Question

所以我给了std::tuple<T...>，我想创建一个接受T...的函数指针，目前这就是我所拥有的;

template<typename... Arguments>
using FunctionPointer = void (*)(Arguments...);

using FunctionPtr = FunctionPointer<typename std::tuple_element<0, V>::type,
                                    typename std::tuple_element<1, V>::type,
                                    typename std::tuple_element<2, V>::type>;

但是，如果不手动输入0, ..., tuple_size<V>::value中的每个索引，我似乎无法找到方法。 FunctionPtr在上下文中定义，其中V=std::tuple<T...>（也已经有一个可变参数模板（因此我不能直接传递T...））

我想我需要生成一些索引列表，并做一些黑魔法......

Answer 1

这是一个可能的解决方案：

#include <tuple>

// This is what you already have...
template<typename... Arguments>
using FunctionPointer = void (*)(Arguments...);

// Some new machinery the end user does not need to no about
namespace detail
{
    template<typename>
    struct from_tuple { };

    template<typename... Ts>
    struct from_tuple<std::tuple<Ts...>>
    {
        using FunctionPtr = FunctionPointer<Ts...>;
    };
}

//=================================================================
// This is how your original alias template ends up being rewritten
//=================================================================
template<typename T>
using FunctionPtr = typename detail::from_tuple<T>::FunctionPtr;

以下是您将如何使用它：

// Some function to test if the alias template works correctly
void foo(int, double, bool) { }

int main()
{
    // Given a tuple type...
    using my_tuple = std::tuple<int, double, bool>;

    // Retrieve the associated function pointer type...
    using my_fxn_ptr = FunctionPtr<my_tuple>; // <== This should be what you want

    // And verify the function pointer type is correct!
    my_fxn_ptr ptr = &foo;
}

Answer 2

一个简单的特征可能会起到作用：

#include <tuple>

template <typename> struct tuple_to_function;

template <typename ...Args>
struct tuple_to_function<std::tuple<Args...>>
{
    typedef void (*type)(Args...);
};

用法：

typedef std::tuple<Foo, Bar, int> myTuple;

tuple_to_function<myTuple>::type fp; // is a void (*)(Foo, Bar, int)

将std :: tuple <t ...>转换为T </t ...>

2 个答案: