No operator << matches these operands

I've been reading questions here for an hour or two regarding this error I'm getting and most of them forgot to #include string (which I had already done), or to overload the << operator.

Here's the code in question:

void Student::getCoursesEnrolled(const vector<Course>& c) { for (int i = 0; i < c.size(); i++) { cout << c[i] << endl; } } 

And the error I'm getting:

Error: No operator matches these operands operand types are: std::ostream << const Course 

All I'm trying to do is return the vector. I read about overloading the << operator but we haven't learned any of that in class so I'm assuming there is another way of doing it?

I appreciate your time!

6 Answers

All I'm trying to do is return the vector.

Not quite; you're trying to print it using cout. And cout has no idea how to print a Course object, unless you provide an overloaded operator<< to tell it how to do so:

std::ostream& operator<<(std::ostream& out, const Course& course) { out << course.getName(); // for example return out; } 

See the operator overloading bible here on StackOverflow for more information.

The problem is that operator << is not overload for type Course objects of which you are trying to output in statement

cout << c[i] << endl; 

You need to overload this operator or write your own function that will output an object of type Course in std::ostream

For example let assume that below is a definition of class Course

class Course { private: std::string name; unsigned int duration; public: Course() : duration( 0 ) {} Course( const std::string &s, unsigned int n ) : name( s ), duration( n ) {} std::ostream & out( std::ostream &os ) const { return ( os << "Course name = " << name << ", course duration = " << duration ); } }; 

When you can write

std::vector<Course> v = { { "A", 1 }, { "B", 2 }, { "C", 3 } }; for ( const Course &c : v ) c.out( std::cout ) << std::endl; 

Instead member function out you can overload operator <<. For example

class Course { private: std::string name; unsigned int duration; public: Course() : duration( 0 ) {} Course( const std::string &s, unsigned int n ) : name( s ), duration( n ) {} friend std::ostream & operator <<( std::ostream &os, const Course & c ) { return ( os << "Course name = " << c.name << ", course duration = " << c.duration ); } }; 

and use it as

std::vector<Course> v = { { "A", 1 }, { "B", 2 }, { "C", 3 } }; for ( const Course &c : v ) std::cout << c << std::endl; 

The stream operator << is used to "output" some representation of that object. If you don't want to overload the operator yet just pick some property to output instead:

for (int i = 0; i < c.size(); i++) { cout << c[i].Name << endl; // assuming Name is a property of Course } 

When you DO overload the operator you just decide then what the proper representation of a Course is:

ostream& operator<< (ostream &out, Course &c) { out << c.Name "(" << c.Description << ")"; return out; } 

Your Course class needs to implement an operator:

class Course { public: /* * Your code here */ // Probably missing this: friend std::ostream& operator << (std::ostream& os, const Course& course) { os << course.name(); // etc.. return os; }; }; // eo class Course 

Since you haven't yet learned to overload operator<<, what you can do instead is to print each member of your Course class. You haven't posted the definition of Course, but perhaps it's something like this:

class Course { public: int get_number() { return _number; } const std::string& get_name() { return _name; } private: int _number; std::string _name; }; 

then you can say:

void Student::getCoursesEnrolled(const vector<Course>& c) { for (int i = 0; i < c.size(); i++) { cout << c[i].get_number() << " " << c[i].get_name() << std::endl; } } 

Your problem is this particular part:

cout << c[i]

In your case c[i] is an object of type Course as dvnrrs correctly pointed out. So either:

1. implement the overloaded << operator for your object OR
2. if your Course object is in someway a typedef to a primitive try explicitly casting it to a string type (or similar)