Overloading Operator<< for a Templated Class

Operator overloading on class templates

// In MyClass.h
MyClass<T>& operator+=(const MyClass<T>& classObj);

// In MyClass.cpp
template <class T>
MyClass<T>& MyClass<T>::operator+=(const MyClass<T>& classObj) {
  // ...
  return *this;
}

This is invalid for templates. The full source code of the operator must be in all translation units that it is used in. This typically means that the code is inline in the header.

Edit: Technically, according to the Standard, it is possible to export templates, however very few compilers support it. In addition, you CAN also do the above if the template is explicitly instantiated in MyClass.cpp for all types that are T- but in reality, that normally defies the point of a template.

More edit: I read through your code, and it needs some work, for example overloading operator[]. In addition, typically, I would make the dimensions part of the template parameters, allowing for the failure of + or += to be caught at compile-time, and allowing the type to be meaningfully stack allocated. Your exception class also needs to derive from std::exception. However, none of those involve compile-time errors, they're just not great code.

Overloading friend operator for template class

You declare operator<< as returning an ostream&, but there is no return statement at all in the method. Should be:

template <class T, class U>
ostream& operator<<(ostream& out, Pair<T,U>& v)
{
    return out << v.val1 << " " << v.val2;
}

Other than that, I have no problems or warnings compiling your code under Visual Studio 2008 with warnings at level 4. Oh, there are the classical linker errors, but that is easily bypassed by moving the template function definition to the class declaration, as explained in the C++ FAQ.

My test code:

#include <iostream>
using namespace std;

template <class T, class U>
class Pair{ 
public:
    Pair(T v1, U v2) : val1(v1), val2(v2){}
    ~Pair(){}
    Pair& operator=(const Pair&);
    friend ostream& operator<<(ostream& out, Pair<T,U>& v)
    {
        return out << v.val1 << " " << v.val2;
    }
private:    
    T val1;
    U val2;
};

int main() {
    Pair<int, int> a(3, 4);
    cout << a;      
}

overloading operator ostream for template class in c++ not working

As the error message said, the friend declaration declares a non-template operator<<, but it's defined as a template, they don't match.

You can make the friend declaration referring to the operator template, e.g.

// forward declaration
template<class T = int, unsigned int SIZE =2>
class FixedPoint;

// declaration
template<class T,unsigned int SIZE>
std::ostream& operator<<(std::ostream& os ,const FixedPoint<T,SIZE>& price);

template<class T, unsigned int SIZE>
class FixedPoint {
   public:
            ...
            friend std::ostream& operator<< <T, SIZE> (std::ostream& os ,const FixedPoint<T, SIZE>& price);
            // or just
            // friend std::ostream& operator<< <> (std::ostream& os ,const FixedPoint& price);
            ...
};

// definition
template<class T,unsigned int SIZE>
inline std::ostream& operator<<(std::ostream& os ,const FixedPoint<T,SIZE>& price){
    os << price.dollars << "." << price.cents;
    return os;
}

Implementing overloads for and operators in a template class

Turning up the warning level of the compiler helps. By using -Wall with g++, I get the following warnings before the linker error.

socc.cc:13:58: warning: friend declaration ‘std::istream& operator>>(std::istream&, MyClass<MYTYPE>&                    )’ declares a non-template function [-Wnon-template-friend]
     friend istream& operator>>(istream&, MyClass<MYTYPE>&);
                                                          ^
socc.cc:13:58: note: (if this is not what you intended, make sure the function template has already                     been declared and add <> after the function name here)
socc.cc:14:57: warning: friend declaration ‘std::ostream& operator<<(std::ostream&, MyClass<MYTYPE>                     ’ declares a non-template function [-Wnon-template-friend]
     friend ostream& operator<<(ostream&, MyClass<MYTYPE>);

You need to use function templates for the operator>> and operator<< functions. You can declare them before the definition of the class with:

// Forward the class template.
template <class MYTYPE> class MyClass;

// Declare the function templates.
template <class MYTYPE> 
std::istream& operator>>(std::istream&, MyClass<MYTYPE>&);

template <class MYTYPE> 
std::ostream& operator<<(st::ostream&, MyClass<MYTYPE>);

Then, you'll have to use the friend declaration with the appropriate template parameter.

// This makes sure that operator>><int> is not a friend of MyClass<double>
// Only operator>><double> is a friend of MyClass<double>
friend std::istream& operator>><MYTYPE>(std::istream&, MyClass<MYTYPE>&);
friend std::ostream& operator<<<MYTYPE>(std::ostream&, MyClass<MYTYPE>);

Here's an updated version of your code that builds for me. I haven't tried to run it.

#include <iostream>
#include <string>
using namespace std;

template <class MYTYPE> class MyClass;

template <class MYTYPE> 
std::istream& operator>>(std::istream&, MyClass<MYTYPE>&);

template <class MYTYPE> 
std::ostream& operator<<(std::ostream&, MyClass<MYTYPE>);

template <class MYTYPE>
class MyClass {
    MYTYPE *myVector;
    int dim;
    string name;
public:
    MyClass(int, string);
    MyClass() {};

    friend std::istream& operator>><MYTYPE>(std::istream&, MyClass<MYTYPE>&);
    friend std::ostream& operator<<<MYTYPE>(std::ostream&, MyClass<MYTYPE>);
};

template <class MYTYPE>
MyClass<MYTYPE>::MyClass(int x, string y) {
    dim = x;
    name = y;
    myVector = new MYTYPE[dim];
}

template <class MYTYPE>
std::istream& operator>>(std::istream& X, MyClass<MYTYPE>& a){
    cout<<"Reading vector: "<<a.name<<endl;
    for(int indice = 0; indice < a.dim; indice++){
        cout<<a.name<<'['<<indice<<"]= ";
        X >> a.myVector[indice];
    }
    return X;
}

template <class MYTYPE>
std::ostream& operator<<(std::ostream& X, MyClass<MYTYPE> a){
    X<<"Vector: "<<a.name<<endl;
    for(int indice = 0; indice < a.dim; indice++)
        X<<a.myVector[indice]<<' ';
        X<<endl;
    return X;
}

int main() {
    MyClass<int> object(4, "Ints vector");
    MyClass<string> object2(5, "String vector");
    cin >> object;
    cin >> object2;
    cout << object;
    cout << object2;
    system("pause");
    return 0;
}

overloading friend operator for template class

This is one of those frequently asked questions that have different approaches that are similar but not really the same. The three approaches differ in who you are declaring to be a friend of your function --and then on how you implement it.

The extrovert

Declare all instantiations of the template as friends. This is what you have accepted as answer, and also what most of the other answers propose. In this approach you are needlessly opening your particular instantiation D<T> by declaring friends all operator<< instantiations. That is, std::ostream& operator<<( std::ostream &, const D<int>& ) has access to all internals of D<double>.

template <typename T>
class Test {
   template <typename U>      // all instantiations of this template are my friends
   friend std::ostream& operator<<( std::ostream&, const Test<U>& );
};
template <typename T>
std::ostream& operator<<( std::ostream& o, const Test<T>& ) {
   // Can access all Test<int>, Test<double>... regardless of what T is
}

The introverts

Only declare a particular instantiation of the insertion operator as a friend. D<int> may like the insertion operator when applied to itself, but it does not want anything to do with std::ostream& operator<<( std::ostream&, const D<double>& ).

This can be done in two ways, the simple way being as @Emery Berger proposed, which is inlining the operator --which is also a good idea for other reasons:

template <typename T>
class Test {
   friend std::ostream& operator<<( std::ostream& o, const Test& t ) {
      // can access the enclosing Test. If T is int, it cannot access Test<double>
   }
};

In this first version, you are not creating a templated operator<<, but rather a non-templated function for each instantiation of the Test template. Again, the difference is subtle but this is basically equivalent to manually adding: std::ostream& operator<<( std::ostream&, const Test<int>& ) when you instantiate Test<int>, and another similar overload when you instantiate Test with double, or with any other type.

The third version is more cumbersome. Without inlining the code, and with the use of a template, you can declare a single instantiation of the template a friend of your class, without opening yourself to all other instantiations:

// Forward declare both templates:
template <typename T> class Test;
template <typename T> std::ostream& operator<<( std::ostream&, const Test<T>& );

// Declare the actual templates:
template <typename T>
class Test {
   friend std::ostream& operator<< <T>( std::ostream&, const Test<T>& );
};
// Implement the operator
template <typename T>
std::ostream& operator<<( std::ostream& o, const Test<T>& t ) {
   // Can only access Test<T> for the same T as is instantiating, that is:
   // if T is int, this template cannot access Test<double>, Test<char> ...
}

Taking advantage of the extrovert

The subtle difference between this third option and the first is in how much you are opening to other classes. An example of abuse in the extrovert version would be someone that wants to get access into your internals and does this:

namespace hacker {
   struct unique {}; // Create a new unique type to avoid breaking ODR
   template <> 
   std::ostream& operator<< <unique>( std::ostream&, const Test<unique>& )
   {
      // if Test<T> is an extrovert, I can access and modify *any* Test<T>!!!
      // if Test<T> is an introvert, then I can only mess up with Test<unique> 
      // which is just not so much fun...
   }
}

Overloading operator for a templated class

The problem is that the compiler is not trying to use the templated operator<< you provided, but rather a non-templated version.

When you declare a friend inside a class you are injecting the declaration of that function in the enclosing scope. The following code has the effect of declaring (and not defining) a free function that takes a non_template_test argument by constant reference:

class non_template_test
{
   friend void f( non_template_test const & );
};
// declares here:
// void f( non_template_test const & ); 

The same happens with template classes, even if in this case it is a little less intuitive. When you declare (and not define) a friend function within the template class body, you are declaring a free function with that exact arguments. Note that you are declaring a function, not a template function:

template<typename T>
class template_test
{
    friend void f( template_test<T> const & t );
};
// for each instantiating type T (int, double...) declares:
// void f( template_test<int> const & );
// void f( template_test<double> const & );

int main() {
    template_test<int> t1;
    template_test<double> t2;
}

Those free functions are declared but not defined. The tricky part here is that those free functions are not a template, but regular free functions being declared. When you add the template function into the mix you get:

template<typename T> class template_test {
   friend void f( template_test<T> const & );
};
// when instantiated with int, implicitly declares:
// void f( template_test<int> const & );

template <typename T>
void f( template_test<T> const & x ) {} // 1

int main() {
   template_test<int> t1;
   f( t1 );
}

When the compiler hits the main function it instantiates the template template_test with type int and that declares the free function void f( template_test<int> const & ) that is not templated. When it finds the call f( t1 ) there are two f symbols that match: the non-template f( template_test<int> const & ) declared (and not defined) when template_test was instantiated and the templated version that is both declared and defined at 1. The non-templated version takes precedence and the compiler matches it.

When the linker tries to resolve the non-templated version of f it cannot find the symbol and it thus fails.

What can we do? There are two different solutions. In the first case we make the compiler provide non-templated functions for each instantiating type. In the second case we declare the templated version as a friend. They are subtly different, but in most cases equivalent.

Having the compiler generate the non-templated functions for us:

template <typename T>
class test 
{
   friend void f( test<T> const & ) {}
};
// implicitly

This has the effect of creating as many non-templated free functions as needed. When the compiler finds the friend declaration within the template test it not only finds the declaration but also the implementation and adds both to the enclosing scope.

Making the templated version a friend

To make the template a friend we must have it already declared and tell the compiler that the friend we want is actually a template and not a non-templated free function:

template <typename T> class test; // forward declare the template class
template <typename T> void f( test<T> const& ); // forward declare the template
template <typename T>
class test {
   friend void f<>( test<T> const& ); // declare f<T>( test<T> const &) a friend
};
template <typename T> 
void f( test<T> const & ) {}

In this case, prior to declaring f as a template we must forward declare the template. To declare the f template we must first forward declare the test template. The friend declaration is modified to include the angle brackets that identify that the element we are making a friend is actually a template and not a free function.

Back to the problem

Going back to your particular example, the simplest solution is having the compiler generate the functions for you by inlining the declaration of the friend function:

template <typename T>
class BinaryTree {
   friend std::ostream& operator<<( std::ostream& o, BinaryTree const & t ) {
      t.dump(o);
      return o;
   }
   void dump( std::ostream& o ) const;
};

With that code you are forcing the compiler into generating a non-templated operator<< for each instantiated type, and that generated function delegates on the dump method of the template.

Overloading operator on template classes

Your code declares the operator<< as a member function, so it would actually take the this pointer as first argument and ostream as second. Instead it needs to be a free function:

template<class NType> class Node {
public:
    NType data;
    Node *prev, *next;
};
//Note how this is declared outside of the class body, so it is a free function instead of a memberfunction
template<class NType> inline std::ostream& operator<<(std::ostream& out, const Node<NType>& val){
    out << val.data;
    return out;
}

however if your operator<< needs access to private data you need to declare it as a friend function instead:

template<class NType> class Node {
public:
    NType data;
    Node *prev, *next;
    friend std::ostream& operator<<(std::ostream& out, const Node& val){
        out << val.data;
        return out;
    }
};

Now for your output: If your operator<< was invoked the compiler would know the type of NType and do the right thing when streaming the data member. However since your operator<< should not have worked (as written) and it seems to give you memoryaddresses as output I would assume you have something like the following:

Node* n = new Node();
std::cout<<n;
//When it should be:
std::cout<<*n;

Now just for curiosity sake: Why are you implementing what looks curiously like a linked list yourself instead of simply using std::list?

Edit:
Now that we can see the testcase it seems the assumptions about how the operator<< was called was correct. The output needs to be changed to:

std::cout << "Node 1: " << *n1 << std::endl;
std::cout << "Node 2: " << *n2 << std::endl;

to actually invoke the operator<< for Node, instead of the generic one for T*

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