模板中的Typedef导致阴影模板parm错误

Question

我有一个模板类，其中使用typedef声明一个映射，如下所示：

#include <map>

template <typename T> class LocalStub {
  typedef std::map<T, T> QueryMap;
  typedef std::pair<T, T> QueryPair;
  typedef QueryMap::iterator QueryMapIterator;

public:
  LocalStub();
  ~LocalStub();
  void AddQuery(const T &, const T &);
  const T &Answer(const T &) const;
private:
  QueryMap _queryMap;
};

---

编译错误

../src/LocalStub/include/localstub.hpp:12: error: declaration of 'class T'
../src/LocalStub/include/localstub.hpp:11: error:  shadows template parm 'class T'
../src/LocalStub/include/localstub.hpp:13: error: template declaration of 'typedef'
../src/LocalStub/include/localstub.hpp:14: error: declaration of 'class T'
../src/LocalStub/include/localstub.hpp:11: error:  shadows template parm 'class T'
../src/LocalStub/include/localstub.hpp:15: error: template declaration of 'typedef'
../src/LocalStub/include/localstub.hpp:16: error: declaration of 'class T'
../src/LocalStub/include/localstub.hpp:11: error:  shadows template parm 'class T'
../src/LocalStub/include/localstub.hpp:17: error: template declaration of 'typedef'
../src/LocalStub/include/localstub.hpp:26: error: 'QueryMap' does not name a type

---

我在做什么错？我不明白为什么会收到该错误。

Answer 1

您忘了用typename写关键字QueryMapIterator。这是更新的版本：

template <typename T> class LocalStub {
  typedef std::map<T, T> QueryMap;
  typedef std::pair<T, T> QueryPair;
  typedef typename QueryMap::iterator QueryMapIterator;

public:
  LocalStub();
  ~LocalStub();
  void AddQuery(const T &, const T &);
  const T &Answer(const T &) const;
private:
  QueryMap _queryMap;
};

之所以有此要求，是因为编译器此时不知道QueryMapIterator是描述成员变量还是嵌套类型

Answer 2

TL; DR-typedef typename QueryMap::iterator QueryMapIterator;

更长的版本： QueryMap::iterator是dependant name，因此要求typename之前使用typedef。从属名称问题来自以下事实：在模板构造type::something中，something可以引用类型，值或函数，如下例所示：

template<class T>
struct foo{
    void bar(){ T::something; }
};

struct baz{
    using something = int;
};

struct bez{
    static const int something = 0;
};

因此，如果您传递的名称涉及类型，则需要typename才能向编译器提供其他信息。