C ++模板：返回list :: iterator [英] C++ templates: Returning list::iterator

问题描述

如何使以下代码工作？在编译期间，我得到一个错误，告诉我 searchForResource 函数没有返回类型。

How can I make the following code work? During compilation I get an error telling me that the searchForResource function has no return type.

template<class T>
class ResourceManager
{
  private:
    struct ResourceWrapper;
    std::list<ResourceWrapper*> resources_; // This compiles fine

    std::list<ResourceWrapper*>::iterator  // Error occurs here
        searchForResource(const std::string& file);
};

此外，这是我如何定义 searchForResource function？

Also, is this how I would define the searchForResource function?

template<class t>
std::list<typename ResourceManager<T>::ResourceWrapper*>::iterator
    ResourceManager<T>::searchForResource(const std::string& file)
{
    // ...
}

推荐答案

std :: list :: iterator很难让编译器理解。

std::list::iterator is hard for the compiler to understand. Prefix it with 'typename' in both the implementation and the declaration to let the compiler know that it's a type.

像这样：

typename std::list<ResourceWrapper*>::iterator searchForResource(const std::string& file);

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