On 01/09/2018 05:57 AM, Prathamesh Kulkarni wrote:
>
> As Jakub pointed out for the case:
> void *f()
> {
> return __builtin_malloc (0);
> }
>
> The malloc propagation would set f() to malloc.
> However AFAIU, malloc(0) returns NULL (?) and the function shouldn't
> be marked as malloc ?
This seems like a pretty significant concern. Given:
return n ? 0 : __builtin_malloc (n);
Is the function malloc-like enough to allow it to be marked?
If not, then ISTM we have to be very conservative in what we mark.
foo (n, m)
{
return n ? 0 : __builtin_malloc (m);
}
Is that malloc-like enough to mark?
Jeff
