问题描述

可能的重复:
如何将元组扩展为可变参数模板函数争论?
“解包”调用匹配函数指针的元组

在 C++11 模板中，有没有办法使用元组作为(可能是模板)函数的单个参数?

In C++11 templates, is there a way to use a tuple as the individual args of a (possibly template) function?

示例:
假设我有这个功能:

Example:
Let's say I have this function:

void foo(int a, int b)  
{  
}

我有元组 auto bar = std::make_tuple(1, 2).

我可以使用它以模板方式调用 foo(1, 2) 吗?

Can I use that to call foo(1, 2) in a templaty way?

我的意思不是简单的 foo(std::get<0>(bar), std::get<1>(bar)) 因为我想在一个模板中做到这一点不知道参数的数量.

I don't mean simply foo(std::get<0>(bar), std::get<1>(bar)) since I want to do this in a template that doesn't know the number of args.

更完整的例子:

template<typename Func, typename... Args>  
void caller(Func func, Args... args)  
{  
    auto argtuple = std::make_tuple(args...);  
    do_stuff_with_tuple(argtuple);  
    func(insert_magic_here(argtuple));  // <-- this is the hard part  
}

我应该注意，我不希望创建一个适用于一个参数的模板，另一个适用于两个参数等......

I should note that I'd prefer to not create one template that works for one arg, another that works for two, etc…

推荐答案

试试这个:

// implementation details, users never invoke these directly
namespace detail
{
    template <typename F, typename Tuple, bool Done, int Total, int... N>
    struct call_impl
    {
        static void call(F f, Tuple && t)
        {
            call_impl<F, Tuple, Total == 1 + sizeof...(N), Total, N..., sizeof...(N)>::call(f, std::forward<Tuple>(t));
        }
    };

    template <typename F, typename Tuple, int Total, int... N>
    struct call_impl<F, Tuple, true, Total, N...>
    {
        static void call(F f, Tuple && t)
        {
            f(std::get<N>(std::forward<Tuple>(t))...);
        }
    };
}

// user invokes this
template <typename F, typename Tuple>
void call(F f, Tuple && t)
{
    typedef typename std::decay<Tuple>::type ttype;
    detail::call_impl<F, Tuple, 0 == std::tuple_size<ttype>::value, std::tuple_size<ttype>::value>::call(f, std::forward<Tuple>(t));
}

示例:

#include <cstdio>
int main()
{
    auto t = std::make_tuple("%d, %d, %d
", 1,2,3);
    call(std::printf, t);
}

通过一些额外的魔法并使用 std::result_of，您可能还可以使整个事物返回正确的返回值.

With some extra magic and using std::result_of, you can probably also make the entire thing return the correct return value.

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