当通过基指针访问时,编译器如何确定成员的位置

试试这个:

C* x = new C();
B* y = x;
cout << x << " " << y << endl;

我的输出:

0x1ae2010 0x1ae2074

C* 到 B* (或在另一个方向)包括必要的指针算法。

x和y指针实际上指向B和C对象的B子对象(使用最派生的类型)。

Example:

int main() {
  C c;
  B *pb = &c; // converting the pointer here actually adjusts the address

  void *pvc = &c; // doesn't adjust, merely type conversion
  void *pvb = pb; // doesn't adjust, merely type conversion

  cout << boolalpha;
  cout << (pvc == pvb) << '\n'; // prints false
  cout << (&c == pb) << '\n'; // prints true!
                              // (&c is converted to a B* for the comparison)

  C *pc = static_cast<C*>(pb); // this static_cast, which is UB if pb doesn't
    // actually point to a C object, also adjusts the address

  cout << (pc == &c) << '\n'; // prints true

  return 0;
}

不 重新解释强制转换),并且只能从void转换为最初用于转换为void的完全相同类型:

reinterpret_cast<C*>(pvb)  // undefined behavior