在C ++中不匹配'operator <<'（包括字符串和ostream以及重载<<）

Question

I'm trying to build a template for a list with priorities (the template parameters are T for the data and Priority for the priority (for example, if I have a line of students with grades, the T would be the students and their Priority would be their grades). The list contains a Node class, and each Node contains data , priority and a pointer to the next Node.

I tried to overload the << operator for Node and for the list so i could use << for the list (printing each node). for example: if I want to print the list named receptionHour, using this line: cout << endl << "containing: " << receptionHour << endl;

The problem is that the complier doesnt recognize the operators I implemented so it doesnt use them and the line won't compile. the error i'm getting for each line is:

no match for 'operator<<' (operand types are 'std::basic_ostream' and 'mtm::PriorityQueue::Node')

here are my implements for the << operator, for the list (called PriorityQueue) and for the Node. Node(insode the node class witch is inside the list class:

    template<class P, class TT>
    friend ostream& operator<<(ostream& os, Node node){
        os << "[";
        os << node.priority;
        os << ",";
        os << node.data;
        os << "]";
        return os;
    }

list(called PriorityQueue):

template<class P, class TT>
friend ostream& operator<<(PriorityQueue<Priority, T>& queue, std::ostream& os){
    Node* nodePtr = queue.head;
    Node node;
    for(int i = 1; i < queue.sizePQ; i++) {
        node = *nodePtr;
        os << node;
        nodePtr = node.next;
    }
    return os;
}

thanks!

Answer 1

You are not using the templates nor the operator<< overload correctly.

(1) When you declare template<class P, class TT> , your function declaration to use those templates should be function(P first, Class<TT> second) . Even though your class definition (and other functions) is something like template<class Name> Class(Name name) , your cannot use template<class T> function(Class<Name>) because the template function name must match the template class name.

And the overload for Node does not need to use template, you know which type your are using ( Node ).

(2) In the overloading function for operator << , the first argument must be the ostream object.