
--
pinskia at gcc dot gnu
