On 11 January 2018 at 04:50, Jeff Law <l...@redhat.com> wrote:
> 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?
Not sure. Should I make it more conservative by marking it as malloc
only if the argument to __builtin_malloc
is constant or it's value-range is known not to include 0? And
similarly for __builtin_calloc ?
Thanks,
Prathamesh
> Jeff
