Earlier quoted context omitted.
> It does matter, for malloc-returned storage. You can put whatever objects you want into that storage as long as it fits and the pointer is properly aligned. You can certainly store an object of arbitrary type, but here it is done through a pointer to an object with pointer arithmetic going beyond the allowed bounds. > memcpy takes a void pointer as its destination, sets the effective type of the storage behind it A…
> but here it is done through a pointer to an object with pointer arithmetic going beyond the allowed bounds. When you do "void *x = malloc(sizeof(struct node))", the returned storage doesn't have struct node object in it, it has an object of no effective type in it, with size "sizeof(struct node)". Taking the pointer to payload[] field is in no way different from doing "(char*) x + offsetof(struct node, payload)" —…
It may differ, depending on the precise notion of provenance being applicable. If provenance only has allocation granularity, I suppose that there is no difference. I know that there were some discussions about provenance and subobjects. I do not know whether the question is resolved.
Where this gets complicated is that zero-sized arrays are non-standard. So even if we could build a convincing argument from standard notions of provenance, how would it transfer to a subobject that is excluded from the standard?
Last but not least, this is not only about creating the effective type through memcpy. The question is also whether this destroys the effective type of the structure. See my previous point about possible padding after 'payload'.
Please also consider that flexible array members are here for a reason. If I follow your argument, then they bring nothing that arrays of length 0 or 1 do not already cover.