Given the code snippet of a template type with a friend operator (or function);
template <typename T>
class Foo {
friend std::ostream& operator<< (std::ostream& os, const Foo<T>& a);
};
The operator<< is being declared as a non-template function. For every type T used with Foo, there needs to be a non-templated operator<<. For example, if there is a type Foo<int> declared, then there must be an operator implementation as follows;
std::ostream& operator<< (std::ostream& os, const Foo<int>& a) {/*...*/}
Since it is not implemented, the linker fails to find it and results in the error.
To correct this, you can declare a template operator before the Foo type and then declare as a friend, the appropriate instantiation. The syntax is a little awkward, but is looks as follows;
// forward declare the Foo
template <typename>
class Foo;
// forward declare the operator <<
template <typename T>
std::ostream& operator<<(std::ostream&, const Foo<T>&);
template <typename T>
class Foo {
friend std::ostream& operator<< <>(std::ostream& os, const Foo<T>& a);
// note the required <> ^^^^
// ...
};
template <typename T>
std::ostream& operator<<(std::ostream&, const Foo<T>&)
{
// ... implement the operator
}
The above code limits the friendship of the operator to the corresponding instantiation of Foo, i.e. the operator<< <int> instantiation is limited to access the private members of the instantiation of Foo<int>.
Alternatives include;
Allowing the friendship to extend to all instantiations of the templates, as follows;
template <typename T>
class Foo {
template <typename T1>
friend std::ostream& operator<<(std::ostream& os, const Foo<T1>& a);
// ...
};
Or, the implementation for the operator<< can be done inline inside the class definition;
template <typename T>
class Foo {
friend std::ostream& operator<<(std::ostream& os, const Foo& a)
{ /*...*/ }
// ...
};
Note, when the declaration of the operator (or function) only appears in the class, the name is not available for "normal" lookup, only for argument dependent lookup, from cppreference;
A name first declared in a friend declaration within class or class template X becomes a member of the innermost enclosing namespace of X, but is not accessible for lookup (except argument-dependent lookup that considers X) unless a matching declaration at the namespace scope is provided...
There is further reading on template friends at cppreference and the C++ FAQ.
Code listing showing the techniques above.
As a side note to the failing code sample; g++ warns about this as follows
warning: friend declaration 'std::ostream& operator<<(...)' declares a non-template function [-Wnon-template-friend]
note: (if this is not what you intended, make sure the function template has already been declared and add <> after the function name here)