Template Classes and Friends
Template class declarations can have friends, too. You can classify friends of templates into three categories:
• Non-template friends
• Bound template friends, meaning the type of the friend is determined by the type of the class when a class is instantiated
• Unbound template friends, meaning that all specializations of the friend are friends to each specialization of the class
Let’s look at examples of each.
Non-Template Friend Functions to Template Classes
Let’s declare an ordinary function in a template class as a friend:
template <class T>
class HasFriend
{
public:
friend void counts(); // friend to all HasFriend instantiations
...
};
This declaration makes the counts() function a friend to all possible instantiations of the template. For example, it would be a friend to the HasFriend<int> class and the HasFriend<string> class.
The counts() function is not invoked by an object (it’s a friend, not a member function), and it has no object parameters, so how does it access a HasFriend object? There are several possibilities. It could access a global object; it could access nonglobal objects by using a global pointer; it could create its own objects; and it could access static data members of a template class, which exist separately from an object.
Suppose you want to provide a template class argument to a friend function. Can you have a friend declaration like this, for example?
friend void report(HasFriend &); // possible?
The answer is no. The reason is that there is no such thing as a HasFriend object. There are only particular specializations, such as HasFriend<short>. To provide a template class argument, then, you have to indicate a specialization. For example, you can use this:
template <class T>
class HasFriend
{
friend void report(HasFriend<T> &); // bound template friend
...
};
To understand what this does, imagine the specialization produced if you declare an object of a particular type:
HasFriend<int> hf;
The compiler would replace the template parameter T with int, giving the friend declaration this form:
class HasFriend<int>
{
friend void report(HasFriend<int> &); // bound template friend
...
};
That is, report() with a HasFriend<int> parameter becomes a friend to the HasFriend<int> class. Similarly, report() with a HasFriend<double> parameter would be an overloaded version of report() that is a friend to the HasFriend<double> class.
Note that report() is not itself a template function; it just has a parameter that is a template. This means that you have to define explicit specializations for the friends you plan to use:
void report(HasFriend<short> &) {...}; // explicit specialization for short
void report(HasFriend<int> &) {...}; // explicit specialization for int
Listing 14.22 illustrates these points. The HasFriend template has a static member ct. Note that this means that each particular specialization of the class has its own static member. The counts() method, which is a friend to all HasFriend specializations, reports the value of ct for two particular specializations: HasFriend<int> and HasFriend<double>. The program also provides two report() functions, each of which is a friend to one particular HasFriend specialization.
// frnd2tmp.cpp -- template class with non-template friends
#include <iostream>
using std::cout;
using std::endl;
template <typename T>
class HasFriend
{
private:
T item;
static int ct;
public:
HasFriend(const T & i) : item(i) {ct++;}
~HasFriend() {ct--; }
friend void counts();
friend void reports(HasFriend<T> &); // template parameter
};
// each specialization has its own static data member
template <typename T>
int HasFriend<T>::ct = 0;
// non-template friend to all HasFriend<T> classes
void counts()
{
cout << "int count: " << HasFriend<int>::ct << "; ";
cout << "double count: " << HasFriend<double>::ct << endl;
}
// non-template friend to the HasFriend<int> class
void reports(HasFriend<int> & hf)
{
cout <<"HasFriend<int>: " << hf.item << endl;
}
// non-template friend to the HasFriend<double> class
void reports(HasFriend<double> & hf)
{
cout <<"HasFriend<double>: " << hf.item << endl;
}
int main()
{
cout << "No objects declared: ";
counts();
HasFriend<int> hfi1(10);
cout << "After hfi1 declared: ";
counts();
HasFriend<int> hfi2(20);
cout << "After hfi2 declared: ";
counts();
HasFriend<double> hfdb(10.5);
cout << "After hfdb declared: ";
counts();
reports(hfi1);
reports(hfi2);
reports(hfdb);
return 0;
}
Some compilers will warn about using a non-template friend. Here is the output of the program in Listing 14.22:
No objects declared: int count: 0; double count: 0
After hfi1 declared: int count: 1; double count: 0
After hfi2 declared: int count: 2; double count: 0
After hfdb declared: int count: 2; double count: 1
HasFriend<int>: 10
HasFriend<int>: 20
HasFriend<double>: 10.5
Bound Template Friend Functions to Template Classes
You can modify the preceding example by making the friend functions templates themselves. In particular, you can set things up for bound template friends, so each specialization of a class gets a matching specialization for a friend. The technique is a bit more complex than for non-template friends and involves three steps.
For the first step, you declare each template function before the class definition:
template <typename T> void counts();
template <typename T> void report(T &);
Next, you declare the templates again as friends inside the function. These statements declare specializations based on the class template parameter type:
template <typename TT>
class HasFriendT
{
...
friend void counts<TT>();
friend void report<>(HasFriendT<TT> &);
};
The <> in the declarations identifies these as template specializations. In the case of report(), the <> can be left empty because the following template type argument can be deduced from the function argument:
HasFriendT<TT>
You could, however, use this instead:
report<HasFriendT<TT> >(HasFriendT<TT> &)
However, the counts() function has no parameters, so you have to use the template argument syntax (<TT>) to indicate its specialization. Note, too, that TT is the parameter type for the HasFriendT class.
Again, the best way to understand these declarations is to imagine what they become when you declare an object of a particular specialization. For example, suppose you declare this object:
HasFriendT<int> squack;
Then the compiler substitutes int for TT and generates the following class definition:
class HasFriendT<int>
{
...
friend void counts<int>();
friend void report<>(HasFriendT<int> &);
};
One specialization is based on TT, which becomes int, and the other is based on HasFriendT<TT>, which becomes HasFriendT<int>. Thus, the template specializations counts<int>() and report<HasFriendT<int> >() are declared as friends to the HasFriendT<int> class.
The third requirement the program must meet is to provide template definitions for the friends. Listing 14.23 illustrates these three aspects. Note that Listing 14.22 has one count() function that is a friend to all HasFriend classes, whereas Listing 14.23 has two count() functions, one of which is a friend to each of the instantiated class types. Because the count() function calls have no function parameter from which the compiler can deduce the desired specialization, these calls use the count<int>() and count<double>() forms to indicate the specialization. For the calls to report(), however, the compiler can use the argument type to deduce the specialization. You could use the <> form to the same effect:
report<HasFriendT<int> >(hfi2); // same as report(hfi2);
// tmp2tmp.cpp -- template friends to a template class
#include <iostream>
using std::cout;
using std::endl;
// template prototypes
template <typename T> void counts();
template <typename T> void report(T &);
// template class
template <typename TT>
class HasFriendT
{
private:
TT item;
static int ct;
public:
HasFriendT(const TT & i) : item(i) {ct++;}
~HasFriendT() { ct--; }
friend void counts<TT>();
friend void report<>(HasFriendT<TT> &);
};
template <typename T>
int HasFriendT<T>::ct = 0;
// template friend functions definitions
template <typename T>
void counts()
{
cout << "template size: " << sizeof(HasFriendT<T>) << "; ";
cout << "template counts(): " << HasFriendT<T>::ct << endl;
}
template <typename T>
void report(T & hf)
{
cout << hf.item << endl;
}
int main()
{
counts<int>();
HasFriendT<int> hfi1(10);
HasFriendT<int> hfi2(20);
HasFriendT<double> hfdb(10.5);
report(hfi1); // generate report(HasFriendT<int> &)
report(hfi2); // generate report(HasFriendT<int> &)
report(hfdb); // generate report(HasFriendT<double> &)
cout << "counts<int>() output:
";
counts<int>();
cout << "counts<double>() output:
";
counts<double>();
return 0;
}
Here is the output of the program in Listing 14.23:
template size: 4; template counts(): 0
10
20
10.5
counts<int>() output:
template size: 4; template counts(): 2
counts<double>() output:
template size: 8; template counts(): 1
As you can see, counts<double> reports a different template size from counts<int>, demonstrating that each T type now gets its own count() friend.
Unbound Template Friend Functions to Template Classes
The bound template friend functions in the preceding section are template specializations of a template declared outside a class. An int class specialization gets an int function specialization, and so on. By declaring a template inside a class, you can create unbound friend functions for which every function specialization is a friend to every class specialization. For unbound friends, the friend template type parameters are different from the template class type parameters:
template <typename T>
class ManyFriend
{
...
template <typename C, typename D> friend void show2(C &, D &);
};
Listing 14.24 shows an example that uses an unbound friend. In it, the function call show2(hfi1, hfi2) gets matched to the following specialization:
void show2<ManyFriend<int> &, ManyFriend<int> &>
(ManyFriend<int> & c, ManyFriend<int> & d);
Because it is a friend to all specializations of ManyFriend, this function has access to the item members of all specializations. But it only uses access to ManyFriend<int> objects.
Similarly, show2(hfd, hfi2) gets matched to this specialization:
void show2<ManyFriend<double> &, ManyFriend<int> &>
(ManyFriend<double> & c, ManyFriend<int> & d);
It, too, is a friend to all ManyFriend specializations, and it uses its access to the item member of a ManyFriend<int> object and to the item member of a ManyFriend<double> object.
// manyfrnd.cpp -- unbound template friend to a template class
#include <iostream>
using std::cout;
using std::endl;
template <typename T>
class ManyFriend
{
private:
T item;
public:
ManyFriend(const T & i) : item(i) {}
template <typename C, typename D> friend void show2(C &, D &);
};
template <typename C, typename D> void show2(C & c, D & d)
{
cout << c.item << ", " << d.item << endl;
}
int main()
{
ManyFriend<int> hfi1(10);
ManyFriend<int> hfi2(20);
ManyFriend<double> hfdb(10.5);
cout << "hfi1, hfi2: ";
show2(hfi1, hfi2);
cout << "hfdb, hfi2: ";
show2(hfdb, hfi2);
return 0;
}
Here’s the output of the program in Listing 14.24:
hfi1, hfi2: 10, 20
hfdb, hfi2: 10.5, 20