c++ - ostream operator<< 调用父 ostream

Question

代码:

cout << "11122333" << endl;

期待:
11122333\n
结果:
11122333\n
好的。
代码:

cout.operator<<("11122333");
cout.operator<<(endl);

期待:
11122333\n
结果:
00B273F8\n
(或其他一些地址，它被转换为 void* :( )

麻烦:想要写派生自 ostream 的类

class SomeStream : public ostream
{
  public:
  explicit SomeStream(streambuf* sb) : ostream(sb) { }
  template <typename T> SomeStream &operator <<(const T &val) 
  {
    std::ostream::operator<<(val); //Trouble in there!
    std::cout << "<<" << " " << typeid(T).name() << " " << val << std::endl;
    /*some other behavior*/
    return *this;
  }
  SomeStream &operator <<(ostream& (*val) (ostream&))
  {
    std::ostream::operator<<(val);
    /*some other behavior*/
    return *this;
  }
  SomeStream &operator <<(ios_base& (*val) (ios_base&))
  {
    std::ostream::operator<<(val);
    /*some other behavior*/
    return *this;
  }
};

当我调用父运营商时 std::ostream::operator<<(val); val 转换为 void*而不是正常工作。怎么做才对？以及为什么直接调用 operator<<对于 ostream工作方式与间接调用不同。

Answer 1

输出 operator <<对于 const char*不是 ostream 类型的成员. Only those重载是成员函数，其中之一用于 void* .还有non-member overloads .

有解决方法:

  template <typename T> SomeStream &operator <<(const T &val) 
  {
    static_cast<std::ostream&>(*this) << val; //Trouble in there!
    std::cout << "<<" << " " << typeid(T).name() << " " << val << std::endl;
    /*some other behavior*/
    return *this;
  }