What is wrong with these types?

Question

In main.cpp:

#include <iostream>
#include "MyType.h"

int main(int argc, const char * argv[]) {
    MyType myobject = MyType();
    std::cout << myobject; //error here
    return 0;
}

in MyType.h:

#include <ostream>

class MyType {
public:
    MyType();
    std::ostream& operator<<(std::ostream &os);
};

in MyType.cpp:

#include "MyType.h"
#include <iostream>

MyType::MyType() {
}

std::ostream& MyType::operator<<(std::ostream &os){
    return os;
}

I am trying to overload << for a custom type. When I do, I get this error: /main.cpp:14:15: Invalid operands to binary expression ('ostream' (aka 'basic_ostream<char>') and 'MyType')

What am I doing wrong here?

Have you searched the FAQ? Did you search StackOverflow for "c++ overloading <<"? — Thomas Matthews, Jan 24 '15 at 02:18

score 7 · Accepted Answer · edited May 23 '17 at 12:12

You declare operator<< as a member function. So it only works if the invoking object is on its left hand side, as

MyType m;
m << (std::cout);

which looks like a monstrosity.

Best is to declare it as a friend,

friend std::ostream& operator<<(std::ostream &os, const MyType& rhs)
{
   os << ...; // do something with rhs here
   return os;
}

then you can "naturally" invoke it

std::cout << myobject;

which is translated as

operator<<(std::cout, myobject);

StackOverflow has a very nice guide about Operator overloading where you can find more details about operator overloading best practices.

Unrelated:

MyType myobject = MyType(); is redundant and inefficient, as you are copy constructing the left hand side from a temporary MyType() that has to be also constructed (although the compiler usually optimizes this via copy elision). It is enough and more efficient to just write MyType myobject.

What is wrong with these types?

1 Answers1