实现这一点有点复杂(尽管这是可能的)。我建议您使用一个已经实现了这一点的库，即Boost。融合(调用函数)。作为奖励，Boost Fusion也可以与c++ 03编译器一起工作。

这是一个c++ 14的解决方案。

template <typename ...Args>
struct save_it_for_later
{
  std::tuple<Args...> params;
  void (*func)(Args...);

  template<std::size_t ...I>
  void call_func(std::index_sequence<I...>)
  { func(std::get<I>(params)...); }
  void delayed_dispatch()
  { call_func(std::index_sequence_for<Args...>{}); }
};

这仍然需要一个辅助函数(call_func)。由于这是一种常见的习惯用法，也许标准应该直接支持它作为std::调用和可能的实现

// helper class
template<typename R, template<typename...> class Params, typename... Args, std::size_t... I>
R call_helper(std::function<R(Args...)> const&func, Params<Args...> const&params, std::index_sequence<I...>)
{ return func(std::get<I>(params)...); }

// "return func(params...)"
template<typename R, template<typename...> class Params, typename... Args>
R call(std::function<R(Args...)> const&func, Params<Args...> const&params)
{ return call_helper(func,params,std::index_sequence_for<Args...>{}); }

然后我们的延迟调度就变成了

template <typename ...Args>
struct save_it_for_later
{
  std::tuple<Args...> params;
  std::function<void(Args...)> func;
  void delayed_dispatch()
  { std::call(func,params); }
};

在给出答案的基础上再思考一下这个问题，我发现了另一种解决同样问题的方法:

template <int N, int M, typename D>
struct call_or_recurse;

template <typename ...Types>
struct dispatcher {
  template <typename F, typename ...Args>
  static void impl(F f, const std::tuple<Types...>& params, Args... args) {
     call_or_recurse<sizeof...(Args), sizeof...(Types), dispatcher<Types...> >::call(f, params, args...);
  }
};

template <int N, int M, typename D>
struct call_or_recurse {
  // recurse again
  template <typename F, typename T, typename ...Args>
  static void call(F f, const T& t, Args... args) {
     D::template impl(f, t, std::get<M-(N+1)>(t), args...);
  }
};

template <int N, typename D>
struct call_or_recurse<N,N,D> {
  // do the call
  template <typename F, typename T, typename ...Args>
  static void call(F f, const T&, Args... args) {
     f(args...);
  }
};

这需要将delayed_dispatch()的实现更改为:

  void delayed_dispatch() {
     dispatcher<Args...>::impl(func, params);
  }

这是通过递归地将std::tuple转换为自己的参数包来实现的。需要Call_or_recurse作为专门化来终止使用实际调用的递归，实际调用只是解包完成的参数包。

我不确定这是否是一个“更好的”解决方案，但这是另一种思考和解决问题的方式。

作为另一个替代解决方案，你可以使用enable_if，形成一些可以说比我之前的解决方案更简单的东西:

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

void f(int a, double b, void* c) {
  std::cout << a << ":" << b << ":" << c << std::endl;
}

template <typename ...Args>
struct save_it_for_later {
  std::tuple<Args...> params;
  void (*func)(Args...);

  template <typename ...Actual>
  typename std::enable_if<sizeof...(Actual) != sizeof...(Args)>::type
  delayed_dispatch(Actual&& ...a) {
    delayed_dispatch(std::forward<Actual>(a)..., std::get<sizeof...(Actual)>(params));
  }

  void delayed_dispatch(Args ...args) {
    func(args...);
  }
};

int main() {
  int a=666;
  double b = -1.234;
  void *c = NULL;

  save_it_for_later<int,double,void*> saved = {
                                 std::tuple<int,double,void*>(a,b,c), f};
  saved.delayed_dispatch();
}

第一个重载只是从元组中再取一个参数，并将其放入参数包中。第二个重载接受一个匹配的参数包，然后进行真正的调用，只有在第二个重载可行的情况下才禁用第一个重载。

Johannes使用c++ 14 std::index_sequence(和函数返回类型作为模板参数RetT)的解决方案的变化:

template <typename RetT, typename ...Args>
struct save_it_for_later
{
    RetT (*func)(Args...);
    std::tuple<Args...> params;

    save_it_for_later(RetT (*f)(Args...), std::tuple<Args...> par) : func { f }, params { par } {}

    RetT delayed_dispatch()
    {
        return callFunc(std::index_sequence_for<Args...>{});
    }

    template<std::size_t... Is>
    RetT callFunc(std::index_sequence<Is...>)
    {
        return func(std::get<Is>(params) ...);
    }
};

double foo(int x, float y, double z)
{
  return x + y + z;
}

int testTuple(void)
{
  std::tuple<int, float, double> t = std::make_tuple(1, 1.2, 5);
  save_it_for_later<double, int, float, double> saved (&foo, t);
  cout << saved.delayed_dispatch() << endl;
  return 0;
}

c++ 17的解决方案是简单地使用std::apply:

auto f = [](int a, double b, std::string c) { std::cout<<a<<" "<<b<<" "<<c<< std::endl; };
auto params = std::make_tuple(1,2.0,"Hello");
std::apply(f, params);

只是觉得应该在这个帖子的回答中声明一次(在它已经出现在其中一个评论中之后)。

c++ 14的基本解决方案在这个线程中仍然缺失。编辑:不，这实际上是沃尔特的回答。

这个函数是:

void f(int a, double b, void* c)
{
      std::cout << a << ":" << b << ":" << c << std::endl;
}

用下面的代码片段调用它:

template<typename Function, typename Tuple, size_t ... I>
auto call(Function f, Tuple t, std::index_sequence<I ...>)
{
     return f(std::get<I>(t) ...);
}

template<typename Function, typename Tuple>
auto call(Function f, Tuple t)
{
    static constexpr auto size = std::tuple_size<Tuple>::value;
    return call(f, t, std::make_index_sequence<size>{});
}

例子:

int main()
{
    std::tuple<int, double, int*> t;
    //or std::array<int, 3> t;
    //or std::pair<int, double> t;
    call(f, t);    
}

DEMO

a lot of answers have been provided but I found them too complicated and not very natural. I did it another way, without using sizeof or counters. I used my own simple structure (ParameterPack) for parameters to access the tail of parameters instead of a tuple. Then, I appended all the parameters from my structure into function parameters, and finnally, when no more parameters were to be unpacked, I run the function. Here is the code in C++11, I agree that there is more code than in others answers, but I found it more understandable.

template <class ...Args>
struct PackParameters;

template <>
struct PackParameters <>
{
    PackParameters() = default;
};

template <class T, class ...Args>
struct PackParameters <T, Args...>
{
    PackParameters ( T firstElem, Args... args ) : value ( firstElem ), 
    rest ( args... ) {}

    T value;
    PackParameters<Args...> rest;
};

template <class ...Args>
struct RunFunction;

template <class T, class ...Args>
struct RunFunction<T, Args...>
{
    template <class Function>
    static void Run ( Function f, const PackParameters<T, Args...>& args );

    template <class Function, class... AccumulatedArgs>
    static void RunChild ( 
                          Function f, 
                          const PackParameters<T, Args...>& remainingParams, 
                          AccumulatedArgs... args 
                         );
};

template <class T, class ...Args>
template <class Function>
void RunFunction<T, Args...>::Run ( 
                                   Function f, 
                                   const PackParameters<T, Args...>& remainingParams 
                                  )
{
    RunFunction<Args...>::template RunChild ( f, remainingParams.rest,
                                              remainingParams.value );
}

template <class T, class ...Args>
template<class Function, class ...AccumulatedArgs>
void RunFunction<T, Args...>::RunChild ( Function f, 
                                         const PackParameters<T, Args...>& remainingParams, 
                                         AccumulatedArgs... args )
{
    RunFunction<Args...>:: template RunChild ( f, remainingParams.rest,
                                               args..., remainingParams.value );
}


template <>
struct RunFunction<>
{
    template <class Function, class... AccumulatedArgs>
    static void RunChild ( Function f, PackParameters<>, AccumulatedArgs... args )
    {
        f ( args... );
    }

    template <class Function>
    static void Run ( Function f, PackParameters<> )
    {
        f ();
    }
};

struct Toto
{
    std::string k = "I am toto";
};

void f ( int i, Toto t, float b, std::string introMessage )
{
    float res = i * b;

    std::cerr << introMessage << " " << res << std::endl;
    std::cerr << "Toto " << t.k << std::endl;
}

int main(){
    Toto t;
    PackParameters<int, Toto, float, std::string> pack ( 3, t, 4.0, " 3 * 4 =" );

    RunFunction<int, Toto, float, std::string>::Run ( f, pack );
    return 0;
}