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模拟 std::function 功能

[英]Emulate std::function functionality

I'm trying to emulate the std::function functionality below, but I'm running into the following error我正在尝试模拟下面的std::function功能,但我遇到了以下错误

class Player
{
public:
void move_to(Point location);
};
std::function<void(Player&, Point)> fun = &Player::move_to;
Player hero;
fun(hero, point{ 2, 4 });
fp.cc:32:33: error: variable ‘fun<void(Player&, Point)> f’ has initializer but incomplete type
   32 |    fun<void(Player &, Point p)> f = &Player::move_to;

#include <iostream>

using namespace std;
template <typename T>
struct fun;

template <typename Ret, typename T, typename ...Args>
struct fun <Ret(*)(T&, Args...)>{
   char *data;
   using fptr = Ret(T::*)(Args...);
   fun(fptr p) : data(p) {}
   Ret operator()(T &t, Args... args) {
      if (std::is_same_v<Ret, void>) {
         (t.*((fptr)data))(args...);
      }
      return (t.*((fptr)data))(args...); 
   }
};

struct Point {
   int x;
   int y;
};

struct Player {
   void move_to(Point p) {
      cout << __PRETTY_FUNCTION__ << endl;
   }
};

int main() {
   fun<void(Player &, Point p)> f = &Player::move_to;
   Point p{1,2};
   Player pl;
//   f(pl, p);
}

Your code is close but has a couple small errors:您的代码很接近,但有几个小错误:

  • You have only provided a definition of fun for Ret(*)(T&, Args...) , but fun<void(Player &, Point p)> does not match this template.您只为Ret(*)(T&, Args...)提供了fun的定义,但fun<void(Player &, Point p)>与此模板不匹配。 So I removed the (*) which is not really required in your specialization.因此,我删除了您的专业中并不真正需要的(*)

  • A function pointer is not convertible to/from char * , so I changed the type of data to be fptr so it matches the argument type. function 指针不能与char *相互转换,因此我将data类型更改为fptr以使其与参数类型匹配。

godbolt link 螺栓链接

#include <iostream>

using namespace std;
template <typename T>
struct fun;

template <typename Ret, typename T, typename ...Args>
struct fun <Ret(T&, Args...)>{
   using fptr = Ret(T::*)(Args...);
   fptr data;
   fun(fptr p) : data(p) {}
   Ret operator()(T &t, Args... args) {
      if (std::is_same_v<Ret, void>) {
         (t.*((fptr)data))(args...);
      }
      return (t.*((fptr)data))(args...); 
   }
};

struct Point {
   int x;
   int y;
};

struct Player {
   void move_to(Point p) {
      cout << __PRETTY_FUNCTION__ << endl;
   }
};

int main() {
   fun<void(Player &, Point p)> f = &Player::move_to;
   Point p{1,2};
   Player pl;
   f(pl, p);
}
struct unused_t {};
// can store any of a function pointer, member function pointer, or other
// pointer.  It is a bit annoying, and not complete even.
union state_t {
  void(unused_t::*pm)();
  void(*pf)();
  void*pv;
  state_t():pv(nullptr){}
  template<class R0, class...A0s>
  state_t(R0(*f)(A0s...)):pf(reinterpret_cast<void(*)()>(f)) {}
  template<class X, class R0, class...A0s>
  state_t(R0(X::*f)(A0s...)):pm(reinterpret_cast<void(unused_t::*)()>(f)) {}
  template<class X, class R0, class...A0s>
  state_t(R0(X::*f)(A0s...) const):pm(reinterpret_cast<void(unused_t::*)()>(f)) {}
  // add in & and && overloads of member function pointers here

  template<class T>
  state_t(T* p):pv((void*)p) {}
};
template<class Sig>
struct fun;

template <class R, class...As>
struct fun <R(As...)>{
  using pf_t = R(*)(state_t, As&&...);

  state_t state;
  pf_t pf = nullptr;

  fun() = default;
  explicit operator bool() const { return pf; }
  template<class R0, class...A0s>
  fun( R0(*f)(A0s...) ):
    state(f),
    pf([](state_t state, As&&...as)->R{
      auto f = (R0(*)(A0s...))(state.pf);
      return std::invoke( f, std::forward<As>(as)... );
    })
  {}
  template<class T, class R0, class...A0s>
  fun( R0(T::*f)(A0s...) ):
    state(f),
    pf([](state_t state, As&&...as)->R{
      auto f = (R0(T::*)(A0s...))(state.pm);
      return std::invoke( f, std::forward<As>(as)... );
    })
  {}
  template<class T, class R0, class...A0s>
  fun( R0(T::*f)(A0s...) const ):
    state(f),
    pf([](state_t state, As&&...as)->R{
      auto f = (R0(T::*)(A0s...) const)(state.pm);
      return std::invoke( f, std::forward<As>(as)... );
    })
  {}
  R operator()(As... args) const {
    return pf( state, std::forward<As>(args)... );
  }
};

this does use std::invoke .这确实使用std::invoke

Test code:测试代码:

struct bob {
  int foo() const { return 3; };
};
int main() {
  fun<int(bob&)> f = &bob::foo;
  bob b;
  std::cout << f(b);
}

Live example .活生生的例子

Here is a version:这是一个版本:

struct unused_t {};
// can store any of a function pointer, member function pointer, or other
// pointer.  It is a bit annoying, and not complete even.
union state_t {
  void(unused_t::*pm)();
  void(*pf)();
  void*pv;
  state_t():pv(nullptr){}
  template<class R0, class...A0s>
  state_t(R0(*f)(A0s...)):pf(reinterpret_cast<void(*)()>(f)) {}
  template<class X, class R0, class...A0s>
  state_t(R0(X::*f)(A0s...)):pm(reinterpret_cast<void(unused_t::*)()>(f)) {}
  template<class X, class R0, class...A0s>
  state_t(R0(X::*f)(A0s...) const):pm(reinterpret_cast<void(unused_t::*)()>(f)) {}
  // add in & and && overloads of member function pointers here

  template<class T>struct tag_t {};
  template<class T>using type_t = T;

  template<class R, class...As>
  type_t<R(*)(As...)> extract( tag_t<R(*)(As...)>) const {
    return (R(*)(As...))pf;
  }
  template<class T, class R, class...As>
  type_t<R(T::*)(As...)> extract( tag_t<R(T::*)(As...)>) const {
    return (R(T::*)(As...))pm;
  }
  template<class T, class R, class...As>
  type_t<R(T::*)(As...) const> extract( tag_t<R(T::*)(As...) const>) const {
    return (R(T::*)(As...) const)pm;
  }
  template<class T>
  type_t<T*> extract( tag_t<T*>) const {
    return (T*)pv;
  }
  template<class T>
  state_t(T* p):pv((void*)p) {}
};
template<class Sig>
struct fun;

template <class R, class...As>
struct fun <R(As...)>{
  using pf_t = R(*)(state_t, As&&...);

  state_t state;
  pf_t pf = nullptr;

  fun() = default;
  explicit operator bool() const { return pf; }
  template<std::invocable<As...> F>
  requires std::is_convertible_v< std::invoke_result_t<F, As...>, R >
  fun( F f ):
    state(std::forward<F>(f)),
    pf([](state_t state, As&&...as)->R{
      auto f = state.extract(state_t::tag_t<F>{});
      return std::invoke( f, std::forward<As>(as)... );
    })
  {}
  R operator()(As... args) const {
    return pf( state, std::forward<As>(args)... );
  }
};

the annoying part is the storage of a pointer to member function, pointer or function pointer in a uniform way.烦人的部分是以统一的方式存储指向成员 function 指针或 function 指针的指针。 The C++ standard states that those 3 types of pointers are not guaranteed compatible. C++ 标准规定这 3 种类型的指针不保证兼容。

So the state_t type can store any one of the 3 in a union, but there is a bunch of boilerplate to treat them uniformly.所以state_t类型可以将 3 个中的任何一个存储在一个 union 中,但是有一堆样板可以统一对待它们。

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