CODEWITHSUNDEEP
c++boostshared-ptr
John
John

Reputation: 821

Strip boost::shared_ptr from a template parameter

I am developing a C++ template library that uses both primitive types and pointer types stored in boost::shared_ptr. I am having a problem with a helper class that is used to pack primitive types into a container class if necessary before being passed into a lower layer of the library. The following shows the basic type that just passes the pointer on, and the implementation for std::string which is one of the primitives.

template <class T> class RRPrimUtil 
{ 
public:
    static rr_DataType GetTypeID() {return rr_void_t;} 


    static boost::shared_ptr<RRObject> PrePack(T val)     {return rr_cast<RRObject>(val);} 

    static T PreUnpack(boost::shared_ptr<RRObject> val) {return rr_cast<T>(val);} 
};

template<> class RRPrimUtil<std::string>
{
public:
    static rr_DataType GetTypeID() {return rr_string_t;} 
    static boost::shared_ptr<RRObject> PrePack(std::string val) {return rr_cast<RRObject>(stringToRRArray(val));} 
    static std::string PreUnpack(boost::shared_ptr<RRObject> val) {return RRArrayToString(rr_cast<RRArray<char>>(val));} 

};

The rr_cast<>() function is an alias to dynamic_pointer_cast. The problem I am having is that for the general case, the template "T" includes the "boost::shared_ptr" prefix because this type may or may not interact with a shared_ptr. The prefix messes up the dynamic_pointer_cast, because it expects just the pointer type. Is there a clean way to work around this?

Upvotes: 0

Views: 418

Answers (1)

K-ballo
K-ballo

Reputation: 81349

A template metafunction will strip that for you:

template<typename T> struct primitive_type
{
    typename T type;
};

template<typename T> struct primitive_type< boost::shared_ptr< T > >
{
    typename T* type;
};

In case the type passed to it is in the form boost::shared_ptr< T > it will return T*; otherwise it will return the type it was instantiated with.

Upvotes: 4

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