Referring to template parameters to define types for function parameter

Question

I would like to do something like this:

template <typename T, int Size>
struct config {
 T field;
 /* various other things */
}

template <typename C>  //  where C is a specific version of the 'config' object. 
void func(C arg0, C::T arg1, std::array<int, C::Size> &arg2) {
   int i = C::Size;
}

ie, can I choose the types my function parameters based on the full type of config ?

I could achieve this by just listing out all the individual types:

template <typename T, int Size>
void func(config<T,Size> config, T arg1, std::array<int, Size> &arg2) { ... }

But my full use case is acquiring quite a large number of parameters, both compile-time and run-time, and I'd like to package them up into my config object and pass it around, without having to replicate many template definitions at dozens of function declarations. Is there a way to do this?

Answer 1

You can of course add a type alias/static constexpr to config , but the only way to guarantee the first parameter of the function is instantiation of config , is the second alternative.

A solution using type alias/ static const:

template <class T, int N>
struct config
{
    using FieldType = T;
    static constexpr int Size = N;

    FieldType field;
};

template <typename C> 
void func(C arg0, typename C::FieldType arg1, std::array<int, C::Size> &arg2) {
   int i = C::Size;
}