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I am trying to create a variadic template function which would call a function to consecutive pairs of arguments.
The desired function signature would be:
template <typename ...Ts>
void apply(Ts &...args);
When called with apply(t1, t2, t3) the function should make a sequence of calls func(t1, t2)and func(t2, t3), where func is a function with signature:
template <typename L, typename R>
void func(L &left, R &right);
The order of operations is not really relevant in my context. The function has to be able to modify objects left and right, hence passed by reference. I cannot simply use polymorphic access through a base class pointer, since the objects have different class templates, a shared class cannot really be taken out.
Is it possible to achieve such a sequence of calls via a variadic template function? None of the pack expansion and fold expression examples that I've seen seem to cover such a scenario. Or should I pass my objects in a different fashion?
My initial attempt, included below (with some details omitted), packed all template parameters into a tuple, and then used a ‘const for-loop’ to ‘loop’ through the tuple elements. However, I soon came to realize that this approach would not work, because the lambda in the const-for loop invokes operator() const and therefore cannot modify the passed objects.
The code I was using does make the desired sequence of calls, but the objects are not modified (set_something() is not a const function). I had to resort to using a wrapper function with different numbers of template parameters, and making the calls to func manually.
template <std::size_t Begin, typename Callable, std::size_t... I>
constexpr void const_for_impl(Callable &&func, std::index_sequence<I...>) {
(func(std::integral_constant<std::size_t, Begin + I>{}), ...);
}
template <std::size_t Begin, std::size_t End, typename Callable>
constexpr void const_for(Callable &&func) {
const_for_impl<Begin>(std::forward<Callable>(func),
std::make_index_sequence<End - Begin>{});
};
template <typename... Ts>
void apply(Ts *... args) {
auto tuple = std::make_tuple(std::forward<Ts>(args)...);
const_for<0, sizeof...(args) - 1>(
[&](auto I) { func((std::get<I>(tuple)), (std::get<I + 1>(tuple))); });
};
template <typename L, typename R>
void func(L &l, R &r) {
// Validate with some type traits
static_assert(has_some_property<L>::value);
static_assert(has_another_property<R>::value);
// Get a shared pointer to something common
auto common = std::make_shared<typename something_common<L, R>::type>();
l.set_something(common);
r.set_something(common);
};
// Application scenario
int main() {
ComplexObjectA<SomeType, SomeParameter> a;
ComplexObjectB<AnotherType, AnotherParameter> b;
ComplexObjectC c;
apply(a, b, c);
return 0;
}
So, what's the problem? Simple fold-like template (and remember that template pattern-matching goes in reverse!)
template<typename T1, typename T2>
void apply(T1 &&t1, T2 &&t2) { func(t1, t2); }
template<typename T1, typename T2, typename... Ts>
void apply(T1 &&t1, T2 &&t2, Ts &&...ts) {
func(t1, t2);
return apply(t2, ts...);
}
Or, more precise, it should actually look as (thanks #MaxLanghof):
void apply(T1 &&t1, T2 &&t2) {
func(std::forward<T1>(t1), std::forward<T2>(t2));
}
template<typename T1, typename T2, typename... Ts>
void apply(T1 &&t1, T2 &&t2, Ts &&...ts) {
func(std::forward<T1>(t1), t2);
return apply(std::forward<T2>(t2), std::forward<TS>(ts)...);
}
An alternative (c++14) approach:
#include <utility>
#include <utility>
#include <tuple>
#include <iostream>
template <typename L, typename R>
void func(L &left, R &right) {
std::cout << left << " " << right << std::endl;
}
template <typename Tup, std::size_t... Is>
void apply_impl(Tup&& tup, std::index_sequence<Is...>) {
int dummy[] = { 0, (static_cast<void>(func(std::get<Is>(tup), std::get<Is + 1>(tup))), 0)... };
static_cast<void>(dummy);
}
template <typename ...Ts>
void apply(Ts &...args) {
apply_impl(std::forward_as_tuple(args...), std::make_index_sequence<sizeof...(Ts) - 1>{});
}
int main() {
int arr[] = {0, 1, 2, 3};
apply(arr[0], arr[1], arr[2], arr[3]);
}
Output:
0 1
1 2
2 3
[online example]
To make it c++11 compliant one would need to use one of the available integer_sequence implementations.
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?
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/.
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 )
{
}
}
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...
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);
}
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).
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://stackoverflow.com/a/9288547/1559666
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.
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);
}
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
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.
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));
}
Extending on #David's solution, you can write a recursive template that
Doesn't use the (overly-verbose, imo) integer_sequence semantics
Doesn't use an extra temporary template parameter int N to count recursive iterations
(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.
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);
}
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
}
I'd like to write a function template, apply, which receives some function f, an integer i, and a parameter pack. apply needs to unpack the parameters and apply f to them, except for the ith parameter, pi. For pi, it needs to call some other function g before passing it as a parameter to f.
It seems that I need a way to partition the parameter pack into a left side, the ith parameter, and the right side. Is this possible? In code:
template<int i, typename Function, typename... Parms>
void apply(Function f, Parms... parms)
{
auto lhs = // what goes here?
auto pi = // what goes here?
auto rhs = // what goes here?
f(lhs..., g(pi), rhs...);
}
OK, here we go! It really ugly but I couldn't come up with a nicer version in a hurry ;) Most of the stuff is bog standard template specialization. The biggest issue is creating a list of integers of the proper size. I seem to recall that I came up with a nice version but somehow I can't recall what I did. Enjoy!
#include <iostream>
#include <utility>
// printing the values
void print_args() {}
template <typename F> void print_args(F f) { std::cout << f; }
template <typename F, typename... T>
void print_args(F f, T... args)
{
std::cout << f << ", ";
print_args(args...);
}
// the function object to be called:
struct Functor
{
template <typename... T>
void operator()(T... args)
{
std::cout << "f(";
print_args(args...);
std::cout << ")\n";
}
};
// conditionally apply g():
template <typename T> T g(T value) { return 1000 + value; }
template <int i, int j, typename T>
typename std::enable_if<i != j, T>::type forward(T t) { return t; }
template <int i, int j, typename T>
typename std::enable_if<i == j, T>::type forward(T t) { return g(t); }
// create a series of integers:
template <int... Values> struct values {};
template <int Add, typename> struct combine_values;
template <int Add, int... Values>
struct combine_values<Add, values<Values...>>
{
typedef values<Values..., Add> type;
};
template <int Size> struct make_values;
template <> struct make_values<0> { typedef values<> type; };
template <int Size>
struct make_values
{
typedef typename combine_values<Size, typename make_values<Size -1>::type>::type type;
};
// applying f(t...) except for ti where g(ti) is called
template <int i, int... Values, typename Function, typename... T>
void apply_aux(values<Values...>, Function f, T... t)
{
f(forward<i, Values>(t)...);
}
template <int i, typename Function, typename... T>
void apply(Function f, T... t)
{
apply_aux<i>(typename make_values<sizeof...(T)>::type(), f, t...);
}
int main()
{
apply<3>(Functor(), 1, 2, 3, 4, 5, 6, 7, 8);
apply<4>(Functor(), 1, 2, 3, 4, 5, 6, 7, 8);
apply<5>(Functor(), 1, 2, 3, 4, 5, 6, 7, 8);
}
Iactually did code something similar a little while ago. So try the following code:
template<unsigned N, unsigned M>
struct call_up_impl{
template<class Func, class Mutator, class Tuple, class... Args>
static void do_call(const Func& func, const Mutator& mutator, const Tuple& args, Args&&... unpacked_args) {
call_up_impl<N-1, M>::do_call(func, mutator, args, std::get<N-1>(args), std::forward<Args>(unpacked_args)...);
}
};
template<unsigned M>
struct call_up_impl<0, M> {
template<class Func, class Mutator, class Tuple, class... Args>
static void do_call(const Func& func, const Mutator&, const Tuple&, Args&&... unpacked_args) {
func(std::forward<Args>(unpacked_args)...);
}
};
template<unsigned M>
struct call_up_impl<M, M> {
template<class Func, class Mutator, class Tuple, class... Args>
static void do_call(const Func& func, const Mutator& mutator, const Tuple& args, Args&&... unpacked_args) {
call_up_impl<M-1, M>::do_call(func, mutator, args, mutator(std::get<M-1>(args)), std::forward<Args>(unpacked_args)...);
}
};
template<int i, typename Function, typename... Parms>
void apply(Function f, Parms... parms) {
std::tuple<Parms...> t(parms...);
call_up_impl<std::tuple_size<decltype(t)>::value, i + 1>::do_call(f, &g, t);
}
This is a quick adaption of my original code, so it isn't thoroughly tested and maybe not the not optimal way to do this, but it should work at least (at least according to a quick test and depending what exactly you want). It should be possible to do this without the tuple, but I haven't gotten that to compile with g++ (it doesn't seem to like the nested variadic templates needed). However changing apply to:
template<int i, typename Function, typename... Parms>
void apply(Function f, Parms&&... parms) {
std::tuple<Parms&&...> t(std::forward<Parms>(parms)...);
call_up_impl<std::tuple_size<decltype(t)>::value, i + 1>::do_call(f, &g, t);
}
will probably avoid most of the overhead introduced by the tuple. It would be even better to make correct forwarding of the results of the std::get calls, but I'm too tired to work that out write now.
I would like to have a type, that would be in effect POD but I would like to be able to specify how and which types are in it, for example:
template<Args...>
struct POD
{
//here I would like to create depending on Args appropriate types as a members.
};
Is it possible to do it with this new variadic templates feature in C++0x?
I’ve never yet used the C++0x variadic templates feature but the following code compiles on G++ 4.5:
template <typename... Args>
struct tuple;
template <typename T, typename... Args>
struct tuple<T, Args...> {
T value;
tuple<Args...> inner;
};
template <typename T>
struct tuple<T> {
T value;
};
However, initializing them is … weird because we need to nest the inner values:
int main() {
tuple<int> n1 = { 23 };
tuple<int, float> n2 = { 42, { 0.5f } };
tuple<std::string, double, int> n3 = { "hello, world", { 3.14, { 97 } } };
}
Retrieving the values is of course a bit tedious. The simplest method is probably to provide a get<N>() function template.
But we cannot implement get directly since function templates cannot be partially specialized. Either we need to use SFINAE (read: boost::enable_if) or we need to delegate the actual function of get to a type that can be partially specialized.
In the following, I did the latter. But first, we need another helper type trait: nth_type, which returns the appropriate return type of the get function:
template <unsigned N, typename... Args>
struct nth_type;
template <unsigned N, typename T, typename... Args>
struct nth_type<N, T, Args...> : nth_type<N - 1, Args...> { };
template <typename T, typename... Args>
struct nth_type<0, T, Args...> {
typedef T type;
};
Easy-peasy. Just returns the nth type in a list of types.
Now we can write our get function:
template <unsigned N, typename... Args>
inline typename nth_type<N, Args...>::type get(tuple<Args...>& tup) {
return get_t<N, Args...>::value(tup);
}
Like I said, this just delegates the task. No biggie. In practice, we probably want to have another overload for const tuples (but then, in practice we would use an existing tuple type).
Now for the killing, followed by a light salad:
template <unsigned N, typename... Args>
struct get_t;
template <unsigned N, typename T, typename... Args>
struct get_t<N, T, Args...> {
static typename nth_type<N, T, Args...>::type value(tuple<T, Args...>& tup) {
return get_t<N - 1, Args...>::value(tup.inner);
}
};
template <typename T, typename... Args>
struct get_t<0, T, Args...> {
static T value(tuple<T, Args...>& tup) {
return tup.value;
}
};
And that’s it. We can test this by printing some values in our previously defined variables:
std::cout << get<0>(n1) << std::endl; // 23
std::cout << get<0>(n2) << std::endl; // 42
std::cout << get<0>(n3) << std::endl; // hello, world
std::cout << get<1>(n2) << std::endl; // 0.5
std::cout << get<1>(n3) << std::endl; // 3.14
std::cout << get<2>(n3) << std::endl; // 97
Man, it’s fun messing with variadic templates.
Are you familiar with std::tuple?
AFAIK it's a POD if all it's members are PODs, if I'm wrong then I'm guessing it isn't possible.
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?
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/.
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 )
{
}
}
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...
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);
}
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).
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://stackoverflow.com/a/9288547/1559666
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.
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);
}
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
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.
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));
}
Extending on #David's solution, you can write a recursive template that
Doesn't use the (overly-verbose, imo) integer_sequence semantics
Doesn't use an extra temporary template parameter int N to count recursive iterations
(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.
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);
}
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
}