1

我有以下代码:

#include <map>
#include <string>

class policy1
{
  public:
    struct data
    {
    };
};

template<typename policy>
class policy_user : public policy
{
  typedef std::map<std::string, typename policy::data> mymap;        // good 
  typedef std::map<std::string, 
           typename policy::data >::iterator myiterator;              // bad
  typedef mymap::iterator myseconditerator;                      // also bad
};

失败了:

der.cpp:17: error: type ‘std::map<std::basic_string<char, std::char_traits<char>, std::allocator<char> >, typename policy::data, std::less<std::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::allocator<std::pair<const std::basic_string<char, std::char_traits<char>, std::allocator<char> >, typename policy::data> > >’ is not derived from type ‘policy_user<policy>’
der.cpp:17: error: expected ‘;’ before ‘myiterator’
der.cpp:18: error: type ‘std::map<std::basic_string<char, std::char_traits<char>, std::allocator<char> >, typename policy::data, std::less<std::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::allocator<std::pair<const std::basic_string<char, std::char_traits<char>, std::allocator<char> >, typename policy::data> > >’ is not derived from type ‘policy_user<policy>’
der.cpp:18: error: expected ‘;’ before ‘myseconditerator’

我不知道为什么。有什么帮助我可以完成我尝试做的事情(创建一个迭代器)吗?

4

1 回答 1

7

您忘记typename了 typedef。

写:

typedef typename std::map<std::string, typename policy::data >::iterator myiterator;      
typedef typename mymap::iterator myseconditerator; 

现在。没关系。typename是必需的,因为iterator它是一个从属名称。另请参阅

于 2011-02-15T19:16:00.230 回答