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C++0x ：在 std::map 中存储任何类型的 std::function

发布于 2024-12-07 08:23:35 字数 2021 浏览 0 评论 0原文

我试图在地图中存储一组 std::function （在 GCC 4.5 下）

我想得到两种东西：

  • 存储已传递参数的函数；那么你就拥有了 调用 f()
  • 存储不带参数的函数；那么你必须打电话 f(...)

我想我用类 Command 和 Manager 实现了第一个：

class Command
{
  std::function<void()> f_;
  public:
    Command() {}
    Command(std::function<void()> f) : f_(f) {}

    void execute() { if(f_) f_(); }

};

class CommandManager
{
  typedef map<string, Command*> FMap;

  public :

  void add(string name, Command* cmd)
  {
     fmap1.insert(pair<string, Command*>(name, cmd));
  }

  void execute(string name)
  {
    FMap::const_iterator it = fmap1.find(name);
    if(it != fmap1.end())
    {
      Command* c = it->second;
      c->execute();
    }
  }

  private :

    FMap fmap1;

};

可以像这样使用：

class Print{

   public:
   void print1(string s, string s1){ cout<<"print1 : "<<"s : "<<s<<" s1 : "<<s1<<endl; }
   int print2(){ cout<<"print2"<<endl; return 2;}

};

#include <string>
#include <functional>

int main()
{
  Print p = Print();

  function<void()> f1(bind(&Print::print1, &p, string("test1"), string("test2")));

  function<int()> f2(bind(&Print::print2, &p));

  CommandManager cmdMgr = CommandManager();
  cmdMgr.add("print1", new Command(f1));
  cmdMgr.execute("print1");

  cmdMgr.add("print2", new Command(f2));
  cmdMgr.execute("print2");

  return 0;
}

现在我希望能够做到这一点：

 int main()
 {
      Print p = Print();

      function<void(string, string)> f1(bind(&Print::print1, &p, placeholders::_1, placeholders::_2));

      CommandManager cmdMgr = CommandManager();
      cmdMgr.add("print1", new Command(f1));
      cmdMgr.execute("print1", string("test1"), string("test2"));

      return 0;
    }

有没有办法，例如使用类型擦除？

原文

I'm trying to store a set of std::function in a map (under GCC 4.5)

I'd like to get 2 kind of things :

  • storing functions with arguments already passed; then you just have
    to call f()
  • storing functions without arguments; then you have to call
    f(...)

I think I achieved the first one with a class Command and a Manager :

class Command
{
  std::function<void()> f_;
  public:
    Command() {}
    Command(std::function<void()> f) : f_(f) {}

    void execute() { if(f_) f_(); }

};

class CommandManager
{
  typedef map<string, Command*> FMap;

  public :

  void add(string name, Command* cmd)
  {
     fmap1.insert(pair<string, Command*>(name, cmd));
  }

  void execute(string name)
  {
    FMap::const_iterator it = fmap1.find(name);
    if(it != fmap1.end())
    {
      Command* c = it->second;
      c->execute();
    }
  }

  private :

    FMap fmap1;

};

can be used like this :

class Print{

   public:
   void print1(string s, string s1){ cout<<"print1 : "<<"s : "<<s<<" s1 : "<<s1<<endl; }
   int print2(){ cout<<"print2"<<endl; return 2;}

};

#include <string>
#include <functional>

int main()
{
  Print p = Print();

  function<void()> f1(bind(&Print::print1, &p, string("test1"), string("test2")));

  function<int()> f2(bind(&Print::print2, &p));

  CommandManager cmdMgr = CommandManager();
  cmdMgr.add("print1", new Command(f1));
  cmdMgr.execute("print1");

  cmdMgr.add("print2", new Command(f2));
  cmdMgr.execute("print2");

  return 0;
}

Now I'd like to be able to do this :

 int main()
 {
      Print p = Print();

      function<void(string, string)> f1(bind(&Print::print1, &p, placeholders::_1, placeholders::_2));

      CommandManager cmdMgr = CommandManager();
      cmdMgr.add("print1", new Command(f1));
      cmdMgr.execute("print1", string("test1"), string("test2"));

      return 0;
    }

Is there a way, using type-erasure for example ?

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评论（3）

触ぅ动初心 2024-12-14 08:23:35

您可以使用动态转换来在运行时确定列表中函数的类型。
请注意，我添加了shared_ptr来消除原始示例中的内存泄漏。如果使用错误的参数调用execute方法（如果dynamic_cast产生0），也许您想抛出异常。

用法：

void x() {}
void y(int ) {}
void main() {
    CommandManager m;
    m.add("print", Command<>(x));
    m.add("print1", Command<int>(y));
    m.execute("print");
    m.execute("print1", 1);
}

代码（带可变参数模板支持，例如 gcc-4.5）：

#include <functional>
#include <map>
#include <string>
#include <memory>

using namespace std;

class BaseCommand
{
public:
    virtual ~BaseCommand() {}
};

template <class... ArgTypes>
class Command : public BaseCommand
{
  typedef std::function<void(ArgTypes...)> FuncType;
  FuncType f_;
  public:
    Command() {}
    Command(FuncType f) : f_(f) {}
    void operator()(ArgTypes... args) { if(f_) f_(args...); }
};

class CommandManager
{
  typedef shared_ptr<BaseCommand> BaseCommandPtr;
  typedef map<string, BaseCommandPtr> FMap;
  public :

  template <class T>
  void add(string name, const T& cmd)
  {
     fmap1.insert(pair<string, BaseCommandPtr>(name, BaseCommandPtr(new T(cmd))));
  }

  template <class... ArgTypes>
  void execute(string name, ArgTypes... args)
  {
    typedef Command<ArgTypes...> CommandType;
    FMap::const_iterator it = fmap1.find(name);
    if(it != fmap1.end())
    {
      CommandType* c = dynamic_cast<CommandType*>(it->second.get());
      if(c)
      {
    (*c)(args...);
      }
    }
  } 

  private :
    FMap fmap1;
};

不带可变参数模板支持（例如 VS2010）：

#include <functional>
#include <map>
#include <string>
#include <memory>

using namespace std;
class Ignored;

class BaseCommand
{
public:
    virtual ~BaseCommand() = 0 {};
};

template <class A1 = Ignored>
class Command : public BaseCommand
{
  typedef std::function<void(A1)> FuncType;
  FuncType f_;
  public:
    Command() {}
    Command(FuncType f) : f_(f) {}
    void operator()(const A1& a1) { if(f_) f_(a1); }
};

template <>
class Command<Ignored> : public BaseCommand
{
  typedef std::function<void()> FuncType;
  FuncType f_;
  public:
    Command() {}
    Command(FuncType f) : f_(f) {}
    void operator()() { if(f_) f_(); }
};

class CommandManager
{
  typedef shared_ptr<BaseCommand> BaseCommandPtr;
  typedef map<string, BaseCommandPtr> FMap;
  public :

  template <class T>
  void add(string name, const T& cmd)
  {
     fmap1.insert(pair<string, BaseCommandPtr>(name, BaseCommandPtr(new T(cmd))));
  }

  template <class A1>
  void execute(string name, const A1& a1)
  {
    typedef Command<A1> CommandType;
    FMap::const_iterator it = fmap1.find(name);
    if(it != fmap1.end())
    {
      CommandType* c = dynamic_cast<CommandType*>(it->second.get());
      if(c)
      {
        (*c)(a1);
      }
    }
  } 

  void execute(string name)
  {
    typedef Command<> CommandType;
    FMap::const_iterator it = fmap1.find(name);
    if(it != fmap1.end())
    {
      CommandType* c = dynamic_cast<CommandType*>(it->second.get());
      if(c)
      {
        (*c)();
      }
    }
  }
  private :
    FMap fmap1;
};

You could use dynamic cast to determine the type of the function in the list at runtime.
Please note that I added shared_ptr to remove the memory leak in the original sample. Perhaps you want to throw a exception if the execute method is called with the wrong arguments (if the dynamic_cast yields 0).

Usage:

void x() {}
void y(int ) {}
void main() {
    CommandManager m;
    m.add("print", Command<>(x));
    m.add("print1", Command<int>(y));
    m.execute("print");
    m.execute("print1", 1);
}

Code (with variadic template support for example gcc-4.5):

#include <functional>
#include <map>
#include <string>
#include <memory>

using namespace std;

class BaseCommand
{
public:
    virtual ~BaseCommand() {}
};

template <class... ArgTypes>
class Command : public BaseCommand
{
  typedef std::function<void(ArgTypes...)> FuncType;
  FuncType f_;
  public:
    Command() {}
    Command(FuncType f) : f_(f) {}
    void operator()(ArgTypes... args) { if(f_) f_(args...); }
};

class CommandManager
{
  typedef shared_ptr<BaseCommand> BaseCommandPtr;
  typedef map<string, BaseCommandPtr> FMap;
  public :

  template <class T>
  void add(string name, const T& cmd)
  {
     fmap1.insert(pair<string, BaseCommandPtr>(name, BaseCommandPtr(new T(cmd))));
  }

  template <class... ArgTypes>
  void execute(string name, ArgTypes... args)
  {
    typedef Command<ArgTypes...> CommandType;
    FMap::const_iterator it = fmap1.find(name);
    if(it != fmap1.end())
    {
      CommandType* c = dynamic_cast<CommandType*>(it->second.get());
      if(c)
      {
    (*c)(args...);
      }
    }
  } 

  private :
    FMap fmap1;
};

without variadic template support (example VS2010):

#include <functional>
#include <map>
#include <string>
#include <memory>

using namespace std;
class Ignored;

class BaseCommand
{
public:
    virtual ~BaseCommand() = 0 {};
};

template <class A1 = Ignored>
class Command : public BaseCommand
{
  typedef std::function<void(A1)> FuncType;
  FuncType f_;
  public:
    Command() {}
    Command(FuncType f) : f_(f) {}
    void operator()(const A1& a1) { if(f_) f_(a1); }
};

template <>
class Command<Ignored> : public BaseCommand
{
  typedef std::function<void()> FuncType;
  FuncType f_;
  public:
    Command() {}
    Command(FuncType f) : f_(f) {}
    void operator()() { if(f_) f_(); }
};

class CommandManager
{
  typedef shared_ptr<BaseCommand> BaseCommandPtr;
  typedef map<string, BaseCommandPtr> FMap;
  public :

  template <class T>
  void add(string name, const T& cmd)
  {
     fmap1.insert(pair<string, BaseCommandPtr>(name, BaseCommandPtr(new T(cmd))));
  }

  template <class A1>
  void execute(string name, const A1& a1)
  {
    typedef Command<A1> CommandType;
    FMap::const_iterator it = fmap1.find(name);
    if(it != fmap1.end())
    {
      CommandType* c = dynamic_cast<CommandType*>(it->second.get());
      if(c)
      {
        (*c)(a1);
      }
    }
  } 

  void execute(string name)
  {
    typedef Command<> CommandType;
    FMap::const_iterator it = fmap1.find(name);
    if(it != fmap1.end())
    {
      CommandType* c = dynamic_cast<CommandType*>(it->second.get());
      if(c)
      {
        (*c)();
      }
    }
  }
  private :
    FMap fmap1;
};
回复 原文
自由如风 2024-12-14 08:23:35

如果没有一些认真的运行时工作和相关成本，您想要做的事情是不可能的。最简单的解决方案当然是只存储 boost::any< /a> (any_function never made it into boost) 在你的地图中并进行必要的转换（或添加一些运行时数据来告诉你要进行哪个转换），尽管你应该不惜一切代价避免这种情况，并且使用固定参数或不使用参数。
然后，您的用户可以使用 bind 修改其函数以匹配您所需的签名。

编辑：在您当前的方案中，我认为 CommandManager 没有理由在地图中存储 Command* 。

Edit2：您还可以删除返回类型。这对于您的用例来说可能没问题，但使其不那么通用。

Edit3：我使用 any 编写了一些代码的工作示例。我觉得存在一些缺陷，我真的不知道这应该实现什么，但这里是：

#include <iostream>
#include <string>
#include <map>
#include <functional>

#include <boost/any.hpp>

class AnyCaller
{
  std::map<std::string, boost::any> calls;
public:
  AnyCaller() {}

  void add(const std::string& name, const boost::any& fun) {
    calls[name] = fun;
  }

  // we always use the throwing version of any_cast
  // icbb by error checking

  // no arg version
  template<typename Ret>
  Ret call(const std::string& s) {
    const boost::any& a = calls[s];
    return boost::any_cast< std::function<Ret(void)> >(a)();
  }

  // this should be a variadic template to be actually usable
  template<typename Ret, typename T>
  Ret call(const std::string& s, T&& arg) {
    // we have to assume that our users know what we are actually returning here
    const boost::any& a = calls[s];
    return boost::any_cast< std::function<Ret(T)> >(a)(std::forward<T>(arg));
  }

  virtual ~AnyCaller() {}
};

int foo() { std::cout << "foo" << std::endl; return 1; }
double foo2(int i) { std::cout << "foo2" << std::endl; return double(i); }

int main()
{
  AnyCaller c; 
  c.add("foo", std::function<int(void)>(foo));
  c.add("foo2", std::function<double(int)>(foo2));

  c.call<int>("foo");
  c.call<double, int>("foo2", 1);
  // this should throw
  c.call<double, int>("foo", 1);
  return 0;
}

至于使用固定签名的示例。想想你要存储的函数的最自然表示是什么（看看你的 Command 示例，我假设它是 std::function; 存储这种类型的函数，每当您的用户尝试使用它时，他都必须绑定他想要使用的任何函数，以便它与此签名匹配。

What you are trying to do is not possible without some serious runtime work and the associated cost. The simplest solution would of course to just store a boost::any (any_function never made it into boost) inside your map and do the necessary casts (or add some runtime data that tells you which cast to make), although you should avoid that at any cost and go with fixed arguments or no arguments.
Your users can then modify their functions using bind to match the signature you require.

Edit: In your current scheme I see no reason for CommandManager to store Command* in the map.

Edit2: Also you drop the return type. This could be OK for your use-case but makes this a lot less generic.

Edit3: I worked out some working example of your code using any. I feel that there is some flaw and I really don't see what this should achieve but here it goes:

#include <iostream>
#include <string>
#include <map>
#include <functional>

#include <boost/any.hpp>

class AnyCaller
{
  std::map<std::string, boost::any> calls;
public:
  AnyCaller() {}

  void add(const std::string& name, const boost::any& fun) {
    calls[name] = fun;
  }

  // we always use the throwing version of any_cast
  // icbb by error checking

  // no arg version
  template<typename Ret>
  Ret call(const std::string& s) {
    const boost::any& a = calls[s];
    return boost::any_cast< std::function<Ret(void)> >(a)();
  }

  // this should be a variadic template to be actually usable
  template<typename Ret, typename T>
  Ret call(const std::string& s, T&& arg) {
    // we have to assume that our users know what we are actually returning here
    const boost::any& a = calls[s];
    return boost::any_cast< std::function<Ret(T)> >(a)(std::forward<T>(arg));
  }

  virtual ~AnyCaller() {}
};

int foo() { std::cout << "foo" << std::endl; return 1; }
double foo2(int i) { std::cout << "foo2" << std::endl; return double(i); }

int main()
{
  AnyCaller c; 
  c.add("foo", std::function<int(void)>(foo));
  c.add("foo2", std::function<double(int)>(foo2));

  c.call<int>("foo");
  c.call<double, int>("foo2", 1);
  // this should throw
  c.call<double, int>("foo", 1);
  return 0;
}

As for the example using a fixed signature. Just think of what would be the most natural representation of a function you are going to store (looking at your Command example I'd assume it is std::function<void(void)>. Store functions of this type and whenever one your users tries to use it, he has to bind whatever function he wants to use, so it matches this signature.

回复 原文
独守阴晴ぅ圆缺 2024-12-14 08:23:35

您的 Command 类构造函数需要一个 function。您正在尝试为其提供一个function。这不会进行类型检查。

如果您需要接受可变参数的函数（例如 printf），则需要接受可变参数的 function<> 和 execute()。您需要知道如何使用它（特别是，您需要一个固定的第一个参数）。然后，您负责类型安全，就像 printf 一样。

如果您只需要可变数量的字符串参数，请使用接受字符串向量等的函数。

所有这些与 std::map 无关。无论您可以存储在普通旧变量中，您也可以存储在 std::map 中。

Your Command class constructor needs a function<void()>. You are trying to feed it a function<void(string,string)>. This is not going to typecheck.

If you need functions that accept variable arguments (like printf), you will need function<> and execute() that accept variable arguments. You need to know how to work with that (in particular, you need a fixed first argument). You are then responsible for type safety, much like with printf.

If you just need a variable number of string arguments, use functions that accept e.g. vectors of strings.

All this has nothing to do whatsoever with std::map. Whatever you can store in a plain old variable, you can store in std::map too.

回复 原文
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