Reputation: 23
template <typename T> class Queue
{
template<typename T> ostream& operator<< (ostream& print, const Queue <T>& x)
{
print<<"\nThe elements are as : \n";
if(q.f!=-1)
{
int fr=q.f,rr=q.r;
while(fr<=rr)
print<<q.q[fr++]<<" <- ";
}
print<<endl;
}
//other stuffs
};
In main():
Queue<int> q(n); //object creation
cout<<q; //calling the overloaded << function
It is giving me the following error:
C:\Users\user\Desktop\PROGRAMS\queue_using_classes.cpp|16|error: declaration of 'class T'|
C:\Users\user\Desktop\PROGRAMS\queue_using_classes.cpp|3|error: shadows template parm 'class T'|
C:\Users\user\Desktop\PROGRAMS\queue_using_classes.cpp|16|error: 'std::ostream& Queue<T>::operator<<(std::ostream&, const Queue<T>&)' must take exactly one argument
Upvotes: 1
Views: 60
Reputation: 206567
In order to use:
Queue<int> q(n);
cout << q;
The function
ostream& operator<<(ostream& print, const Queue <T>& x)
needs to be defined as a non-member function. See my answer to another question for additional information on this particular overload.
Declaring a friend
function is tricky for class templates. Here's a bare bones program that shows the concept.
// Forward declaration of the class template
template <typename T> class Queue;
// Declaration of the function template.
template<typename T> std::ostream& operator<< (std::ostream& print, const Queue <T>& x);
// The class template definition.
template <typename T> class Queue
{
// The friend declaration.
// This declaration makes sure that operator<<<int> is a friend of Queue<int>
// but not a friend of Queue<double>
friend std::ostream& operator<<<T> (std::ostream& print, const Queue& x);
};
// Implement the function.
template<typename T>
std::ostream& operator<< (std::ostream& print, const Queue <T>& x)
{
print << "Came here.\n";
return print;
}
int main()
{
Queue<int> a;
std::cout << a << std::endl;
}
Upvotes: 1