4 ms·

If you do this Derived* derived = (Derived*)base; and then use both base and derived, is that not violating aliasing rules?

• haberman 11y ago
  Pretty sure it's safe! Take this program: typedef struct { int x; } Base; typedef struct { Base base; int y; } Derived; int f(Base* b, Derived* d) { b->x = 0; d->base.x = 1; return b->x; } Notice that if we are accessing the base members of "d", we are still accessing them through a struct of type "Base" (d->base.x). If we compile this with strict aliasing, you can see the output is allowing that the two might alias (while this isn't a proof, it's a strong indication that this is aliasing-correct). 0000000000000000 <f>: 0: c7 07 00 00 00 00 mov DWORD PTR [rdi],0x0 6: c7 06 01 00 00 00 mov DWORD PTR [rsi],0x1 c: 8b 07 mov eax,DWORD PTR [rdi] e: c3 ret
  • vc98mvco 11y ago
    It is defined. 6.5.7. An object shall have its stored value accessed only by an lvalue expression that has one of the following types: - an aggregate or union type that includes one of the aforementioned types among its members (including, recursively, a member of a subaggregate or contained union) This means Derived is allowed to alias Base.
  • TorKlingberg 11y ago
    Thanks, this is good to know.