我对 C++ 的使用有什么问题吗？标准库的发现？

Question

我正在尝试使用 C++ 标准库的 find 算法，如下所示：

  template<class T>
  const unsigned int AdjacencyList<T>::_index_for_node(
      const std::vector<T>& list, const T& node
  ) throw(NoSuchNodeException)
  {
    std::vector<T>::iterator iter = std::find(list.begin(), list.end(), node);
  }

当我尝试编译时，出现以下错误：

In file included from ../AdjacencyList.cpp:8:
../AdjacencyList.h: In member function ‘const unsigned int Graph::AdjacencyList<T>::_index_for_node(const std::vector<T, std::allocator<_Tp1> >&, const T&)’:
../AdjacencyList.h:99: error: expected ‘;’ before ‘iter’
../AdjacencyList.h:100: error: ‘iter’ was not declared in this scope
In file included from ../AdjacencyListTest.cpp:9:
../AdjacencyList.h: In member function ‘const unsigned int Graph::AdjacencyList<T>::_index_for_node(const std::vector<T, std::allocator<_Tp1> >&, const T&)’:
../AdjacencyList.h:99: error: expected ‘;’ before ‘iter’
../AdjacencyList.h:100: error: ‘iter’ was not declared in this scope
../AdjacencyList.h: In member function ‘const unsigned int Graph::AdjacencyList<T>::_index_for_node(const std::vector<T, std::allocator<_Tp1> >&, const T&) [with T = int]’:
../AdjacencyList.h:91:   instantiated from ‘const std::vector<T, std::allocator<_Tp1> > Graph::AdjacencyList<T>::neighbours(const T&) [with T = int]’
../AdjacencyListTest.cpp:18:   instantiated from here
../AdjacencyList.h:99: error: dependent-name ‘std::vector::iterator’ is parsed as a non-type, but instantiation yields a type
../AdjacencyList.h:99: note: say ‘typename std::vector::iterator’ if a type is meant

我感觉像“dependent-name 'std::vector::iterator'被解析为非类型，但实例化产生类型”位是理解我做错了什么的关键，但我的豌豆大脑无法提取含义。

更新：我需要根据接受的答案添加一个typename，并且还使用const_iterator，因此有问题的代码行变成：

    typename std::vector<T>::const_iterator iter = std::find(list.begin(), list.end(), node);

Answer 1

std::vector<T>::iterator iter = /* .... */; 

iterator 是一个从属名称（实际上，它取决于类型参数T）。除非您使用 typename，否则假定从属名称不会命名类型：

typename std::vector<T>::iterator iter = /* .... */;

有关更多信息，请考虑阅读 Stack Overflow C++ 常见问题解答条目 “我必须在何处以及为什么必须将“template”和“typename”放在依赖名称上？”

您还需要使用 const_iterator，因为 list 是 const 限定的。您可能还应该删除异常规范；最好“永远不要编写异常规范。”