operator<<(ostream&, X) for class X 嵌套在 class 模板中

Question

This one compiles and works like it should (non-nested template):

#include <iostream>

template<typename T> class Z;

template <typename T> 
std::ostream& operator<< (std::ostream& os, const Z<T>&) {
    return (os << "Z");
}

template<typename T> class Z {
    friend std::ostream& operator<< <> (std::ostream& os, const Z&);
};

int main () {
    Z<int> z;
    std::cout << z << std::endl;
}

This one doesn't compile (gcc 4.4 and gcc 4.6, in both 03 and 0x mode):

#include <iostream>

template<typename T> class Z;

template<typename T> 
std::ostream& operator<< (std::ostream& os, const typename Z<T>::ZZ&) {
    return (os << "ZZ!");
}

template <typename T> class Z {
  public:
    class ZZ {
        friend std::ostream& operator<< <> (std::ostream& os, const ZZ&);
    };
};


int main () {
    Z<int>::ZZ zz;
    std::cout << zz << std::endl;
}

The error message looks like this:

error: template-id ‘operator<< <>’ for ‘std::ostream& operator<<(std::ostream&,
const Z<int>::ZZ&)’ does not match any template declaration
error: no match for ‘operator<<’ in ‘std::cout << zz’

In the 0x mode the second error message is different, but the meaning is the same.

Is it possible to do what I want to do?

EDIT Apparently, there's an instance of non-deduced context here, which explains the error messages. The question, however, still stands: can I have a working operator<< for a class nested in a class template?

Answer 1

This is a general problem for functions:

template <typename C>
void func(typename C::iterator i);

Now, if I call func(int*) , which value of C should I use?

In general, you cannot work backward ! Many different C could have defined an internal type iterator that happens to be a int* for some set of parameters.

In your case, you are complicating the situation a bit:

template <typename T>
void func(typename Z<T>::ZZ const&);

But fundamentally this is the same issue, Z<T> is a template, not a full class, and you are asking to create a function for the inner type ZZ of this template.

Suppose I do:

template <typename T>
struct Z { typedef T ZZ; };

template <typename T>
struct Z<T const> { typedef T ZZ; };

Note: typical of iterators, the value_type is not const-qualified

Then, when invoking func(int) , should I use Z<int> or Z<int const> ?

It is non-deducible.

And thus the whole thing is referred to as a non-deducible context , and the Standard forbids it because there is no sensible answer.

Rule of Thumb: be suspicious of typename in the parameters of a function.

Note: they are OK if another argument already pinned down the type, example typename C::iterator find(C&, typename C::const_reference); because once C is deduced, then C::const_reference may be used without trouble

Answer 2

Apart from the problem that the friend declaration doesn't match the operator template (perhaps fixable as)

class ZZ {
    template<class U>
    friend std::ostream& operator<<(std::ostream& os, const ZZ&);
};

you also have a problem with a "non-deduced context", which is what Matthieu links to.

In this template

template<typename T> 
std::ostream& operator<< (std::ostream& os, const typename Z<T>::ZZ&) {
    return (os << "ZZ!");
}

the compiler isn't able to figure out for what T's you parameter will match. There could be several matches, if you specialize for some types

template<>
class Z<long>
{
public:
    typedef double   ZZ;
};

template<>
class Z<bool>
{
 public:
    typedef double   ZZ;
};

Now if I try to print a double , T could be either bool or long .

The compiler cannot know this for sure without checking for all possible T's, and it doesn't have to do that. It just skips your operator instead.

Answer 3

Matthieu explained the problem very well, but a simple work-around can be used in this case. You can implement the friend function inside the class with the friend declaration:

#include <iostream>

template <typename T> class Z {
public:
  class ZZ {
    friend std::ostream& operator<< (std::ostream& os, const ZZ&) {
      return os << "ZZ!";
    }
  };
};


int main () {
  Z<int>::ZZ zz;
  std::cout << zz << std::endl;
}

Answer 4

A different technique is to provide an out of line class template which works like a hook.

template <typename T> class ZZZ { 
  T &getDerived() { return static_cast<T&>(*this); }
  T const &getDerived() const { return static_cast<T const&>(*this); }

protected:
  ~ZZZ() { }
};

template <typename T> class Z {
  public:
    class ZZ : public ZZZ<ZZ> {
    };
};

template<typename ZZ> 
std::ostream& operator<< (std::ostream& os, const ZZZ<ZZ> &zzz) {
    ZZ const& zz = zzz.getDerived();
    /* now you can use zz */
    return (os << "ZZ!");
}

Using a friend function definition is preferable though. But it's good to know about alternatives.