Question

假设我有一个有100个孩子的基类：

class Base { 
  virtual void feed();
  ...   
};
class Child1 : public Base {
  void feed();  //specific procedure for feeding Child1
  ... 
};
...
class Child100 : public Base { 
  void feed();  //specific procedure for feeding Child100
  ...
};

在运行时，我想读取一个文件，其中包含要创建和提供的子项。假设我已经读过文件，字符串“names”的向量包含子类的名称（即Child1，Child4，Child99）。现在我将迭代这些字符串，创建特定子项的实例，并使用其特定的提供程序来提供它：

vector<Base *> children;    
for (vector<string>::iterator it = names.begin(); it != names.end(); ++it) {
  Base * child = convert_string_to_instance(*it)       
  child->feed()
  children.push_back(child);
}

如何创建convert_string_to_instance（）函数，如果它接受字符串“Child1”，则返回“new Child1”，如果字符串参数为“Child4”则返回“new Child4”等

<class C *> convert_string_to_instance(string inName) {
  // magic happens
  return new C;  // C = inName

  // <brute force?>
  // if (inName == "Child1")
  //   return new Child1;
  // if (inName == "Child2")
  //   return new Child2;    
  // if (inName == "Child3")
  //   return new Child3;    
  // </brute force>
  }

Answer 1

C ++没有提供像这样动态构造类实例的方法。但是，您可以使用代码生成从类列表中生成“强力”代码（如上所示）。然后，#include convert_string_to_instance方法生成的代码。{/ p>

您还可以设置项目构建系统，以便在类列表发生更改时重建生成的代码。

Answer 2

我问了一个题为automatic registration of object creator function with a macro的问题，该问题包含以下运行的示例程序：

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

struct Object{ virtual ~Object() {} }; // base type for all objects

struct ObjectFactory {
  static Object* create(const std::string& id) { // creates an object from a string
    const Creators_t::const_iterator iter = static_creators().find(id);
    return iter == static_creators().end() ? 0 : (*iter->second)(); // if found, execute the creator function pointer
  }

 private:
  typedef Object* Creator_t(); // function pointer to create Object
  typedef std::map<std::string, Creator_t*> Creators_t; // map from id to creator
  static Creators_t& static_creators() { static Creators_t s_creators; return s_creators; } // static instance of map
  template<class T = int> struct Register {
    static Object* create() { return new T(); };
    static Creator_t* init_creator(const std::string& id) { return static_creators()[id] = create; }
    static Creator_t* creator;
  };
};

#define REGISTER_TYPE(T, STR) template<> ObjectFactory::Creator_t* ObjectFactory::Register<T>::creator = ObjectFactory::Register<T>::init_creator(STR)

namespace A { struct DerivedA : public Object { DerivedA() { std::cout << "A::DerivedA constructor\n"; } }; }
REGISTER_TYPE(A::DerivedA, "A");

namespace B { struct DerivedB : public Object { DerivedB() { std::cout << "B::DerivedB constructor\n"; } }; }
REGISTER_TYPE(B::DerivedB, "Bee");

namespace C { struct DerivedC : public Object { DerivedC() { std::cout << "C::DerivedC constructor\n"; } }; }
REGISTER_TYPE(C::DerivedC, "sea");

namespace D { struct DerivedD : public Object { DerivedD() { std::cout << "D::DerivedD constructor\n"; } }; }
REGISTER_TYPE(D::DerivedD, "DEE");

int main(void)
{
  delete ObjectFactory::create("A");
  delete ObjectFactory::create("Bee");
  delete ObjectFactory::create("sea");
  delete ObjectFactory::create("DEE");
  return 0;
}

编译并运行输出是：

> g++ example2.cpp && ./a.out
A::DerivedA constructor
B::DerivedB constructor
C::DerivedC constructor
D::DerivedD constructor

Answer 3

如果你有很多课程，你通常会选择一种不那么暴力的方法。类名和工厂函数之间的trie或hash_map是一种很好的方法。

您可以使用Greg建议的codegen方法来构建此工厂表，例如doxygen可以解析您的源代码并输出xml格式的所有类的列表以及继承关系，因此您可以轻松找到所有派生类来自一个共同的“接口”基类。

Answer 4

您可以滥用预处理器并设置一些静态类成员，这些成员通过像Ben描述的hash_map向工厂注册您的类。如果您有visual studio，请查看在MFC中如何实现DECLARE_DYNCREATE。我做过类似实现类工厂的事情。非标准肯定，但由于C ++不对这种机制提供任何形式的支持，任何解决方案都可能是非标准的。

修改

我之前在评论中说过，我正在努力记录我所做过的缩小版本。按比例缩小的版本仍然相当大I posted it here。如果有足够的兴趣，我可以在这个网站上复制/粘贴它。让我知道。

Answer 5

听起来您可能正在使用子类来处理应该编码为字段的内容。

不要在100个类中编写不同的行为，而是考虑使用规则/常量/函数指针构建一个查找表，以便从一个类中实现正确的行为。

例如，而不是：

class SmallRedSquare  : public Shape {...};
class SmallBlueSquare : public Shape {...};
class SmallBlueCircle : public Shape {...};
class SmallRedCircle  : public Shape {...};
class BigRedSquare    : public Shape {...};
class BigBlueSquare   : public Shape {...};
class BigBlueCircle   : public Shape {...};
class BigRedCircle    : public Shape {...};

尝试：

struct ShapeInfo
{
   std::string type;
   Size size;
   Color color;
   Form form;
};

class Shape
{
public:
    Shape(std::string type) : info_(lookupInfoTable(type)) {}

    void draw()
    {
        // Use info_ to draw shape properly.
    }

private:
    ShapeInfo* lookupInfoTable(std::string type) {info_ = ...;}

    ShapeInfo* info_;
    static ShapeInfo infoTable_[];
};

const ShapeInfo Shape::infoTable_[] =
{
    {"SmallRedSquare",  small,  red, &drawSquare},
    {"SmallBlueSquare", small, blue, &drawSquare},
    {"SmallRedCircle",  small,  red, &drawCircle},
    {"SmallBlueCircle", small, blue, &drawCircle},
    {"BigRedSquare",      big,  red, &drawSquare},
    {"BigBlueSquare",     big, blue, &drawSquare},
    {"BigBlueCircle",     big,  red, &drawCircle},
    {"BigRedCircle",      big, blue, &drawCircle}
}

int main()
{
    Shape s1("SmallRedCircle");
    Shape s2("BigBlueSquare");
    s1.draw();
    s2.draw();
}

这个想法可能不适用于您的问题，但我认为无论如何都不会有任何伤害。： - ）

我的想法就像Replace Subclass with Fields重构一样，但我会更进一步。

Answer 6

这是一种可怕的，可怕的方式的骨架：

class Factory {
  public:
    virtual Base * make() = 0;
};

template<typename T> class TemplateFactory : public Factory {
  public:
    virtual Base * make() {
      return dynamic_cast<Base *>(new T());
    }
};

map<string, Factory *> factories;

#define REGISTER(classname) factories[ #classname ] = new TemplateFactory<classname>()

然后为REGISTER(classname);的每个相关派生类调用Base，并使用factories["classname"]->make()获取类型为classname的新对象。上面编写的代码有明显的缺陷，包括内存泄漏的巨大可能性，以及组合宏和模板的一般可怕性。

Answer 7

看到强大的助推器。

要使用我的解决方案，您必须做的一件事是向所有类添加一个新成员，这是一个包含该类名称的static const string。可能还有其他方法可以做到，但这就是我现在所拥有的。

#include <iostream>
#include <vector>
#include <string>

#include <boost/fusion/container/list/cons.hpp>
#include <boost/fusion/algorithm/iteration/for_each.hpp>
#include <boost/fusion/view/iterator_range.hpp>

using namespace std;
using boost::fusion::cons;


class Base { virtual void feed(){ } };

class Child1 : public Base{
  void feed(){ }

public:
  static const string name_;
};
const string Child1::name_ = "Child1";

class Child3 : public Base{
  void feed(){ }

public:
  static const string name_;
};
const string Child3::name_ = "Child3";

//...
class Child100 : public Base{

  void feed(){ }

public:
  static const string name_;
};
const string Child100::name_ = "Child100";

// This is probably the ugliest part, but I think it's worth it.
typedef cons<Child1, cons<Child3, cons<Child100> > > MyChildClasses;

typedef vector<Base*> Children;
typedef vector<string> Names;

struct CreateObjects{      // a.k.a convert_string_to_instance() in your example.

  CreateObjects(Children& children, string name) : children_(&children), name_(name){ }

  template <class T>
  void operator()(T& cs) const{

    if( name_ == cs.name_ ){
      cout << "Created " << name_ << " object." << endl;
      (*children_).push_back(new T);
    }else{
      cout << name_ << " does NOT match " << cs.name_ << endl;
    }
  }

  Children* children_;
  string name_;
};

int main(int argc, char* argv[]){

  MyChildClasses myClasses;

  Children children;
  Names names;
  names.push_back("Child1");
  names.push_back("Child100");
  names.push_back("Child1");
  names.push_back("Child100");

  // Extra test.
  // string input;
  // cout << "Enter a name of a child class" << endl;
  // cin >> input;
  // names.push_back(input);

  using namespace boost::fusion;
  using boost::fusion::begin;
  using boost::fusion::for_each;

  for(Names::iterator namesIt = names.begin(); namesIt != names.end(); ++namesIt){

    // You have to know how many types there are in the cons at compile time.
    // In this case I have 3; Child1, Child3, and Child100
    boost::fusion::iterator_range<
      result_of::advance_c<result_of::begin<MyChildClasses>::type, 0>::type,
      result_of::advance_c<result_of::begin<MyChildClasses>::type, 3>::type
      > it(advance_c<0 >(begin(myClasses)),
       advance_c<3>(begin(myClasses)));
    for_each(it, CreateObjects(children, *namesIt));
  }

  cout << children.size() << " objects created." << endl;
  return 0;
}

从包含C ++中类名的字符串动态创建类的实例

7 个答案:

修改