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我需要帮助来定义专业map，我无法获得专业的 map::iterator 来正确编译。

我如何需要为find()调用定义这个迭代器？

代码：

// My case-insensitive-less functor, which has historically worked fine for me:
struct ci_less : std::binary_function<std::string, std::string, bool>
{
  // case-independent (ci) compare_less binary function
  struct nocase_compare : public std::binary_function<unsigned char,unsigned char,bool>
  {
    bool operator() (const unsigned char& c1, const unsigned char& c2) const {
      return tolower (c1) < tolower (c2);
    }
  };
  bool operator() (const std::string & s1, const std::string & s2) const {
    return std::lexicographical_compare (s1.begin (), s1.end (),
                                         s2.begin (), s2.end (),
                                         nocase_compare());  // comparison
  }
};

//My offending code:
template <class T>
class CaseInsensitiveMap : public map<string, T, ci_less>
{
 public:
// This one actually works, but it requires two "find()" calls.
// I can't ethically call find twice.
  const T* lookup(const T& key) const {
    if (find(key) == map<string, T, ci_less>::end()) return 0;
    else                                             return &find(key)->first;
  }
// This one complains with errors shown below.
  T* lookup(const T& key) {
    CaseInsensitiveMap<T>::iterator itr = find(key);
    if (itr == map<string, T, ci_less>::end()) return 0;
    else              return itr->second;
  }
};

错误：

在成员函数中'T* CaseInsensitiveMap<T>::lookup(const T&)'：错误：错误之前
    预期     ：未在此范围内声明';''itr'
'itr'

4

2 回答 2

4

将typename关键字添加到变量的类型：

typename CaseInsensitiveMap<T>::iterator itr = find(key);

无论如何，您不应该继承 STL 容器。阅读为什么你不应该在这里这样做。

编辑：由于您要实现的只是一个不区分大小写的映射，因此您可以通过这种方式实现它，无需继承std::map，只需提供您自己的比较对象：

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

using namespace std;

struct nocase_compare {
    bool operator() (const unsigned char& c1, const unsigned char& c2) const {
      return tolower (c1) < tolower (c2);
    }
};

struct map_comparer {
  bool operator() (const std::string & s1, const std::string & s2) const {
    return std::lexicographical_compare (s1.begin (), s1.end (),
                                         s2.begin (), s2.end (),
                                         nocase_compare());  // comparison
  }
};

template<class T>
struct CaseInsensitiveMap {
    typedef std::map<std::string, T, map_comparer> type;
};

int main() {
    CaseInsensitiveMap<int>::type my_map;
    my_map["foo"] = 12;
    std::cout << my_map["FoO"] << "\n";
    my_map["FOO"] = 100;
    std::cout << my_map["fOo"] << "\n";
}

这输出：

12
100
于 2012-05-03T19:38:50.520 回答
0

类型名 CaseInsensitiveMap::iterator itr = find(key) ;

在第 31 行

于 2012-05-03T19:39:34.313 回答