c++ - 以模板容器作为参数的模板函数-奇怪的错字错误

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Where and why do I have to put the “template” and “typename” keywords?
                                
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                                5年前关闭。
            
                    
我收到奇怪的错别字错误，没有任何意义。我担心这可能是C ++编译器问题（在具有10.6.8和Xcode 3.x的Mac上）。如果有人能真正发现问题，我将不胜感激：

template<typename T> int getIdxInVector(const std::vector<T>&  vec, const T& toMatch)
{
std::vector<T>::const_iterator cit = std::find(vec.begin(),vec.end(),toMatch);
return( cit != vec.end() ? cit - vec.begin() : -1 );
}

这是我遇到的错误：

LooseFunctions.h:27: error: expected `;' before 'cit'
LooseFunctions.h:28: error: 'cit' was not declared in this scope
LooseFunctions.h:27: error: dependent-name 'std::vector<T,std::allocator<_CharT> >::const_iterator' is parsed as a non-type, but instantiation yields a type
LooseFunctions.h:27: note: say 'typename std::vector<T,std::allocator<_CharT> >::const_iterator' if a type is meant

谢谢你的帮助！

最佳答案

const_iterator是从属名称，因此您需要使用typename来指定它引用一种类型：

typename std::vector<T>::const_iterator = ...

请注意，C ++ 11使此操作更容易：

auto cit = std::find(vec.begin(),vec.end(),toMatch);