Considering the code below, is it possible to acquire the function pointer of operator() for Test2 given that you have all its template arguments? For example, say I want a function pointer that points to operator()<float, double> where Args={float} and Type2=double.
template<class Type1>
struct Test1
{
Type1 value;
template<class Type2, class ... Args>
bool operator()(const Type2& type)
{
//... Do something with Args... //
return type == value;
}
};
template<class Type1>
struct Test2
{
Type1 value;
template<class ... Args, class Type2>
bool operator()(const Type2& type)
{
//... Do something with Args... //
return type == value;
}
};
int main()
{
//OK!
auto ptr1 = &Test1<int>::template operator()<float, double>;
//Not ok! Assuming here that Args = {float, double}, Type2 = ?. Want: Args = {float}, Type2 = double
auto ptr2 = &Test2<int>::template operator()<float, double>;
}
I can see why just providing the template arguments to the function, as usual, doesn't work here. The template arguments are being deduced to be Args={float, double} and Type2 is left, hence a compilation error. This, I think, is evident from the fact that the Test1 version compiles perfectly fine. Is there a way around this? Thank you.