G++编译旧C++回调代码时的数组越界警告是否合理？

我在测试Rich Hickey提出的一款旧C回调实现，使用G 14.2以g++ -c -Wall -O2编译以下简化代码时，触发了数组越界警告。该警告从G++ 11版本开始出现，旧版本无此提示，且在-O0或-O1编译选项下会消失。

测试代码 #

#include <cstddef>
#include <cstring>

struct FunctorBase
{
   typedef void (FunctorBase::*MethodPtr)();

   template<typename Method>
   Method get_method() const
   {
      return *static_cast<Method const*>(static_cast<void const*>(_method));
   }

   FunctorBase(const void* method, std::size_t sz) : _method{}
   {
      std::memcpy(_method, method, sz);
   }

   alignas (MethodPtr) unsigned char _method[sizeof(MethodPtr)];
};

template<class P1>
struct Functor1 : FunctorBase
{
   typedef void (*Thunk)(const FunctorBase&, P1);

   Functor1(Thunk t, const void* mf, size_t sz) : FunctorBase(mf, sz), thunk(t) {}

   void operator()(P1 p1) const
   {
      thunk(*this, p1);
   }

   Thunk thunk;
};

template<class Callee, class Method>
struct MemberOf1stArgTranslator1 : Functor1<Callee>
{
   MemberOf1stArgTranslator1(const Method& m) : Functor1<Callee>(thunk, &m, sizeof(Method)) {}

   static void thunk(const FunctorBase& ftor, Callee callee)
   {
      (callee.*ftor.get_method<Method>())();
   }
};

template<class P1, class TRT, class CallType>
inline MemberOf1stArgTranslator1<P1, TRT (CallType::*)()> create_functor(Functor1<P1>*, TRT (CallType::*const& f)())
{
   typedef TRT (CallType::*Method)();
   return MemberOf1stArgTranslator1<P1, Method>(f);
}

struct MyCallee
{
   void my_f() {}
};

void f()
{
   typedef Functor1<MyCallee&> F;
   F f = create_functor((F*)0, &MyCallee::my_f);
   MyCallee c;
   f(c);
}

编译器警告信息 #

In static member function 'static void MemberOf1stArgTranslator1<Callee, Method>::thunk(const FunctorBase&, Callee) [with Callee = MyCallee&; Method = void (MyCallee::*)()]',
    inlined from 'void Functor1<P1>::operator()(P1) const [with P1 = MyCallee&]' at <source>:31:7,
    inlined from 'void f()' at <source>:65:5:
<source>:44:42: warning: array subscript 'int (**)(...)[0]' is partly outside array bounds of 'MyCallee [1]' [-Warray-bounds=]
   44 |       (callee.*ftor.get_method<Method>())();
      |       ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^~
<source>: In function 'void f()':
<source>:64:13: note: object 'c' of size 1
   64 |    MyCallee c;
      |             ^

疑问 #

我完全无法理解这个警告——代码中并没有MyCallee类型的数组。想请教：这个编译器提示的具体含义是什么？这是代码存在真实问题，还是编译器的误报？

内容的提问来源于stack exchange，提问作者Markus