更新时间:2023-11-14 20:39:52
好的,所以我以为我会死, do itç_ç
Ok, so I thought I would die, but I finally managed to do it ç_ç
首先,我使用常用的索引。因为我没有官方的,我使用了几个月前写的旧索引:
First, I used the usual indices. Since I do not have the official ones, I used old indices I wrote some months ago:
template<std::size_t...>
struct indices {};
template<std::size_t N, std::size_t... Ind>
struct make_indices:
make_indices<N-1, N-1, Ind...>
{};
template<std::size_t... Ind>
struct make_indices<0, Ind...>:
indices<Ind...>
{};
然后,我使用了一些 function traits 在***上找到。他们很好,我认为他们相当于Boost库链接在注释:
Then, I used some function traits found somewhere on ***. They are nice, and I think that they are equivalent to the Boost library linked in the comments:
template<typename T>
struct function_traits:
function_traits<decltype(&T::operator())>
{};
template<typename C, typename Ret, typename... Args>
struct function_traits<Ret(C::*)(Args...) const>
{
enum { arity = sizeof...(Args) };
using result_type = Ret;
template<std::size_t N>
using arg = typename std::tuple_element<N, std::tuple<Args...>>::type;
};
然后,我能够写一个合适的 make_foo
函数及其实现函数,因为它们都需要使用索引。注意,这很简单:
Then, I was able to write a proper make_foo
function and it implementation function, since both are required to use indices. Be careful, it's plain ugly:
template<typename Function, std::size_t... Ind>
auto make_foo_(Function&& func, indices<Ind...>)
-> Foo<
typename function_traits<typename std::remove_reference<Function>::type>::result_type,
typename function_traits<typename std::remove_reference<Function>::type>::template arg<Ind>...>
{
using Ret = typename function_traits<typename std::remove_reference<Function>::type>::result_type;
return { std::function<Ret(typename function_traits<typename std::remove_reference<Function>::type>::template arg<Ind>...)>(func) };
}
template<typename Function, typename Indices=make_indices<function_traits<typename std::remove_reference<Function>::type>::arity>>
auto make_foo(Function&& func)
-> decltype(make_foo_(std::forward<Function>(func), Indices()))
{
return make_foo_(std::forward<Function>(func), Indices());
}
代码不知怎么丑陋和不可读,但它绝对有效。希望它不依赖于一些实现定义的行为。此外,感谢所有的您的建议,它帮助! :)
The code is somehow ugly and unreadable, but it definitely works. Hope it does not rely on some implementation-defined behaviour now. Also, thanks all for your advice, it helped! :)
int main()
{
auto lambda = [](int i, float b, long c)
{
return long(i*10+b+c);
};
auto foo = make_foo(lambda);
std::cout << foo(5, 5.0, 2) << std::endl; // 57, it works!
}
a href =http://coliru.stacked-crooked.com/a/f3c026d881096320 =nofollow> live example :)