operator overload <<

Code:

std::ostream& operator<<(std::ostream& os, const BmvMessage& bm);

I don't see anything incorrect, but it gives the following error:

error: `std::ostream& BMV::BmvMessage::operator<<(std::ostream&, const BMV::BmvMessage&)' must take exactly one argument.

I do not know why this happens. Any suggestions are welcome. I have done this before and never came across this error. I have also checked online and it looks like:

ostream& operator<< (ostream& out, char c );`

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Take operator<< outside the class, making it a free function. Make it a friend of the class if it needs access to private parts.

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The operator has to be a free function, because its first argument is not of the same type as your class. In general, when you overload a binary operator Foo , the member function version only takes a single argument, and FOO(a, b) means a.Foo(b) .

Since a << b would invoke a.operator<<(b) , but a is the stream, this is of no use for us.

So make a free function, or perhaps a free friend function. Having a public toString member function can help:

class Foo {
public:
  std::string toString() const;
  // ...
};

std::ostream & operator<<(std::ostream & o, const Foo & x) {
  return o << x.toString(); 
}

---

You are using the free form signature to define a member function. Member functions have an implicit this argument, so in your case your member function attempt at overloading operator << would result in a function that takes 3 arguments: implicit this , std::ostream& os and BmvMessage const& bm.

You can't define streaming operators as members, since the first argument needs to be of stream class. Instead, you define them as free functions, possibly friended if needed.

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