我需要帮助来定义专业 map
,我无法获得专业 map::iterator 进行正确编译。
我需要如何为 find()
定义此迭代器?打电话?
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:
In member function
'T* CaseInsensitiveMap<T>::lookup(const T&)'
:
error: expected';'
before'itr'
error:'itr'
was not declared in this scope
最佳答案
将 typename
关键字添加到变量的类型中:
typename CaseInsensitiveMap<T>::iterator itr = find(key);
无论如何,你不应该继承STL容器。了解为什么不应该这样做 here .
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";
}
这个输出:
12
100
关于c++ - 如何在 std::map 类中定义迭代器,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/10438152/