Nick Hutchinson Nick Hutchinson - 3 months ago 10
C++ Question

Using 'void' template arguments in C++

Take the following minimal example:

using Type1 = std::function<void(void)>;

template <typename T>
using Type2 = std::function<void(T)>;

Type1 whyDoesThisWork;
Type2<void> andYetThisDoesNot;


If the second type alias, I get the error "Argument may not have 'void' type". (I tested with Xcode 4.5, Clang/c++11/libc++, OS X 10.7.)

I find this curious: I would have expected
Type1
and
Type2<void>
to behave identically. What's going on here? And is there a way to rewrite the second type alias so I can write
Type2<void>
and get
std::function<void(void)>
instead of an error?

Edit I should probably add that the reason I want this is to allow for something like the following:

template <typename ... T>
using Continuation = std::function<void(T...)>;

auto someFunc = []() -> void {
printf("I'm returning void!\n");
};

Continuation<decltype(someFunc())> c;


Continuation<decltype(someFunc())>
becomes
Continuation<void>
and I get the error.

Answer

The short answer is "templates are not string substitution". void f(void) has meaning only so far as it is an alias for void f() in C++, in order to be backwards compatible with C.

The first step is to use variadics, as noted elsewhere.

The second step is figuring out how to map void returning functions to ... well, maybe something like std::function<void()>, or maybe something else. I say maybe something else because unlike the other cases, you cannot call std::function<void()> foo; foo( []()->void {} ); -- it isn't a true continuation.

Something like this maybe:

template<typename T>
struct Continuation
{
  typedef std::function<void(T)> type;
};

template<>
struct Continuation<void>
{
  typedef std::function<void()> type;
};

then use it like this:

auto someFunc = []()->void {};
Continuation<decltype(someFunc())>::type c;

which gives you the type you want. You could even add in an apply to continuation:

template<typename T>
struct Continuation
{
  typedef std::function<void(T)> type;

  template<typename func, typename... Args>
  static void Apply( type const& cont, func&& f, Args... args)
  {
    cont( f(args...) );
  }
};

template<>
struct Continuation<void>
{
  typedef std::function<void()> type;
  template<typename func, typename... Args>
  static void Apply( type const& cont, func&& f, Args... args)
  {
    f(args...);
    cont();
  }
};

which lets you apply a continuation to an execution of a function uniformly if the incoming type is a void or if it is a non-void type.

However, I would ask "why would you want to do this"?

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