模板中的 Typedef 导致阴影模板参数错误

Typedef in template is causing shadows template parm error

我有模板 class,我在其中使用 typedef 来声明映射,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

我做错了什么?我不明白为什么会出现该错误。

您忘记用 QueryMapIterator 写关键字 typename。这是更新后的版本:

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;
};

这个要求的原因是编译器此时不知道 QueryMapIterator 描述的是成员变量还是嵌套类型

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

更长的版本: QueryMap::iterator 是一个 dependant name,因此需要您在 typedef 之前使用 typename。从属名称问题来自这样一个事实,即在模板构造 type::something 中, something 可能引用类型、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,如果您传递的名称指的是类型,则需要 typename 向编译器提供额外信息。