C ++服务提供商 [英] C++ Service Providers

问题描述

我一直在学习C ++，来自C＃，我习惯于使用服务提供商：基本上是一个Dictionary< Type，object>。不幸的是，我不知道如何在C ++中做到这一点。因此，问题基本上是：

1. 如何在C ++中创建字典。




2. 如果我知道C ++中没有Type，我该如何使用Type。



3. 与上述相同，但使用'object'。

谢谢！

解决方案

我假设您正在尝试将类型映射到单个对象实例。您可以尝试以下行：

  #include< typeinfo> 
 #include< map> 
 #include< string> 
 using namespace std; 
 
 class SomeClass 
 {
 public：
 virtual〜SomeClass（）{} //获取v表的虚函数
}; 
 
 struct type_info_less 
 {
 bool operator（）（const std :: type_info * lhs，const std :: type_info * rhs）const 
 {
 return lhs-> before（* rhs）！= 0; 
} 
}; 
 
 class TypeMap 
 {
 typedef map< type_info *，void *，type_info_less> TypenameToObject; 
 TypenameToObject ObjectMap; 
 
 public：
 template< typename T> 
 T * Get（）const 
 {
 TypenameToObject :: const_iterator iType = ObjectMap.find（& typeid（T））; 
 if（iType == ObjectMap.end（））
 return NULL; 
 return reinterpret_cast< T *>（iType-> second）; 
} 
 template< typename T> 
 void Set（T * value）
 {
 ObjectMap [& typeid（T）] = reinterpret_cast< void *>（value）; 
} 
}; 
 
 int main（）
 {
 TypeMap Services; 
 Services.Set< SomeClass>（new SomeClass（））; 
 SomeClass * x = Services.Get< SomeClass>（）; 
} 
  

在C ++中，类型不是第一类对象，最少的类型名将是唯一的，所以你可以通过键来。

编辑：名称实际上并不保证是唯一的，所以坚持type_info指针并使用before方法来比较它们。

I've been learning C++, coming from C#, where I've gotten used to using service providers: basically a Dictionary<Type, object>. Unfortunately, I can't figure out how to do this in C++. So the questions are basically:

1. How would I make a dictionary in C++.

2. How would I use 'Type' with it, as far as I know there is no 'Type' in C++.

3. Same as above, but with 'object'.

Thanks!

解决方案

I'm assuming that you're trying to map a type to a single object instance. You could try something along these lines:

#include <typeinfo>
#include <map>
#include <string>
using namespace std;

class SomeClass
{
public:
    virtual ~SomeClass() {} // virtual function to get a v-table
};

struct type_info_less
{
    bool operator() (const std::type_info* lhs, const std::type_info* rhs) const
    {
        return lhs->before(*rhs) != 0;
    }
};

class TypeMap
{
    typedef map <type_info *, void *, type_info_less> TypenameToObject;
    TypenameToObject ObjectMap;

public:
    template <typename T> 
    T *Get () const
    {
    	TypenameToObject::const_iterator iType = ObjectMap.find(&typeid(T));
    	if (iType == ObjectMap.end())
    		return NULL;
    	return reinterpret_cast<T *>(iType->second);
    }
    template <typename T> 
    void Set(T *value) 
    {
    	ObjectMap[&typeid(T)] = reinterpret_cast<void *>(value);
    }
};

int main()
{
    TypeMap Services;
    Services.Set<SomeClass>(new SomeClass());
    SomeClass *x = Services.Get<SomeClass>();
}

In C++ types are not first-class objects in their own right, but at least the type-name is going to be unique, so you can key by that.

Edit: The names aren't actually guaranteed to be unique, so hold on to the type_info pointers and use the before method to compare them.

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