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...
--
Marc Glisse
