Casting a variadic parameter pack to (void)

Question

I've effectively got the following problem: I want to be able to build with -Wall -Wextra -Werror, however, the following code will complain about unused parameters:

struct foo
{
    template <typename... Args>
    static void bar()
    { }

    template <typename T, typename ... Args>
    static void bar(T&& value, Args&& ... args)
    {
    #ifdef DEBUG
        std::cout << value;
        bar(std::forward<Args>(args)...);
    #endif
    }
};

The first unused parameter is easy to fix:

    #ifdef DEBUG
        std::cout << value;
        bar(std::forward<Args>(args)...);
    #else // Shut the compiler up
        (void) value;
    #endif

My question is, how can I do this with the remaining args? Neither

(void)(args...);

Nor

(void)(args)...;

will work, both complain about the parameter pack not being expanded.

(This is under GCC 4.7.3, if that will make any difference to a potential solution).

Answer 1

When working with variadic template, it is more clean to use sink:

struct sink { template<typename ...Args> sink(Args const & ... ) {} };


#ifdef DEBUG
    std::cout << value;
    bar(std::forward<Args>(args)...);
#else 
    sink { value, args ... }; //eat all unused arguments!
#endif

Answer 2

You could really go with conditional naming here.

#ifdef DEBUG
#define DEBUG_NAME(x) x
#else
#define DEBUG_NAME(x)
#endif

static void bar(T&& DEBUG_NAME(value), Args&& DEBUG_NAME(args)) {}