No Operator
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...
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:
- implement the overloaded << operator for your object OR
- if your
Courseobject is in someway atypedefto a primitive try explicitly casting it to a string type (or similar)