CODEWITHSUNDEEP
c++operator-overloading
CodersSC
CodersSC

Reputation: 710

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 );`

Upvotes: 2

Views: 1087

Answers (3)

Kerrek SB
Kerrek SB

Reputation: 477010

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(); 
}

Upvotes: 3

K-ballo
K-ballo

Reputation: 81349

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.

Upvotes: 1

Fred Foo
Fred Foo

Reputation: 363547

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

Upvotes: 3

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