为什么在虚函数中转换为Base类会产生分段错误？

Question

我想用.打印出派生类operator<<.当我打印派生类时,我想首先打印它的基础,然后打印它自己的内容.

但是我遇到了一些麻烦(见下面的segfault):

class Base {
 public:
  friend std::ostream& operator<<(std::ostream&, const Base&);
  virtual void Print(std::ostream& out) const {
    out << "BASE!";
  }
};
std::ostream& operator<<(std::ostream& out, const Base& b) {
  b.Print(out);
  return out;
}

class Derived : public Base {
 public:
  virtual void Print(std::ostream& out) const {
    out << "My base: ";
    //((const Base*)this)->Print(out); // infinite, calls this fct recursively
    //((Base*)this)->Print(out);       // segfault (from infinite loop?)                                                          
    ((Base)*this).Print(out);          // OK
    out << " ... and myself.";
  }
};

int main(int argc, char** argv){
    Derived d;
    std::cout << d;
    return 0;
}

为什么我不能用其中一种方式施展？

    ((const Base*)this)->Print(out); // infinite, calls this fct recursively
    ((Base*)this)->Print(out);       // segfault (from infinite loop?)

Answer 1

试试Base :: Print(out)