ostream运算符<<调用父ostream

Question

code:

cout << "11122333" << endl;

expect:
11122333\\n
result:
11122333\\n
All right.
code:

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

expect:
11122333\\n
result:
00B273F8\\n
(or some other address, it's cast to void* :( )

trouble: Want write class derived from ostream

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;
  }
};

When I call parent operator std::ostream::operator<<(val); val cast to void* and not normal work. How it right to do? and why direct call operator<< for ostream work not same as indirect call.

Answer 1

The output operator << for const char* is not a member of type ostream . Only those overloads are member functions, one of them is for void* . There are also non-member overloads .

There is workaround:

  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;
  }