我使用C++標準庫的find有什麼問題？

Question

我試圖使用C++標準庫的find算法是這樣的：

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); 
    } 

當我嘗試編譯，我得到這些錯誤：

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 

我覺得像「從屬名稱'std :: vector :: iterator'被解析爲非類型，但實例化產生一個類型「位持有理解我做錯了什麼的關鍵，但是我的豌豆腦不能提取意義。

更新：我需要添加一個typename，按照公認的答案，並且還使用了const_iterator，所以有問題的代碼行變成了：

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

Answer 1

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

iterator是一個從屬名稱（實際上，其取決於在類型參數T上）。相關名被假定除非你使用typename未命名類型：

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

欲瞭解更多，可以閱讀堆棧溢出的C++ FAQ條目"Where and why do I have to put 「template」 and 「typename」 on dependent names?"

您還需要使用const_iterator，因爲list是const限定。你也許應該放棄異常規範;最好是"never write an exception specification."