hide function template, declare specializations

Question

This is a followup to C++ templates: prevent instantiation of base template

I use templates to achieve function overloading without the mess of implicit type conversions

Answer 1

When compiling main.cpp, there's no way for the compiler to know which template specializations may exist in some other compilation unit -- so there's no way to flag this error at compile time, you have to wait until link time.

Answer 2

I don't believe it's possible to do what you want. See these FAQs for more info:

How can I avoid linker errors with my template functions?

How can I avoid linker errors with my template classes?

Answer 3

It seems to produce a linker error even if you don't put it in the separate file.

However, to produce a compiler error for other instantiations, implement the function and use a compile-time assertion, e.g

#include <boost/static_assert.hpp>

template <class T> T f(T)
{
    //assert some type-dependent "always-false" condition,
    //so it won't be triggered unless this function is instantiated
    BOOST_STATIC_ASSERT(sizeof(T) == 0 && "Only long or bool are available");
}

template<> long f(long v) { return -v; }
template<> bool f(bool v) { return !v; }

int main()
{
    //f(100);
    f(100L);
    f(false);
}

And just for general information, C++0x has a much more elegant way to deal with it:

template <class T> T f(T) = delete;

template<> long f(long v) { return -v; }
template<> bool f(bool v) { return !v; }

Answer 4

The best way is to implement that basic template with something invalid (not illegal) C++ code. For example,

template<class T> T f(T v) { return v.Default_Implementation_Not_Available; }

This error will be compile time; and it's generated only when you instantiate any version other than 'long' & 'bool'. If you don't instantiate 'int' version, compilation will go fine.