Question

Edit:
The data is as follows

typedef std::shared_ptr<T> Resource;
typedef std::map<std::string, Resource> ResourceMap;

This is the function

const T& Get(const std::string& key)
{
    ResourceMap::iterator itr = mResources.find(key);
    return (itr != mResources.end()) ? itr->second : mDefault;

}

Error:

Error   1   error C2446: ':' : no conversion from 'sf::Texture' to 'std::tr1::shared_ptr<_Ty>'  d:\sanity\trunk\client\src\assetmanager.h   28

Also I’m creating an object like that :

AssetManager<sf::Texture> imgManager;

Okay, sorry for missing information, I was in a rush.

The only problem left is :

return (itr != mResources.end()) ? itr->second.get() : mDefault;

get() returns raw pointer and I need to return reference to what the shared_ptr is pointing.

How should I do that?

Answer 1

Error   1   error C2446: ':' : no conversion from 'sf::Texture' to
'std::tr1::shared_ptr<_Ty>' d:\sanity\trunk\client\src\assetmanager.h 28

I guess your problem is that itr->second is of one type (Resource, if your std::map typedef is correct), while mDefault must be something other.

The ternary operator cannot handle the two different types, so you must correct your code to be sure both items left and right of the : part of the ?: operator are of the same type (or compatible ones).

So confirm this, I need:

• the declaration of the Resource type
• the declaration of the T type
• the declaration of the mDefault member variable

Edit

Let’s assume you have something like:

typedef std::shared_ptr<T> Resource;
typedef std::map<std::string, Resource> ResourceMap;

template <typename T>
class AssetManager
{
    const T& Get(const std::string& key)
    {
        ResourceMap::iterator itr = mResources.find(key);
        return (itr != mResources.end()) ? itr->second : mDefault;
    }

    ResourceMap mResources ;
    ??? mDefault ;

    // etc.
} ;

instanciated like:

AssetManager<sf::Texture> imgManager;

Now, I need the type of mDefault to continue.

My guess: You MUST make sure your code is more like:

    const T& Get(const std::string& key) const
    {
        ResourceMap::const_iterator itr = mResources.find(key);
        return (itr != mResources.end()) ? *(itr->second) : *(mDefault);
    }

    ResourceMap mResources ;
    Resource mDefault ;

As you want to return the Resource, not the shared_ptr of the Resource.

Note that I added const keywords/prefix to be consistent with the const return

Edit 2

As you have:

    ResourceMap mResources ;
    T mDefault ;

So I guess you should write:

    const T& Get(const std::string& key) const
    {
        ResourceMap::const_iterator itr = mResources.find(key);
        return (itr != mResources.end()) ? *(itr->second) : mDefault;
    }

itr->second is a smart pointer, so if you want to get the pointer object, you simply need to dereference the smart pointer: *(itr->second).

As for returning a reference to mDefault, this is indicated by the function’s return type const T &, so you don’t need anything more.