Question

我有2 structs，其构造函数接受传入的不同数量和类型的参数。例如，

struct A:Base
{
   A(int i, char c){}
};

struct B:Base
{
   B(char c){}
};

我还有一个工厂方法来创建这两个structs对象，

struct F
{
   Base* Do(int i)
   {
      if (i==0)
      {
         return new A(i,'c');
      }
      else
      {
         return new B('c'); 
      }
   }  
};

现在我尝试像这样改进Factory F

struct F
{
   template<int i, class X>
   Base* Do();
   template<class X>
   Base* Do<0>()
   {
      return new X(i, 'c');
   }
   template<class X>
   Base* Do<1>()
   {
      return new X('c');
   }
};

我唯一对这种方法不感兴趣的是，我必须编写很多专门的模板，以防我有很多结构A,B,C,...但我只喜欢模板，我不喜欢{ {1}}。

所以我的问题是：

如何最小化长代码但仍然使用模板？
如何停止返回指向基类的指针？动态分配sux，我更喜欢工厂堆栈中缓慢而温柔的经典分配器。
我喜欢使用可变参数模板作为return new X(...)＆＃39;的参数。构建函数。

Answer 1

std::any和可变参数模板将成为您的朋友。 Yuu可以将任何内容存储在std::any中。 any_cast有点问题。

请参见以下工作代码：

#include <iostream>
#include <map>
#include <utility>
#include <any>


// Some demo classes ----------------------------------------------------------------------------------
struct Base {
    Base(int d) : data(d) {};
    virtual ~Base() { std::cout << "Destructor Base\n"; }
    virtual void print() { std::cout << "Print Base\n"; }
    int data{};
};
struct Child1 : public Base {
    Child1(int d, std::string s) : Base(d) { std::cout << "Constructor Child1 " << d << " " << s << "\n"; }
    virtual ~Child1() { std::cout << "Destructor Child1\n"; }
    virtual void print() { std::cout << "Print Child1: " << data << "\n"; }
};
struct Child2 : public Base {
    Child2(int d, char c, long l) : Base(d) { std::cout << "Constructor Child2 " << d << " " << c << " " << l << "\n"; }
    virtual ~Child2() { std::cout << "Destructor Child2\n"; }
    virtual void print() { std::cout << "Print Child2: " << data << "\n"; }
};
struct Child3 : public Base {
    Child3(int d, long l, char c, std::string s) : Base(d) { std::cout << "Constructor Child3 " << d << " " << l << " " << c << " " << s << "\n"; }
    virtual ~Child3() { std::cout << "Destructor Child3\n"; }
    virtual void print() { std::cout << "Print Child3: " << data << "\n"; }
};



using UPTRB = std::unique_ptr<Base>;


template <class Child, typename ...Args>
UPTRB createClass(Args...args) { return std::make_unique<Child>(args...); }

// The Factory ----------------------------------------------------------------------------------------
template <class Key, class Object>
class Factory
{
    std::map<Key, std::any> selector;
public:
    Factory() : selector() {}
    Factory(std::initializer_list<std::pair<const Key, std::any>> il) : selector(il) {}

    template<typename Function>
    void add(Key key, Function&& someFunction) { selector[key] = std::any(someFunction); };

    template <typename ... Args>
    Object create(Key key, Args ... args) {
        if (selector.find(key) != selector.end()) {
            return std::any_cast<std::add_pointer_t<Object(Args ...)>>(selector[key])(args...);
        }
        else return nullptr;
    }
};

int main()
{
    Factory<int, UPTRB> factory{
        {1, createClass<Child1, int, std::string>},
        {2, createClass<Child2, int, char, long>}
    };
    factory.add(3, createClass<Child3, int, long, char, std::string>);


    // Some test values
    std::string s1(" Hello1 "); std::string s3(" Hello3 ");
    int i = 1;  const int ci = 1;   int& ri = i;    const int& cri = i;   int&& rri = 1;

    UPTRB b1 = factory.create(1, 1, s1);
    UPTRB b2 = factory.create(2, 2, '2', 2L);
    UPTRB b3 = factory.create(3, 3, 3L, '3', s3);

    b1->print();
    b2->print();
    b3->print();
    b1 = factory.create(2, 4, '4', 4L);
    b1->print();
    return 0;
}

结束工厂方法

1 个答案: