ostream must take exactly one argument - c++

Question

I am getting a compile error in my output stream operator and I cannot seem to find out how to fix it as I never received this error before:
linkedList.cpp:258: error: ‘std::ostream& linkedList<T>::operator<<(std::ostream&, linkedList<T>)’ must take exactly one argument
make: *** [linkedList.o] Error 1
Here is the definition of my operator<<
template <class T>
ostream& linkedList<T>::operator<<(ostream &output, linkedList<T> list)
{
node *curr;
curr=list.head;
while(curr!=NULL )
{
output << curr->data;
curr = curr->next;
}
return output;
}
Here it is in my header:
//ostream operator for printing the list
template <class T>
ostream &operator<<(ostream &output, linkedList<T> list);
Any help will be appreciated!

Answer 1

You defined it as a member function, you have to define it as a free standing (probably friend) function, either
outside your class
template <class U>
ostream& operator<<(ostream &output, linkedList<U> list){...}
in which case you also have to declare it inside the class as
template <class U> // note the different type name
friend ostream& operator<<(ostream &output, linkedList<U> list)
or
inside the class as
// no need for template, type is passed automatically the function is not templated anymore
friend ostream& operator<<(ostream &output, linkedList list){...}
The difference between these 2 declarations is a bit subtle, but for your purpose both work equally well. And probably you want to pass linkedList by const reference, linkedList<T>& list.
EDIT
A common mistake is to just declare the friend operator inside the class as
template<typename T>
class linkedlist
{
//....
friend ostream& operator<<(ostream& output, const linkList& list); // declaration only
}
then try to define it outside the class as
template<typename T>
ostream& operator<<(ostream& output, const linkList<T>& list){...}
Guess what? The code will compile, but won't link, as the declaration inside the class declares a non-template function, one for each type T you pass to linkList<T>. Then, when you declare e.g. linkList<int> lst, and try cout << lst, the compiler will see the declaration of the friend, which for it looks like
friend ostream& operator<<(ostream& output, const linkList<int>& list);
and will try searching for its definition. However there is no definition in the rest of the header, only the template operator, so the linker isn't able to find the implementation and will spit a linker error. Live example here.

Answer 2

It's all in the error message. This function that you declared:
template <class T>
ostream &operator<<(ostream &output, linkedList<T> list);
is a member function of linkedList<T> that takes two arguments. All the binary operators (+, *, <<, ... all except for the call operator ()), when defined as member functions, must take exactly one argument, hence the error. What you had intended to do was to declare a non-member function, external to the class:
template <typename T>
class linkedList { .. };
template <typename T>
ostream &operator<<(ostream &output, linkedList<T> list) {
// implementation
}
You could also define it as a non-member friend inside the class definition:
template <class T>
class linkedList {
...
friend ostream &operator<<(ostream &output, linkedList list){
...
}
...
};