Unable to declare 2 friend overloaded<< in a template .h

I'm trying to create two overloaded operators in a template BSTree.h and am encountering errors that really don't tell me what the problem is. Running a search on the error codes seperate or in conjunction hasn't yielded anything for me.

The first overloaded<< for the BSTree doesn't cause any errors on compile, but the 2nd overloaded<< I created for my Node struct keeps returning the following errors:

error C4430: missing type specifier - int assumed. Note: C++ does not support default-int

error C2143: syntax error : missing ',' before '*'

#ifndef BSTREE_H
#define BSTREE_H

#include <iostream>
#include <fstream>

template <typename T>
class BSTree{

friend ostream& operator<<(ostream&, const BSTree<T>&); 

public:
    BSTree()开发者_运维百科;
    //BSTree(const BSTree &);
    ~BSTree();

    void buildTree(ifstream&);
    void setType(char);
    bool getType(char);

    bool insert(T*);

    bool isEmpty();


private:
    char type;

    struct Node{
        T* data;

        //subnode[0] == left subtree
        //subnode[1] == right subtree
        Node* subnode[2];
    };

    Node* head;
    void destructorHelper(Node* &);
    bool insertHelper(T*, Node* &);
    friend ostream& operator<<(ostream&, const Node*&); 

};

The compiler says the errors occur at the line where the Node overloaded<< code is.

template <typename T>
ostream& operator<<(ostream &output, const BSTree<T> &out) {
    if(head != NULL)
        output << head;
    return output;
}

template <typename T>
ostream& operator<<(ostream &output, const Node* &out) {
    if(out != NULL){
        output << out->subnode[0];
        output << *out->data;
        output << out->subnode[1];
    }

    return output;
}  

Am I not allowed to declare 2 overloaded<< in the same .h even if they are for different objects? Or am I messing something up in my code?

error C4430: missing type specifier - int assumed. Note: C++ does not support default-int

Usually this means that the compiler doesn't know an identifier as a type, so it goes assuming it's a parameter name, with the type implicitly being int. (In C of old there was a rule that the int in a parameter type could be omitted.) Code like

void foo(bar);

might emit this, if the compiler doesn't know the type bar and assumes void foo(int bar).

This

template <typename T>
std::ostream& operator<<(std::ostream &output, const typename BSTree<T>::Node* &out)
{
  // ...
}

should compile. (Note the qualifications std:: and BSTree::.)

You've got several mistakes in your code:

• you need to add <> after the function names you declare as friend
• even if you're in the body of a friend a function, you need to specify objects you call methods on. Remember that functions are not bound to an instance.
• in the last prototype, Node is not reachable outside the class it's defined in. You must precise BSTree::Node

Possibly you need this:

const BSTree::Node* &out

Node is internal structure.

I suspect the problem is that ostream is not in scope at the time that your operator<<() friend functions are declared.

Either add using std::ostream; just inside class BSTree{, or specify the fully qualified typenames:

friend std::ostream& operator<<(std::ostream&, const BSTree<T>&); 
...
friend std::ostream& operator<<(std::ostream&, const Node*&);

Either way, the actual definitions of these functions will need to be changed similarly. Whatever you do, don't be tempted to use either using std::ostream; or (even worse) using namespace std; in a header file at file scope, because it will affect every subsequent declaration in the translation unit.