How do I expand a tuple into variadic template function's arguments?

Question

Consider the case of a templated function with variadic template arguments:

template<typename Tret, typename... T> Tret func(const T&... t);

Now, I have a tuple t of values. How do I call func() using the tuple values as arguments? I've read about the bind() function object, with call() function, and also the apply() function in different some now-obsolete documents. The GNU GCC 4.4 implementation seems to have a call() function in the bind() class, but there is very little documentation on the subject.

Some people suggest hand-written recursive hacks, but the true value of variadic template arguments is to be able to use them in cases like above.

Does anyone have a solution to is, or hint on where to read about it?

Simons

The C++14 standard has a solution see; open-std.org/jtc1/sc22/wg21/docs/papers/2013/n3658.html – Seeress

The idea is to unpack the tuple in a single variadic blast, using integer_sequence, see en.cppreference.com/w/cpp/utility/integer_sequence – Seeress

Having an integer_sequence S, you simply call your function as func(std::get<S>(tuple)...), and let the compiler handle the rest. – Seeress

If using C++17 or later, ignore this answer and see the one below using std::apply – Psycholinguistics

Answer 1

In C++17 you can do this:

std::apply(the_function, the_tuple);

This already works in Clang++ 3.9, using std::experimental::apply.

Responding to the comment saying that this won't work if the_function is templated, the following is a work-around:

#include <tuple>

template <typename T, typename U> void my_func(T &&t, U &&u) {}

int main(int argc, char *argv[argc]) {

  std::tuple<int, float> my_tuple;

  std::apply([](auto &&... args) { my_func(args...); }, my_tuple);

  return 0;
}

This work around is a simplified solution to the general problem of passing overload sets and function template where a function would be expected. The general solution (one that is taking care of perfect-forwarding, constexpr-ness, and noexcept-ness) is presented here: https://blog.tartanllama.xyz/passing-overload-sets/.

Pinochle

According to the example code at std::apply it does not seem to work if the_function is templated. – Britnibrito

@Britnibrito You can specify the function's template arguments: std::apply(add_generic<float>, std::make_pair(2.0f, 3.0f)); – Ducky

This is the simplest, most elegant solution. And it works wonders. Thanks so much, M. Alaggan!!!!!! +100 votes – Bona

Answer 2

Here's my code if anyone is interested

Basically at compile time the compiler will recursively unroll all arguments in various inclusive function calls <N> -> calls <N-1> -> calls ... -> calls <0> which is the last one and the compiler will optimize away the various intermediate function calls to only keep the last one which is the equivalent of func(arg1, arg2, arg3, ...)

Provided are 2 versions, one for a function called on an object and the other for a static function.

#include <tr1/tuple>

/**
 * Object Function Tuple Argument Unpacking
 *
 * This recursive template unpacks the tuple parameters into
 * variadic template arguments until we reach the count of 0 where the function
 * is called with the correct parameters
 *
 * @tparam N Number of tuple arguments to unroll
 *
 * @ingroup g_util_tuple
 */
template < uint N >
struct apply_obj_func
{
  template < typename T, typename... ArgsF, typename... ArgsT, typename... Args >
  static void applyTuple( T* pObj,
                          void (T::*f)( ArgsF... ),
                          const std::tr1::tuple<ArgsT...>& t,
                          Args... args )
  {
    apply_obj_func<N-1>::applyTuple( pObj, f, t, std::tr1::get<N-1>( t ), args... );
  }
};

//-----------------------------------------------------------------------------

/**
 * Object Function Tuple Argument Unpacking End Point
 *
 * This recursive template unpacks the tuple parameters into
 * variadic template arguments until we reach the count of 0 where the function
 * is called with the correct parameters
 *
 * @ingroup g_util_tuple
 */
template <>
struct apply_obj_func<0>
{
  template < typename T, typename... ArgsF, typename... ArgsT, typename... Args >
  static void applyTuple( T* pObj,
                          void (T::*f)( ArgsF... ),
                          const std::tr1::tuple<ArgsT...>& /* t */,
                          Args... args )
  {
    (pObj->*f)( args... );
  }
};

//-----------------------------------------------------------------------------

/**
 * Object Function Call Forwarding Using Tuple Pack Parameters
 */
// Actual apply function
template < typename T, typename... ArgsF, typename... ArgsT >
void applyTuple( T* pObj,
                 void (T::*f)( ArgsF... ),
                 std::tr1::tuple<ArgsT...> const& t )
{
   apply_obj_func<sizeof...(ArgsT)>::applyTuple( pObj, f, t );
}

//-----------------------------------------------------------------------------

/**
 * Static Function Tuple Argument Unpacking
 *
 * This recursive template unpacks the tuple parameters into
 * variadic template arguments until we reach the count of 0 where the function
 * is called with the correct parameters
 *
 * @tparam N Number of tuple arguments to unroll
 *
 * @ingroup g_util_tuple
 */
template < uint N >
struct apply_func
{
  template < typename... ArgsF, typename... ArgsT, typename... Args >
  static void applyTuple( void (*f)( ArgsF... ),
                          const std::tr1::tuple<ArgsT...>& t,
                          Args... args )
  {
    apply_func<N-1>::applyTuple( f, t, std::tr1::get<N-1>( t ), args... );
  }
};

//-----------------------------------------------------------------------------

/**
 * Static Function Tuple Argument Unpacking End Point
 *
 * This recursive template unpacks the tuple parameters into
 * variadic template arguments until we reach the count of 0 where the function
 * is called with the correct parameters
 *
 * @ingroup g_util_tuple
 */
template <>
struct apply_func<0>
{
  template < typename... ArgsF, typename... ArgsT, typename... Args >
  static void applyTuple( void (*f)( ArgsF... ),
                          const std::tr1::tuple<ArgsT...>& /* t */,
                          Args... args )
  {
    f( args... );
  }
};

//-----------------------------------------------------------------------------

/**
 * Static Function Call Forwarding Using Tuple Pack Parameters
 */
// Actual apply function
template < typename... ArgsF, typename... ArgsT >
void applyTuple( void (*f)(ArgsF...),
                 std::tr1::tuple<ArgsT...> const& t )
{
   apply_func<sizeof...(ArgsT)>::applyTuple( f, t );
}

// ***************************************
// Usage
// ***************************************

template < typename T, typename... Args >
class Message : public IMessage
{

  typedef void (T::*F)( Args... args );

public:

  Message( const std::string& name,
           T& obj,
           F pFunc,
           Args... args );

private:

  virtual void doDispatch( );

  T*  pObj_;
  F   pFunc_;
  std::tr1::tuple<Args...> args_;
};

//-----------------------------------------------------------------------------

template < typename T, typename... Args >
Message<T, Args...>::Message( const std::string& name,
                              T& obj,
                              F pFunc,
                              Args... args )
: IMessage( name ),
  pObj_( &obj ),
  pFunc_( pFunc ),
  args_( std::forward<Args>(args)... )
{

}

//-----------------------------------------------------------------------------

template < typename T, typename... Args >
void Message<T, Args...>::doDispatch( )
{
  try
  {
    applyTuple( pObj_, pFunc_, args_ );
  }
  catch ( std::exception& e )
  {

  }
}

Boleslaw

Is it possible to adapt this to work in a case where the "function" in question is actually a constructor? – Palomino

Could you provide an example of what you want to do and we can go from there. – Boleslaw

This solution proviedes only a compile time overhead and at the end it will be simplified to (pObj->*f)( arg0, arg,1, ... argN); right? – Abiotic

yes, the compiler will compress the multiple function calls into the final one as if you had written it yourself which is the beauty of all this meta programming stuff. – Boleslaw

all the tr1 stuff can get taken out now with c++11 – Weinert

If using C++17 or later, ignore this answer and see the one below using std::apply – Psycholinguistics

Answer 3

In C++ there is many ways of expanding/unpacking tuple and apply those tuple elements to a variadic template function. Here is a small helper class which creates index array. It is used a lot in template metaprogramming:

// ------------- UTILITY---------------
template<int...> struct index_tuple{}; 

template<int I, typename IndexTuple, typename... Types> 
struct make_indexes_impl; 

template<int I, int... Indexes, typename T, typename ... Types> 
struct make_indexes_impl<I, index_tuple<Indexes...>, T, Types...> 
{ 
    typedef typename make_indexes_impl<I + 1, index_tuple<Indexes..., I>, Types...>::type type; 
}; 

template<int I, int... Indexes> 
struct make_indexes_impl<I, index_tuple<Indexes...> > 
{ 
    typedef index_tuple<Indexes...> type; 
}; 

template<typename ... Types> 
struct make_indexes : make_indexes_impl<0, index_tuple<>, Types...> 
{}; 

Now the code which does the job is not that big:

 // ----------UNPACK TUPLE AND APPLY TO FUNCTION ---------
#include <tuple>
#include <iostream> 

using namespace std;

template<class Ret, class... Args, int... Indexes > 
Ret apply_helper( Ret (*pf)(Args...), index_tuple< Indexes... >, tuple<Args...>&& tup) 
{ 
    return pf( forward<Args>( get<Indexes>(tup))... ); 
} 

template<class Ret, class ... Args> 
Ret apply(Ret (*pf)(Args...), const tuple<Args...>&  tup)
{
    return apply_helper(pf, typename make_indexes<Args...>::type(), tuple<Args...>(tup));
}

template<class Ret, class ... Args> 
Ret apply(Ret (*pf)(Args...), tuple<Args...>&&  tup)
{
    return apply_helper(pf, typename make_indexes<Args...>::type(), forward<tuple<Args...>>(tup));
}

Test is shown bellow:

// --------------------- TEST ------------------
void one(int i, double d)
{
    std::cout << "function one(" << i << ", " << d << ");\n";
}
int two(int i)
{
    std::cout << "function two(" << i << ");\n";
    return i;
}

int main()
{
    std::tuple<int, double> tup(23, 4.5);
    apply(one, tup);

    int d = apply(two, std::make_tuple(2));    

    return 0;
}

I'm not big expert in other languages, but I guess that if these languages do not have such functionality in their menu, there is no way to do that. At least with C++ you can, and I think it is not so much complicated...

In

"... and apply those tuple elements to a variadic template function". The test section only contains non template variadic functions though. If I add one like template<class ... T> void three(T...) {} and try to use apply on that it does not compile. – Britnibrito

Unfortunatelly, this technique doesn't work in nvcc (and probably other EDG-based compilers). It fails with error: template parameter pack not at end of parameter list, in the line template<class Ret, class... Args, int... Indexes > Ret apply_helper( Ret (*pf)(Args...), index_tuple< Indexes... >, tuple<Args...>&& tup) . – Hearten

Answer 4

I find this to be the most elegant solution (and it is optimally forwarded):

#include <cstddef>
#include <tuple>
#include <type_traits>
#include <utility>

template<size_t N>
struct Apply {
    template<typename F, typename T, typename... A>
    static inline auto apply(F && f, T && t, A &&... a)
        -> decltype(Apply<N-1>::apply(
            ::std::forward<F>(f), ::std::forward<T>(t),
            ::std::get<N-1>(::std::forward<T>(t)), ::std::forward<A>(a)...
        ))
    {
        return Apply<N-1>::apply(::std::forward<F>(f), ::std::forward<T>(t),
            ::std::get<N-1>(::std::forward<T>(t)), ::std::forward<A>(a)...
        );
    }
};

template<>
struct Apply<0> {
    template<typename F, typename T, typename... A>
    static inline auto apply(F && f, T &&, A &&... a)
        -> decltype(::std::forward<F>(f)(::std::forward<A>(a)...))
    {
        return ::std::forward<F>(f)(::std::forward<A>(a)...);
    }
};

template<typename F, typename T>
inline auto apply(F && f, T && t)
    -> decltype(Apply< ::std::tuple_size<
        typename ::std::decay<T>::type
    >::value>::apply(::std::forward<F>(f), ::std::forward<T>(t)))
{
    return Apply< ::std::tuple_size<
        typename ::std::decay<T>::type
    >::value>::apply(::std::forward<F>(f), ::std::forward<T>(t));
}

Example usage:

void foo(int i, bool b);

std::tuple<int, bool> t = make_tuple(20, false);

void m()
{
    apply(&foo, t);
}

Unfortunately GCC (4.6 at least) fails to compile this with "sorry, unimplemented: mangling overload" (which simply means that the compiler doesn't yet fully implement the C++11 spec), and since it uses variadic templates, it wont work in MSVC, so it is more or less useless. However, once there is a compiler that supports the spec, it will be the best approach IMHO. (Note: it isn't that hard to modify this so that you can work around the deficiencies in GCC, or to implement it with Boost Preprocessor, but it ruins the elegance, so this is the version I am posting.)

GCC 4.7 now supports this code just fine.

Edit: Added forward around actual function call to support rvalue reference form *this in case you are using clang (or if anybody else actually gets around to adding it).

Edit: Added missing forward around the function object in the non-member apply function's body. Thanks to pheedbaq for pointing out that it was missing.

Edit: And here is the C++14 version just since it is so much nicer (doesn't actually compile yet):

#include <cstddef>
#include <tuple>
#include <type_traits>
#include <utility>

template<size_t N>
struct Apply {
    template<typename F, typename T, typename... A>
    static inline auto apply(F && f, T && t, A &&... a) {
        return Apply<N-1>::apply(::std::forward<F>(f), ::std::forward<T>(t),
            ::std::get<N-1>(::std::forward<T>(t)), ::std::forward<A>(a)...
        );
    }
};

template<>
struct Apply<0> {
    template<typename F, typename T, typename... A>
    static inline auto apply(F && f, T &&, A &&... a) {
        return ::std::forward<F>(f)(::std::forward<A>(a)...);
    }
};

template<typename F, typename T>
inline auto apply(F && f, T && t) {
    return Apply< ::std::tuple_size< ::std::decay_t<T>
      >::value>::apply(::std::forward<F>(f), ::std::forward<T>(t));
}

Here is a version for member functions (not tested very much!):

using std::forward; // You can change this if you like unreadable code or care hugely about namespace pollution.

template<size_t N>
struct ApplyMember
{
    template<typename C, typename F, typename T, typename... A>
    static inline auto apply(C&& c, F&& f, T&& t, A&&... a) ->
        decltype(ApplyMember<N-1>::apply(forward<C>(c), forward<F>(f), forward<T>(t), std::get<N-1>(forward<T>(t)), forward<A>(a)...))
    {
        return ApplyMember<N-1>::apply(forward<C>(c), forward<F>(f), forward<T>(t), std::get<N-1>(forward<T>(t)), forward<A>(a)...);
    }
};

template<>
struct ApplyMember<0>
{
    template<typename C, typename F, typename T, typename... A>
    static inline auto apply(C&& c, F&& f, T&&, A&&... a) ->
        decltype((forward<C>(c)->*forward<F>(f))(forward<A>(a)...))
    {
        return (forward<C>(c)->*forward<F>(f))(forward<A>(a)...);
    }
};

// C is the class, F is the member function, T is the tuple.
template<typename C, typename F, typename T>
inline auto apply(C&& c, F&& f, T&& t) ->
    decltype(ApplyMember<std::tuple_size<typename std::decay<T>::type>::value>::apply(forward<C>(c), forward<F>(f), forward<T>(t)))
{
    return ApplyMember<std::tuple_size<typename std::decay<T>::type>::value>::apply(forward<C>(c), forward<F>(f), forward<T>(t));
}

// Example:

class MyClass
{
public:
    void foo(int i, bool b);
};

MyClass mc;

std::tuple<int, bool> t = make_tuple(20, false);

void m()
{
    apply(&mc, &MyClass::foo, t);
}

Untouched

+1 out of the answers listed, yours was the closest I could get to working with arguments whose arguments are vectors... ...but I am still getting compile errors. ideone.com/xH5kBH If you compile this with -DDIRECT_CALL and run it, you will see what the output should be. I get a compile error otherwise (I think decltype is not smart enough to figure out my special case), with gcc 4.7.2. – Staten

The version of gcc on ideaone is to old for this to pass, it doesn't support the mangled decltype return type overloading. I have tested this code relatively thoroughly in gcc 4.7.2, and I haven't run into any problems. With gcc 4.8, you can use the new C++17 automatic return value feature to avoid all the nasty decltype trailing return types. – Untouched

+1 I really like that automatic return value feature - thx for the heads up – Staten

I know, in my book the lack of the automatic return value thing for single statement (non-lambda) functions was the biggest miss of C++11. That and shared_from_this. – Untouched

Out of curiosity, in the non-member apply function, why is f not wrapped with a std::forward call, as it is in the return type? Is it not needed? – Phaedra

That would be because I forgot it :) Fixed – Untouched

Out of curiosity, I tried compiling this in GCC 4.8, and foo('x', true) compiled to the exact same assembly code as apply(foo, ::std::make_tuple('x', true)) with any level of optimization besides -O0. – Untouched

With C++14 integer_sequence you even get an almost correct implementation of apply() in its example. see my answer below. – Larval

This is really short version, no doubts. But what about calling member function? – Intellectual

@tower120: I added a version for member functions. It compiles but I haven't tested it. Should work though! – Hostile

One thing I cannot quite grasp: Why does the functor/function object f in Apply<0> need to be forwarded for the call? – Perr

The functor gets forwarded for the call due the rvalue from *this spec. It allows you to have a different overload if *this is an rvalue. For example, a copy method could actually return the object itself if *this is an rvalue. – Untouched

When called as in the example on a non templated function this works, but OP's question specifically mentioned a variadic template. If I use that I get errors about template argument deduction/substitution failed. Just as the example for the C++17's std::apply which mentions that same error. Is this supposed to also work on variadic templates or am I missing something ? – Britnibrito

Answer 5

template<typename F, typename Tuple, std::size_t ... I>
auto apply_impl(F&& f, Tuple&& t, std::index_sequence<I...>) {
    return std::forward<F>(f)(std::get<I>(std::forward<Tuple>(t))...);
}
template<typename F, typename Tuple>
auto apply(F&& f, Tuple&& t) {
    using Indices = std::make_index_sequence<std::tuple_size<std::decay_t<Tuple>>::value>;
    return apply_impl(std::forward<F>(f), std::forward<Tuple>(t), Indices());
}

This is adapted from the C++14 draft using index_sequence. I might propose to have apply in a future standard (TS).

Larval

Answer 6

All this implementations are good. But due to use of pointer to member function compiler often cannot inline the target function call (at least gcc 4.8 can't, no matter what Why gcc can't inline function pointers that can be determined?)

But things changes if send pointer to member function as template arguments, not as function params:

/// from https://mcmap.net/q/16874/-quot-unpacking-quot-a-tuple-to-call-a-matching-function-pointer
template<int ...> struct seq {};
template<int N, int ...S> struct gens : gens<N-1, N-1, S...> {};
template<int ...S> struct gens<0, S...>{ typedef seq<S...> type; };

template<typename TT>
using makeSeq = typename gens< std::tuple_size< typename std::decay<TT>::type >::value >::type;


// deduce function return type
template<class ...Args>
struct fn_type;

template<class ...Args>
struct fn_type< std::tuple<Args...> >{

    // will not be called
    template<class Self, class Fn>
    static auto type_helper(Self &self, Fn f) -> decltype((self.*f)(declval<Args>()...)){
        //return (self.*f)(Args()...);
        return NULL;
    }
};

template<class Self, class ...Args>
struct APPLY_TUPLE{};

template<class Self, class ...Args>
struct APPLY_TUPLE<Self, std::tuple<Args...>>{
    Self &self;
    APPLY_TUPLE(Self &self): self(self){}

    template<class T, T (Self::* f)(Args...),  class Tuple>
    void delayed_call(Tuple &&list){
        caller<T, f, Tuple >(forward<Tuple>(list), makeSeq<Tuple>() );
    }

    template<class T, T (Self::* f)(Args...), class Tuple, int ...S>
    void caller(Tuple &&list, const seq<S...>){
        (self.*f)( std::get<S>(forward<Tuple>(list))... );
    }
};

#define type_of(val) typename decay<decltype(val)>::type

#define apply_tuple(obj, fname, tuple) \
    APPLY_TUPLE<typename decay<decltype(obj)>::type, typename decay<decltype(tuple)>::type >(obj).delayed_call< \
            decltype( fn_type< type_of(tuple) >::type_helper(obj, &decay<decltype(obj)>::type::fname) ), \
            &decay<decltype(obj)>::type::fname \
            > \
            (tuple);

And ussage:

struct DelayedCall
{  
    void call_me(int a, int b, int c){
        std::cout << a+b+c;
    }

    void fire(){
        tuple<int,int,int> list = make_tuple(1,2,3);
        apply_tuple(*this, call_me, list); // even simpler than previous implementations
    }
};

Proof of inlinable http://goo.gl/5UqVnC

---

With small changes, we can "overload" apply_tuple:

#define VA_NARGS_IMPL(_1, _2, _3, _4, _5, _6, _7, _8, N, ...) N
#define VA_NARGS(...) VA_NARGS_IMPL(X,##__VA_ARGS__, 7, 6, 5, 4, 3, 2, 1, 0)
#define VARARG_IMPL_(base, count, ...) base##count(__VA_ARGS__)
#define VARARG_IMPL(base, count, ...) VARARG_IMPL_(base, count, __VA_ARGS__)
#define VARARG(base, ...) VARARG_IMPL(base, VA_NARGS(__VA_ARGS__), __VA_ARGS__)

#define apply_tuple2(fname, tuple) apply_tuple3(*this, fname, tuple)
#define apply_tuple3(obj, fname, tuple) \
    APPLY_TUPLE<typename decay<decltype(obj)>::type, typename decay<decltype(tuple)>::type >(obj).delayed_call< \
            decltype( fn_type< type_of(tuple) >::type_helper(obj, &decay<decltype(obj)>::type::fname) ), \
            &decay<decltype(obj)>::type::fname \
            /* ,decltype(tuple) */> \
            (tuple);
#define apply_tuple(...) VARARG(apply_tuple, __VA_ARGS__)

...

apply_tuple(obj, call_me, list);
apply_tuple(call_me, list);       // call this->call_me(list....)

---

Plus this is the only one solution which works with templated functions.

Intellectual

Answer 7

1) if you have a readymade parameter_pack structure as function argument, you can just use std::tie like this:

template <class... Args>
void tie_func(std::tuple<Args...> t, Args&... args)
{
 std::tie<Args...>(args...) = t;
}

int main()
{
 std::tuple<int, double, std::string> t(2, 3.3, "abc");

 int i;
 double d;
 std::string s;

 tie_func(t, i, d, s);

 std::cout << i << " " << d << " " << s << std::endl;
}

2) if you don't have a readymade parampack arg, you'll have to unwind the tuple like this

#include <tuple>
#include <functional>
#include <iostream>



template<int N>
struct apply_wrap {
    template<typename R, typename... TupleArgs, typename... UnpackedArgs>
    static R applyTuple( std::function<R(TupleArgs...)>& f, const std::tuple<TupleArgs...>& t, UnpackedArgs... args )
    {
        return apply_wrap<N-1>::applyTuple( f, t, std::get<N-1>( t ), args... );
    }
};


template<>
struct apply_wrap<0>
{
    template<typename R, typename... TupleArgs, typename... UnpackedArgs>
    static R applyTuple( std::function<R(TupleArgs...)>& f, const std::tuple<TupleArgs...>&, UnpackedArgs... args )
    {
        return f( args... );
    }
};



template<typename R, typename... TupleArgs>
R applyTuple( std::function<R(TupleArgs...)>& f, std::tuple<TupleArgs...> const& t )
{
    return apply_wrap<sizeof...(TupleArgs)>::applyTuple( f, t );
}



int fac(int n)
{
    int r=1;
    for(int i=2; i<=n; ++i)
        r *= i;
    return r;
}



int main()
{
    auto t = std::make_tuple(5);
    auto f = std::function<decltype(fac)>(&fac);
    cout << applyTuple(f, t);
}

Amanda

Answer 8

The news does not look good.

Having read over the just-released draft standard, I'm not seeing a built-in solution to this, which does seem odd.

The best place to ask about such things (if you haven't already) is comp.lang.c++.moderated, because some folks involved in drafting the standard post there regularly.

If you check out this thread, someone has the same question (maybe it's you, in which case you're going to find this whole answer a little frustrating!), and a few butt-ugly implementations are suggested.

I just wondered if it would be simpler to make the function accept a tuple, as the conversion that way is easier. But this implies that all functions should accept tuples as arguments, for maximum flexibility, and so that just demonstrates the strangeness of not providing a built-in expansion of tuple to function argument pack.

Update: the link above doesn't work - try pasting this:

http://groups.google.com/group/comp.lang.c++.moderated/browse_thread/thread/750fa3815cdaac45/d8dc09e34bbb9661?lnk=gst&q=tuple+variadic#d8dc09e34bbb9661

Borderland

I wonder why they even bother having separate notions of tuple and function argument pack. Maybe in a conforming compiler they are interchangeable but I haven't spotted an indication of that anywhere I've read about them. – Borderland

Because tuple<int, char, string> is necessary as a separate type; as is the ability to make a function that doesn't require make_type in the middle of every call. – Nutrilite

Also, the best place is not comp.lang.c++.moderated. Questions about C++1x are almost always better directed to comp.std.c++. – Nutrilite

Answer 9

How about this:

// Warning: NOT tested!
#include <cstddef>
#include <tuple>
#include <type_traits>
#include <utility>

using std::declval;
using std::forward;
using std::get;
using std::integral_constant;
using std::size_t;
using std::tuple;

namespace detail
{
    template < typename Func, typename ...T, typename ...Args >
    auto  explode_tuple( integral_constant<size_t, 0u>, tuple<T...> const &t,
     Func &&f, Args &&...a )
     -> decltype( forward<Func>(f)(declval<T const>()...) )
    { return forward<Func>( f )( forward<Args>(a)... ); }

    template < size_t Index, typename Func, typename ...T, typename ...Args >
    auto  explode_tuple( integral_constant<size_t, Index>, tuple<T...> const&t,
     Func &&f, Args &&...a )
     -> decltype( forward<Func>(f)(declval<T const>()...) )
    {
        return explode_tuple( integral_constant<size_t, Index - 1u>{}, t,
         forward<Func>(f), get<Index - 1u>(t), forward<Args>(a)... );
    }
}

template < typename Func, typename ...T >
auto  run_tuple( Func &&f, tuple<T...> const &t )
 -> decltype( forward<Func>(f)(declval<T const>()...) )
{
    return detail::explode_tuple( integral_constant<size_t, sizeof...(T)>{}, t,
     forward<Func>(f) );
}

template < typename Tret, typename ...T >
Tret  func_T( tuple<T...> const &t )
{ return run_tuple( &func<Tret, T...>, t ); }

The run_tuple function template takes the given tuple and pass its elements individually to the given function. It carries out its work by recursively calling its helper function templates explode_tuple. It's important that run_tuple passes the tuple's size to explode_tuple; that number acts as a counter for how many elements to extract.

If the tuple is empty, then run_tuple calls the first version of explode_tuple with the remote function as the only other argument. The remote function is called with no arguments and we're done. If the tuple is not empty, a higher number is passed to the second version of explode_tuple, along with the remote function. A recursive call to explode_tuple is made, with the same arguments, except the counter number is decreased by one and (a reference to) the last tuple element is tacked on as an argument after the remote function. In a recursive call, either the counter isn't zero, and another call is made with the counter decreased again and the next-unreferenced element is inserted in the argument list after the remote function but before the other inserted arguments, or the counter reaches zero and the remote function is called with all the arguments accumulated after it.

I'm not sure I have the syntax of forcing a particular version of a function template right. I think you can use a pointer-to-function as a function object; the compiler will automatically fix it.

Demi

Answer 10

I am evaluating MSVS 2013RC, and it failed to compile some of the previous solutions proposed here in some cases. For example, MSVS will fail to compile "auto" returns if there are too many function parameters, because of a namespace imbrication limit (I sent that info to Microsoft to have it corrected). In other cases, we need access to the function's return, although that can also be done with a lamda: the following two examples give the same result..

apply_tuple([&ret1](double a){ret1 = cos(a); }, std::make_tuple<double>(.2));
ret2 = apply_tuple((double(*)(double))cos, std::make_tuple<double>(.2));

And thanks again to those who posted answers here before me, I wouldn't have gotten to this without it... so here it is:

template<size_t N>
struct apply_impl {
    template<typename F, typename T, typename... A>
    static inline auto apply_tuple(F&& f, T&& t, A&&... a)
    -> decltype(apply_impl<N-1>::apply_tuple(std::forward<F>(f), std::forward<T>(t),
                          std::get<N-1>(std::forward<T>(t)), std::forward<A>(a)...)) {
         return apply_impl<N-1>::apply_tuple(std::forward<F>(f), std::forward<T>(t),
                          std::get<N-1>(std::forward<T>(t)), std::forward<A>(a)...);
    }
    template<typename C, typename F, typename T, typename... A>
    static inline auto apply_tuple(C*const o, F&& f, T&& t, A&&... a)
    -> decltype(apply_impl<N-1>::apply_tuple(o, std::forward<F>(f), std::forward<T>(t),
                          std::get<N-1>(std::forward<T>(t)), std::forward<A>(a)...)) {
         return apply_impl<N-1>::apply_tuple(o, std::forward<F>(f), std::forward<T>(t),
                          std::get<N-1>(std::forward<T>(t)), std::forward<A>(a)...);
    }
};

// This is a work-around for MSVS 2013RC that is required in some cases
#if _MSC_VER <= 1800 /* update this when bug is corrected */
template<>
struct apply_impl<6> {
    template<typename F, typename T, typename... A>
    static inline auto apply_tuple(F&& f, T&& t, A&&... a)
    -> decltype(std::forward<F>(f)(std::get<0>(std::forward<T>(t)), std::get<1>(std::forward<T>(t)), std::get<2>(std::forward<T>(t)),
           std::get<3>(std::forward<T>(t)), std::get<4>(std::forward<T>(t)), std::get<5>(std::forward<T>(t)), std::forward<A>(a)...)) {
         return std::forward<F>(f)(std::get<0>(std::forward<T>(t)), std::get<1>(std::forward<T>(t)), std::get<2>(std::forward<T>(t)),
           std::get<3>(std::forward<T>(t)), std::get<4>(std::forward<T>(t)), std::get<5>(std::forward<T>(t)), std::forward<A>(a)...);
    }
    template<typename C, typename F, typename T, typename... A>
    static inline auto apply_tuple(C*const o, F&& f, T&& t, A&&... a)
    -> decltype((o->*std::forward<F>(f))(std::get<0>(std::forward<T>(t)), std::get<1>(std::forward<T>(t)), std::get<2>(std::forward<T>(t)),
           std::get<3>(std::forward<T>(t)), std::get<4>(std::forward<T>(t)), std::get<5>(std::forward<T>(t)), std::forward<A>(a)...)) {
         return (o->*std::forward<F>(f))(std::get<0>(std::forward<T>(t)), std::get<1>(std::forward<T>(t)), std::get<2>(std::forward<T>(t)),
           std::get<3>(std::forward<T>(t)), std::get<4>(std::forward<T>(t)), std::get<5>(std::forward<T>(t)), std::forward<A>(a)...);
    }
};
#endif

template<>
struct apply_impl<0> {
    template<typename F, typename T, typename... A>
    static inline auto apply_tuple(F&& f, T&&, A&&... a)
    -> decltype(std::forward<F>(f)(std::forward<A>(a)...)) {
         return std::forward<F>(f)(std::forward<A>(a)...);
    }
    template<typename C, typename F, typename T, typename... A>
    static inline auto apply_tuple(C*const o, F&& f, T&&, A&&... a)
    -> decltype((o->*std::forward<F>(f))(std::forward<A>(a)...)) {
         return (o->*std::forward<F>(f))(std::forward<A>(a)...);
    }
};

// Apply tuple parameters on a non-member or static-member function by perfect forwarding
template<typename F, typename T>
inline auto apply_tuple(F&& f, T&& t)
-> decltype(apply_impl<std::tuple_size<typename std::decay<T>::type>::value>::apply_tuple(std::forward<F>(f), std::forward<T>(t))) {
     return apply_impl<std::tuple_size<typename std::decay<T>::type>::value>::apply_tuple(std::forward<F>(f), std::forward<T>(t));
}

// Apply tuple parameters on a member function
template<typename C, typename F, typename T>
inline auto apply_tuple(C*const o, F&& f, T&& t)
-> decltype(apply_impl<std::tuple_size<typename std::decay<T>::type>::value>::apply_tuple(o, std::forward<F>(f), std::forward<T>(t))) {
     return apply_impl<std::tuple_size<typename std::decay<T>::type>::value>::apply_tuple(o, std::forward<F>(f), std::forward<T>(t));
}

Underset

Why you make object argument a const pointer? Not reference, not const reference, not just pointer? What if callable function will not const? – Intellectual

Answer 11

Extending on @David's solution, you can write a recursive template that

1. Doesn't use the (overly-verbose, imo) integer_sequence semantics
2. Doesn't use an extra temporary template parameter int N to count recursive iterations
3. (Optional for static/global functors) uses the functor as a template parameter for compile-time optimizaion

E.g.:

template <class F, F func>
struct static_functor {
    template <class... T, class... Args_tmp>
    static inline auto apply(const std::tuple<T...>& t, Args_tmp... args)
            -> decltype(func(std::declval<T>()...)) {
        return static_functor<F,func>::apply(t, args...,
                std::get<sizeof...(Args_tmp)>(t));
    }
    template <class... T>
    static inline auto apply(const std::tuple<T...>& t, T... args)
            -> decltype(func(args...)) {
        return func(args...);
    }
};

static_functor<decltype(&myFunc), &myFunc>::apply(my_tuple);

Alternatively if your functor is not defined at compile-time (e.g., a non-constexpr functor instance, or a lambda expression), you can use it as a function parameter instead of a class template parameter, and in fact remove the containing class entirely:

template <class F, class... T, class... Args_tmp>
inline auto apply_functor(F&& func, const std::tuple<T...>& t,
        Args_tmp... args) -> decltype(func(std::declval<T>()...)) {
    return apply_functor(func, t, args..., std::get<sizeof...(Args_tmp)>(t));
}
template <class F, class... T>
inline auto apply_functor(F&& func, const std::tuple<T...>& t,
        T... args) -> decltype(func(args...)) {
    return func(args...);
}

apply_functor(&myFunc, my_tuple);

For pointer-to-member-function callables, you can adjust either of the above code pieces similarly as in @David's answer.

Explanation

In reference to the second piece of code, there are two template functions: the first one takes the functor func, the tuple t with types T..., and a parameter pack args of types Args_tmp.... When called, it recursively adds the objects from t to the parameter pack one at a time, from beginning (0) to end, and calls the function again with the new incremented parameter pack.

The second function's signature is almost identical to the first, except that it uses type T... for the parameter pack args. Thus, once args in the first function is completely filled with the values from t, it's type will be T... (in psuedo-code, typeid(T...) == typeid(Args_tmp...)), and thus the compiler will instead call the second overloaded function, which in turn calls func(args...).

The code in the static functor example works identically, with the functor instead used as a class template argument.

Thrice

any comments on the compile-time optimization of the first option would be appreciated, so I can make my answer more complete (and maybe learn something new). – Thrice

Answer 12

Why not just wrap your variadic arguments into a tuple class and then use compile time recursion (see link) to retrieve the index you are interested in. I find that unpacking variadic templates into a container or collection may not be type safe w.r.t. heterogeneous types

template<typename... Args>
auto get_args_as_tuple(Args... args) -> std::tuple<Args...> 
{
    return std::make_tuple(args);
}

Paddie

The question was the other way around. Not Args... -> tuple, but tuple -> Args.... – Grumble

Answer 13

This simple solution works for me:

template<typename... T>
void unwrap_tuple(std::tuple<T...>* tp)
{
    std::cout << "And here I have the tuple types, all " << sizeof...(T) << " of them" << std::endl;
}

int main()
{
    using TupleType = std::tuple<int, float, std::string, void*>;

    unwrap_tuple((TupleType*)nullptr); // trick compiler into using template param deduction
}

Fairman