When writing a class template that offers functions related to the template that support implicit type conversions on all parameters, define those functions as friends inside the class template.

If we want to do implicit type conversion on all arguments, we should use non-member functions (item 24). Now we want a templatized version of Rational class to support the same mixed-mode arithmetic as shown in item 24, and the code could start like this:

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template<typename T>
class Rational {
public:
    Rational(const T& numerator = 0, // see item 20 for why params are passed by ref
             const T& denominator = 1);
    const T numerator() const;    // see item 28 for why return values are passed by value
    const T denominator() const;  // see item 3 for why return values are const
};

template<typename T>
const Rational<T> operator*(const Rational<T>& lhs,
                            const Rational<T>& rhs)
{...}

However, this simple update to templatized version will not compile for following code:

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Rational<int> oneHalf(1, 2);
Rational<int> result = oneHalf * 2;  // error! won't compile

The problem here is that implicit type conversion functions are never considered during template argument deduction. This means it is impossible for the compilers to convert the second parameter 2 into a Rational<int> using non-explicit constructor, so compilers can’t figure out what T is for this function template named operator* taking two parameters of type Rational<T>, can’t deduce parameter types for this function templates, and thus can’t instantiate the appropriate functions. In the end, the function we want to call fails to be declared, before we could apply implicit type conversions during a later function call.

Solve the compiling issue

How to declare the operator* properly? Notice that class templates don’t depend on template argument deduction (this process only applies to function templates), so T is always known at the time the class Rational<T> is instantiated. Combine this knowledge with another fact that a friend declaration in a template class refers to a specific function (not a function template), so as part of the class instantiation process, the friend function will be automatically declared. Taking advantage of these 2 points, we could update the code to a new version¹:

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template<typename T>
class Rational {
public:
    ...
    friend  // declare operator* function
    const Rational operator*(const Rational& lhs, 
                             const Rational& rhs);
};

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template<typename T>  // define operator* function
const Rational<T> operator*(const Rational<T>& lhs,  
                            const Rational<T>& rhs)
{...}

When the oneHalf is declared to be of type Rational<int>, the class Rational<int> is instantiated, and the friend function operator* that takes Rational<int> parameters is then declared automatically. As a declared function, compilers can use implicit conversion functions (such as Rational’s non-explicit constructor) when calling it, making the mixed-mode call compile.

Solve the linking issue

Yes, the code compiles. Yet it won’t link.

This time, the problem is that the target function operator* is only declared inside Rational, not defined there. Indeed, the operator* template outside the class is intended to provide that definition, but things don’t work this way, and linkers can’t find the definition.

To solve it, the simplest thing is to put the function body into its declaration:

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template<typename T>
class Rational {
public:
    ...
    friend 
    const Rational operator*(const Rational& lhs, 
                             const Rational& rhs)
    {
        return Rational(lhs.numerator() * rhs.numerator(),
                        lhs.denominator() * rhs.denominator()); // same impl as item 24
    }

};

This works as intended: mixed-mode calls to operator* now compiles, link, and run.

This design is, in some sense, kind of unconventional, because the use of friendship has nothing to do with a need to access non-public parts of the class. We declared it as friend because it is the only choise we have:

• to make type conversions possible on all arguments, we need a non-member function (item 24)
• to have the proper function automatically instantiated, we need to declare the function inside the class
• the only way to declare a non-member function inside a class is to make it a friend

Have the friend call a helper

Such functions as operator* defined inside a class are implicitly declared inline (item 30), so we may minimize the impact of such inline declarations by having operator* do nothing but call a helper function defined outside of the class, especially when the function body is complex.

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template<typename T> class Raional;  // declare Rational template

template<typename T>  // helper template
const Rational<T> doMultiply(const Rational<T>& lhs,
                             const Rational<T>& rhs); 
template<typename T>
class Rational {
public:
    ...
    friend // have friend call helper
    const Rational operator*(const Rational& lhs, 
                             const Rational& rhs)
    { return doMultiply(lhs, rhs); }
    ...
};

Many compilers force us to put all template definitions in header files, so we may need to define doMultiply in the header as well:

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template<typename T> 
const Rational<T> doMultiply(const Rational<T>& lhs,
                             const Rational<T>& rhs)
{
    return Rational<T>(lhs.numerator() * rhs.numerator(),
                       lhs.denominator() * rhs.denominator());
}

Here, doMultiply is a template, so it won’t support mixed-mode multiplication. The workflow is like this:

• The friend function operator* supports mixed-mode operations and is in charge of necessary type conversions, ending with two Rational objects passed to doMultiply
• These two objects are feed to an appropriate instantiation of the doMultiply helper template, which then do the actual multiplication