尝试制作 STL 地图容器的包装器时出错

发布于 2024-09-27 15:17:13 字数 1087 浏览 3 评论 0原文

我正在尝试为 STL 映射容器创建一个包装器，以便添加一个 const 方法来返回给定键的值。在map中，operator[]不是const，并且find()需要取消引用才能获取值（map.find()->second）。我的一些“研究”基于惯用的 C++ 阅读顺便说一句，来自 const 映射。

到目前为止的代码（全部位于单个头文件中）：

#include <map>
template <typename K, typename V>
class easymap : public std::map<K, V>
{
    //Constructor
    easymap() : std::map<K, V>() {};

    //The get method
    V get(K key)
    {
        std::map<K, V>::const_iterator iter(find(key));
        return iter != end() ? iter->second : V();
    }
};

当我尝试编译此代码时，出现以下错误：

In member function `V easymap::get(K)':
    expected `;' before "iter"
`iter' was not declared in this scope
there are no arguments to `end' that depend on a template parameter, so a declaration of `end' must be available|
(if you use `-fpermissive', G++ will accept your code, but allowing the use of an undeclared name is deprecated)

我尝试执行此操作的方式有意义吗？如果是这样，我该如何进行这项工作？如果没有，我该如何实现我想要的效果？

原文

I'm trying to make a wrapper to the STL map container, in order to add a const method to return the value given the key. In map, operator[] isn't const, and find() requires dereferencing to get the value (map.find()->second). I'm basing some of my "research" off of Idiomatic C++ for reading from a const map by the way.

The code so far (all inside a single header file):

#include <map>
template <typename K, typename V>
class easymap : public std::map<K, V>
{
    //Constructor
    easymap() : std::map<K, V>() {};

    //The get method
    V get(K key)
    {
        std::map<K, V>::const_iterator iter(find(key));
        return iter != end() ? iter->second : V();
    }
};

When I try to compile this, I get the following errors:

In member function `V easymap::get(K)':
    expected `;' before "iter"
`iter' was not declared in this scope
there are no arguments to `end' that depend on a template parameter, so a declaration of `end' must be available|
(if you use `-fpermissive', G++ will accept your code, but allowing the use of an undeclared name is deprecated)

Does how I'm trying to go about doing this make sense? If so, how do I make this work? If not, how would I go about achieving the effect I'm looking for?

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刘备忘录 2024-10-04 15:17:13

不要从 std::map 派生。相反，请在 easymap 中包装一个 std::map 实例，遵循先组合后继承的原则。除了所有技术原因之外，这更好地反映了设计意图：提供一个简化的 API 来映射隐藏默认 API：

template<typename K, typename V>
class easymap {
 std::map<K, V> mMap;
public:
  V Get(K Key) const {
   // ...
  }
};

Do not derive from std::map. Rather, wrap a std::map instance in your easymap, following the composition before inheritance principle. Besides of all the technical reasons, this reflects the design intent much better: provide a simplified API to map hiding the default one:

template<typename K, typename V>
class easymap {
 std::map<K, V> mMap;
public:
  V Get(K Key) const {
   // ...
  }
};

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墟烟 2024-10-04 15:17:13

您缺少映射的模板参数，在声明迭代器时必须指定 typename （请参阅此处），并且由于某种我不知道的原因（可能是命名空间冲突），您在调用 end() 时必须使用 this：

template <typename K, typename V>
class easymap : public std::map<K,V>
{
    //Constructor
    easymap() : std::map<K, V>() {};

    //The get method
    V get(K key)
    {
        typename std::map<K, V>::const_iterator iter(find(key));
        return iter != this->end() ? iter->second : V();
    }
};

You are missing the template parameters for map, you have to specify typename when declaring the iterator (see here), and for some reason unknown to me (probably a namespace conflict) you have to use this when calling end():

template <typename K, typename V>
class easymap : public std::map<K,V>
{
    //Constructor
    easymap() : std::map<K, V>() {};

    //The get method
    V get(K key)
    {
        typename std::map<K, V>::const_iterator iter(find(key));
        return iter != this->end() ? iter->second : V();
    }
};

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