Variadic Templates, Perfect Forwarding to functions with default arguments

Question

I have been using a variadic template that acts as an exception firewall in an interface between C and C++. The template simply takes a function, followed by N arguments and calls the function inside a try catch block. This has been working fine, unfortunately one of the functions I wish to call now takes an additional default argument. As a result the function name is not resolved and the template fails to compile.

The error is:

perfect-forward.cpp: In function ‘void FuncCaller(Func, Args&& ...) [with Func = void (*)(const std::basic_string<char>&, double, const std::vector<int>&), Args = {const char (&)[7], double}]’:
perfect-forward.cpp:69:41: instantiated from here
perfect-forward.cpp:46:4: error: too few arguments to function

A simplified version of the code is as follows:

template< class Func, typename ...Args >
void FuncCaller( Func f, Args&&... params )
{
   try
   {
       cout << __func__ << " called\n";
       f(params...);
   }
   catch( std::exception& ex )
   {
       cout << "Caught exception: " << ex.what() << "\n";
   }
}

void Callee( const string& arg1, double d, const vector<int>&v = vector<int>{} )
{
   cout << __func__ << " called\n";
   cout << "\targ1: " << arg1 << "\n";
   cout << "\td: " << d << "\n";
   cout << "\tv: ";
   copy( v.begin(), v.end(), ostream_iterator<int>( cout, " "  ) );
   cout << "\n";
}

int main()
{
   vector<int> v { 1, 2, 3, 4, 5 };

   FuncCaller( Callee, "string", 3.1415, v );
   FuncCaller( Callee, "string", 3.1415 );  **// Fails to compile**

   return 0;
} 

Should this code work or am I expecting too much from the compiler?

Note: I have tested the use of perfect forwarding with constructors that have default arguments and the code compiles and works as expected,

i.e.:

template<typename TypeToConstruct> struct SharedPtrAllocator 
{
    template<typename ...Args> shared_ptr<TypeToConstruct> 
        construct_with_shared_ptr(Args&&... params) {
        return std::shared_ptr<TypeToConstruct>(new TypeToConstruct(std::forward<Args>(params)...));
    };
};

works when calling the cfollowing constructor with 2 or 3 arguments...

MyClass1( const string& arg1, double d, const vector<int>&v = vector<int>{} )

Answer 1

I don't think there's any way this can be achieved. The default argument values are not part of the function signature. They're only code-generation short-hands that are expanded by the compiler when you call the function literally. Similarly, std::bind won't pick up default arguments, either.