variadic template parameter pack expanding for fun

Question

I am looking for something like that:

template< typename T>  
void func(T t)
{
}

template< typename... Parms> 
void anyFunc( Parms... p)
{
    func<Parms>(p)... ;  //error 
    func(p)... ;         //error 
}

If the parameter pack expansion is done inside another function call it works:

template< typename T>
int some(T t)
{}

template< typename... Parms>
void func(Parms ...p)
{}

template< typename... Parms>
void somemore(Parms... p)
{
   func( some(p)...);
}

int main() 
{
 somemore(1,2,3,4,10,8,7, "Hallo");
}

The parameter pack expansion will also work for a list of base class initializers.

Is there any solution which will also work for functions which will return 'void'. The above workaround will not, while using the function calls returning void inside a parameter list could never work.

Any ideas?

Answer 1

Unfortunately, as you noticed, expanding a parameter pack is only valid in certain contexts where the parser expects a comma-separated list of entries – contexts where the comma is just a syntactic separator, not the comma operator. This is arguably a deficiency in the current text.

An ugly workaround:

func((some(p), 0)...);

Do note that the evaluation order of function arguments, and thus the order of the some invocations, is unspecified, so you have to be careful with any side effects.

Answer 2

How about a small helper class:

template <typename Func, typename A, typename ...Args> struct Caller
{
  static void call(Func & f, A && a, Args && ...args)
  {
    f(std::forward<A>(a));
    Caller<Func, Args...>::call(f, args...);
  }
};

template <typename Func, typename A> struct Caller<Func, A>
{
  static void call(Func & f, A && a)
  {
    f(std::forward<A>(a));
  }
};

template <typename Func, typename ...Args>
void Call(Func & f, Args && ...args)
{
  Caller<Func, Args...>::call(f, std::forward<Args>(args)...);
}

Then you can put the following in your client code:

void foo(A);
Call(foo, a1, a2, a3);