Question

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);
};

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)
{
    // ...
}

Answer 1

std::list<ResourceWrapper*>::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.

Like so:

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