Question

I have been trying to implement a complex number class for fixed point types where the result type of the multiply operation will be a function of the input types. I need to have functions where I can do multiply complex by complex and also complex by real number.

This essentially is a simplified version of the code. Where A is my complex type.

template<typename T1, typename T2> struct rt {};

template<> struct rt<double, double> { 
    typedef double type;
};
//forward declaration
template<typename T> struct A;

template<typename T1, typename T2>
struct a_rt {
    typedef A<typename rt<T1,T2>::type> type;
};

template <typename T>
struct A {
    template<typename T2>
    typename a_rt<T,T2>::type operator*(const T2& val) const {
        typename a_rt<T,T2>::type ret;
        cout << "T2& called" << endl;
        return ret;
    }
    template<typename T2>
    typename a_rt<T,T2>::type operator*(const A<T2>& val) const {
        typename a_rt<T,T2>::type ret;
        cout << "A<T2>& called" << endl;
        return ret;
    }
};

TEST(TmplClassFnOverload, Test) {
    A<double> a;
    A<double> b;
    double c;
    a * b;
    a * c;
}

The code fails to compile because the compiler is trying to instantiate the a_rt template with double and A<double>. I don’t know what is going on under the hood since I imagine the compiler should pick the more specialized operator*(A<double>&) so a_rt will only be instantiated with <double, double> as arguments.

Would you please explain to me why this would not work?
And if this is a limitation, how should I work around this.

Thanks a tonne!

unittest.cpp: In instantiation of 'a_rt<double, A<double> >':
unittest.cpp:198:   instantiated from here 
unittest.cpp:174: error: no type named 'type' in 'struct rt<double, A<double> >' 

Update

The compiler appears to be happy with the following change. There is some subtlety I’m missing here. Appreciate someone who can walk me through what the compiler is doing in both cases.

    template<typename T2>
    A<typename rt<T,T2>::type> operator*(const T2& val) const {
        A<typename rt<T,T2>::type> ret;
        cout << "T2& called" << endl;
        return ret;
    }
    template<typename T2>
    A<typename rt<T,T2>::type> operator*(const A<T2>& val) const {
        A<typename rt<T,T2>::type> ret;
        cout << "A<T2>& called" << endl;
        return ret;
    }

Answer 1

Resolving function calls in C++ proceeds in five phases:

1. name lookup: this finds two versions of operator*
2. template argument deduction: this will be applied to all functions found in step 1)
3. overload resolution: the best match will be selected
4. access control: can the best match in fact be invoked (i.e. is it not a private member)
5. virtuality: if virtual function are involved, a lookup in the vtable might be required

First note that the return type is never ever being deduced. You simply cannot overload on return type. The template arguments to operator* are being deduced and then substituted into the return type template.

So what happens at the call a * b;? First, both versions of operator* have their arguments deduced. For the first overload, T2 is deduced to being A<double>, and for the second overload T2 resolves to double. If there multiple overloads, the Standard says:

14.7.1 Implicit instantiation [temp.inst] clause 9

If a function template or a member function template specialization is
used in a way that involves overload resolution, a declaration of the
specialization is implicitly instantiated (14.8.3).

So at the end of argument deduction when the set of candidate functions are being generated, (so before overload resolution) the template gets instantiated and you get an error because rt does not have a nested type. This is why the more specialized second template will not be selected: overload resolution does not take place. You might have expected that this substitution failure would not be an error. HOwever, the Standard says:

14.8.2 Template argument deduction [temp.deduct] clause 8

If a substitution results in an invalid type or expression, type
deduction fails. An invalid type or expression is one that would be
ill-formed if written using the substituted arguments.
Only invalid types and expressions in the immediate context of the
function type and its template parameter types can result in a
deduction failure. [ Note: The evaluation of the substituted types and
expressions can result in side effects such as the instantiation of
class template specializations and/or function template
specializations, the generation of implicitly-defined functions, etc.
Such side effects are not in the “immediate context” and can result in
the program being ill-formed. — end note ]

In your original code, the typename a_rt<T,T2>::type return type is not an immediate context. Only during template instantiation does it get evaluated, and then the lack of the nested type in rt is an erorr.

In your updated code A<typename rt<T,T2>::type> return type is an immediate context and the Substitution Failure is Not An Erorr (SFINAE) applies: the non-deduced function template is simply removed from the overload resolution set and the remaining one is being called.

With your updated code, output will be:

> A<T2>& called     
> T2& called