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 ;)).
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
void *__attribute__((malloc)) bar();
void *f(size_t n)
{
if (n == 2)
return bar ();
return malloc (n);
}
is malloc.
Richard.
