C++ templates: prevent instantiation of base template

Question

I have an interface

std::string
get_string(Source const &s, std::string const &d);
int
get_int(Source const &s, int const &d);
bool
get_bool(Source const &s, bool const &d);

which I'd like to change to

template<class T>
T
get(Source const &s, T const &d);

But there's no sensible base template, so the actual base definition is a legal but useless (return d;). What can I do to force compile-time failure if the base is instantiated? Is there an idiomatic solution for this situation?

Answer 1

Don't define the template, just declare it and define the three specializations.

template <typename T>
T get(Source const &, T const &);

template<>
std::string get(Source const &s, std::string const &d) {
    return d + s.stringval(); // or whatever
}

[Edit: removed stuff about overloads - just for once, template function specialization does actually seem to be better. Who woulda thunk?]

Answer 2

Declare the baseclass (t) as abstract, that way an instance can never be created of that class.

Answer 3

just do

string get(source, string);
int get (source, int);
bool get(source, bool);

Answer 4

If you are willing to pay for run-time polymorphism, you can do this...

template <typename T>
class Interface
{
   virtual T get(Source const &s, T const &d) = 0;
};

class StringInterface : public Interface<std::string>
{
   virtual std::string get(Source const& s, std::string const& d);
};

// etc.

Since your base is an abstract class, you will get a compile-time failure if you try to instantiate it directly.