推断lambda或任意可调用的“make_function”的调用签名。 [英] Inferring the call signature of a lambda or arbitrary callable for "make_function"

问题描述

在某些情况下，需要能够使用 operator（），lambda，<$ c $来类型擦除可调用（例如函数，函数指针， c> mem_fn ），例如在使用Boost适配器与C ++ 11 lambdas 其中需要一个可复制赋值和默认构造类型。

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 将是理想的，但似乎没有办法自动确定什么签名实例化类模板 std :: function 与。有一个简单的方法来获取任意可调用的函数签名和/或将它包装在适当的 std :: function 实例化实例中（即 make_function 函数模板）？

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

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

> make_function（[]（int i）{return 0;}）返回 std :: function< int（int）> 。显然，如果实例可以使用多个签名（例如具有多个模板或默认参数 operator（）的对象） 。

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是好的，虽然不是非过于复杂的非Boost解决方案是首选。

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

编辑：回答我自己的问题。

answering my own question.

推荐答案

讨厌的非库解决方案，使用lambdas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)); }

任何可以简化或改进的想法？任何明显的错误？

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

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