Specify return type based on other template argument

Question

I'd like to specify my templated function's return type, by an other template argument. All of this inside a class.

In the header file:

class MyClass {
    template<int type, typename RT>
    RT myfunc();
};

In the .cpp something like this:

template<>
int MyClass::myfunc<1, int>() { return 2; }

template<>
double MyClass::myfunc<2, double>() { return 3.14; }

template<>
const char* MyClass::myfunc<3, const char*>() { return "some string"; }

And I would like to be able to use my function like this:

MyClass m;
int i = m.myfunc<1>(); // i will be 2
double pi = m.myfunc<2>(); // pi will be 3.14
const char* str = m.myfunc<3>(); // str == "some string"

So I would like my function to be able to parameterized by one template integer (or enumeration whatsoever), and the return type will be different, based on this integer. I don't want the function to work with any other integer arguments than the specified ones, for example here m.myfunc<4>() would give compile error.

I want to parameterize my function by one template argument only, because m.myfunc<1, int>() would be working, but I don't want to write the typename all the time.

I tried with auto return types, or templating other way around, but always got some compile errors. (Function not found, unresolved externals...)

Is this possible in any way?

Answer 1

Is this what you seek?

template<int n>
struct Typer
{
};

template<>
struct Typer<1>
{
    typedef int Type;
};
template<>
struct Typer<2>
{
    typedef double Type;
};
template<>
struct Typer<3>
{
    typedef const char* Type;
};

class MyClass
{
public:
    template<int typeCode>
    typename Typer<typeCode>::Type myfunc();
};

template<> Typer<1>::Type MyClass::myfunc<1>(){ return 2; } 

template<> Typer<2>::Type MyClass::myfunc<2>() { return 3.14; }

template<> Typer<3>::Type MyClass::myfunc<3>() { return "some string"; }