初始化包含带有lambda的std :: function的类 [英] Initialize class containing a std::function with a lambda

问题描述

我通过以下方式创建了一个包含 std :: function 的模板类：

I created a template class containing a std::function as a member the following way:

template<typename Ret, typename... Args>
class Foo
{
private:
    std::function<Ret(Args...)> _func;

public:
    Foo(const std::function<Ret(Args...)>& func):
        _func(func)
    {}
};

为了不必指定传递函数的参数和返回类型， code> make_foo 重载：

In order not to have to specify the arguments and return type of the passed function, I created some make_foo overloads:

template<typename Ret, typename... Args>
auto make_foo(Ret (&func)(Args...))
    -> Foo<Ret, Args...>
{
    return { std::function<Ret(Args...)>(func) };
}

template<typename Ret, typename... Args>
auto make_foo(const std::function<Ret(Args...)>& func)
    -> Foo<Ret, Args...>
{
    return { func };
}

但是，我无法创建 make_foo 重载，它将lambda作为参数：

However, I was unable to create a make_foo overload that takes a lambda as parameter:

template<typename Ret, typename... Args>
auto make_foo(??? func)
    -> Foo<Ret, Args...>
{
    return { std::function<Ret(Args...)>(func) };
}

我找不到一种方法来获得返回类型和参数类型从lambda自动推导。是否有一个惯用的方法来解决这个问题？

I just can't find a way to have the return type and argument types automatically deduced from the lambda. Is there an idiomatic way to solve such a problem?

推荐答案

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StackOverflow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 StackOverflow. 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 ：）

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