CODEWITHSUNDEEP
c++templatesc++11resolution
David KWH
David KWH

Reputation: 157

template function with nested type cannot be deduced

I am working with the SystemC library which requires all user defined types to have a operator<< and sc_trace() function. However the user defined type is actually a nested type inside a template class, because the "nested type" is computed from the template argument specified in the outer class. 

template<typename T>
class Block {
    typedef typename transform<T>::value NewType;
public:
    struct SomeType {
        SomeType() {}
        SomeType(T val) : member(val) {}
        NewType member;
    };
};

When I define the operator<< for SomeType like so

template<typename T>
std::ostream& operator<<(std::ostream& os, const typename Block<T>::SomeType& type) {
    return os << type.member;
}

The compiler cannot deduce the call inside the systemC library that does attempt to dump the nested defined type using the streaming operator. Since I rather not touch the library code (outside my control). Would any one of you experts out there know a way to work around this?

And if there is no clean workaround, would you know if the C++11 has a solution for this?

Upvotes: 3

Views: 936

Answers (1)

David KWH
David KWH

Reputation: 157

I actually found a solution myself. It is referred to as the Barton-Nackman Trick in Vandevoorde/Josuttis.

The key is to avoid using function template. The standard excludes nested type of a dependent template class from template argument deduction. The operator<< and sc_trace function needs to be defined as a friend function in the template class. This way the function is a non-templated function when the template class is instantiated, but with the friend keyword the function takes on the scope of the enclosing namespace.

Upvotes: 3

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