On 01/11/2018 01:27 AM, Richard Biener wrote:
> On Thu, 11 Jan 2018, Marc Glisse wrote:
>
>> On Tue, 9 Jan 2018, 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 ?
>>
>> Why not? Even for
>> void*f(){return 0;}
>> are any of the properties of the malloc attribute violated? It seems to
>> me that if you reject malloc(0), then even the standard malloc function
>> should be rejected as well...
>
> The constraint for the malloc attribute is that it returns a
> unique pointer for each invocation that cannot alias with any
> other pointer in the program, or NULL. So yes, { return 0; }
> qualifies as 'malloc' (but I wouldn't mark that on its own ;)).
It'd certainly be strange to see something like that marked as malloc.
At some point the result of an allocation has to feed into the return
value.
>
> Also
>
> int x;
>
> void *f(size_t n)
> {
> if (x)
> return malloc (n);
> return 0;
> }
>
> does. The only constraint is really that the value feeding
> the return statement is either literal zero or produced
> by a call with the malloc attribute. So even
Right.
>
> void *__attribute__((malloc)) bar();
>
> void *f(size_t n)
> {
> if (n == 2)
> return bar ();
> return malloc (n);
> }
>
> is malloc.
Also agreed.
jeff
