问题描述

在某些情况下，希望能够对可调用对象进行类型擦除(例如，函数、函数指针、带有 operator()、lambda、mem_fn 的对象实例)，例如在 使用带有 C++11 lambdas 的 Boost 适配器 中，其中复制-需要可赋值和默认构造的类型.

In some situations it's desirable to be able to type-erase a callable (e.g. function, function pointer, object instance with operator(), lambda, mem_fn), for instance in Using Boost adaptors with C++11 lambdas where a copy-assignable and default-constructible type is required.

std::function 是理想的，但似乎没有办法自动确定实例化类模板的签名std::function 与.有没有一种简单的方法来获取任意可调用的函数签名和/或将其包装在适当的 std::function 实例化实例(即 make_function 函数模板)中?

std::function would be ideal, but there seems to be no way to automatically determine what signature to instantiate the class template std::function with. Is there an easy way to get the function signature of an arbitrary callable and/or wrap it in an appropriate std::function instantiation instance (i.e. a make_function function template)?

具体来说，我正在寻找其中之一

Specifically, I'm looking for one or other of

template<typename F> using get_signature = ...;
template<typename F> std::function<get_signature<F>> make_function(F &&f) { ... }

这样 make_function([](int i) { return 0; }) 返回一个 std::function.显然，如果一个实例可以用多个签名调用(例如，具有多个、模板或默认参数 operator()s 的对象)，这将不会起作用.

such that make_function([](int i) { return 0; }) returns a std::function<int(int)>. Obviously this wouldn't be expected to work if an instance is callable with more than one signature (e.g. objects with more than one, template or default-parameter operator()s).

Boost 很好，虽然不是过于复杂的非 Boost 解决方案是首选.

Boost is fine, although non-Boost solutions that aren't excessively complex are preferred.

回答我自己的问题.

推荐答案

我想出了一个相当讨厌的非库解决方案，利用 lambdas 具有 operator() 的事实:

I've come up with a fairly nasty non-library solution, using the fact that lambdas have operator():

template<typename T> struct remove_class { };
template<typename C, typename R, typename... A>
struct remove_class<R(C::*)(A...)> { using type = R(A...); };
template<typename C, typename R, typename... A>
struct remove_class<R(C::*)(A...) const> { using type = R(A...); };
template<typename C, typename R, typename... A>
struct remove_class<R(C::*)(A...) volatile> { using type = R(A...); };
template<typename C, typename R, typename... A>
struct remove_class<R(C::*)(A...) const volatile> { using type = R(A...); };

template<typename T>
struct get_signature_impl { using type = typename remove_class<
    decltype(&std::remove_reference<T>::type::operator())>::type; };
template<typename R, typename... A>
struct get_signature_impl<R(A...)> { using type = R(A...); };
template<typename R, typename... A>
struct get_signature_impl<R(&)(A...)> { using type = R(A...); };
template<typename R, typename... A>
struct get_signature_impl<R(*)(A...)> { using type = R(A...); };
template<typename T> using get_signature = typename get_signature_impl<T>::type;

template<typename F> using make_function_type = std::function<get_signature<F>>;
template<typename F> make_function_type<F> make_function(F &&f) {
    return make_function_type<F>(std::forward<F>(f)); }

有什么可以简化或改进的想法吗?有什么明显的错误吗?

Any ideas where this can be simplified or improved? Any obvious bugs?

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