| My understanding is that this is covered by the "effective type" language in both C99 and C11[1]. malloc itself is specified to return a void pointer, but the effective type established during assignment circumvents what would otherwise be UB via aliasing[2]. Edit: Specifically, the reasoning is: 1. Allocated objects have no declared type; 2. 6.5.6: Objects with no declared type have the type of their accessing lvalue 3. 6.5.7: Access of a stored value is valid for lvalue expressions that are type compatible with the effective type. So there's no UB here. `malloc` itself doesn't need to know about the effective type produced by the lvalue, and C (at least C99 onwards) respects that type for strict aliasing purposes. [1]: https://port70.net/%7Ensz/c/c11/n1570.html#6.5p6 [2]: https://port70.net/%7Ensz/c/c11/n1570.html#6.5p7 |
> The lifetime of an allocated object extends from the allocation until the deallocation. Each such allocation shall yield a pointer to an object disjoint from any other object.
If the implementation of malloc is visible, then it can be inlined etc., and then you have to deal with the effects of this obviously not being true.