C++: Is there a way to instantiate objects from a string holding their class name?

Answer 1

Meaning reflection as in Java. there is some info here: http://msdn.microsoft.com/en-us/library/y0114hz2(VS.80).aspx

Generally speaking, search google for "c++ reflection"

Answer 2

No there isn't. My preferred solution to this problem is to create a dictionary which maps name to creation method. Classes that want to be created like this then register a creation method with the dictionary. This is discussed in some detail in the GoF patterns book.

Answer 3

This is the factory pattern. See wikipedia (and this example). You cannot create a type per se from a string without some egregious hack. Why do you need this?

Answer 4

I have answered in another SO question about C++ factories. Please see there if a flexible factory is of interest. I try to describe an old way from ET++ to use macros which has worked great for me.

ET++ was a project to port old MacApp to C++ and X11. In the effort of it Eric Gamma etc started to think about Design Patterns

Answer 5

The short answer is you can't. See these SO questions for why:

1. Why does C++ not have reflection?
2. How can I add reflection to a C++ application?

Answer 6

Nope, there is none. You have to do the mapping yourself. C++ has no mechanism to create objects whose types are determined at runtime. You can use a map to do that mapping yourself, though:

template<typename T> Base * createInstance() { return new T; }

typedef std::map<std::string, Base*(*)()> map_type;

map_type map;
map["DerivedA"] = &createInstance<DerivedA>;
map["DerivedB"] = &createInstance<DerivedB>;

And then you can do

return map[some_string]();

Getting a new instance. Another idea is to have the types register themself:

// in base.hpp:
template<typename T> Base * createT() { return new T; }

struct BaseFactory {
    typedef std::map<std::string, Base*(*)()> map_type;

    static Base * createInstance(std::string const& s) {
        map_type::iterator it = getMap()->find(s);
        if(it == getMap()->end())
            return 0;
        return it->second();
    }

protected:
    static map_type * getMap() {
        // never delete'ed. (exist until program termination)
        // because we can't guarantee correct destruction order 
        if(!map) { map = new map_type; } 
        return map; 
    }

private:
    static map_type * map;
};

template<typename T>
struct DerivedRegister : BaseFactory { 
    DerivedRegister(std::string const& s) { 
        getMap()->insert(std::make_pair(s, &createT<T>));
    }
};

// in derivedb.hpp
class DerivedB {
    ...;
private:
    static DerivedRegister<DerivedB> reg;
};

// in derivedb.cpp:
DerivedRegister<DerivedB> DerivedB::reg("DerivedB");

You could decide to create a macro for the registration

#define REGISTER_DEC_TYPE(NAME) \
    static DerivedRegister<NAME> reg

#define REGISTER_DEF_TYPE(NAME) \
    DerivedRegister<NAME> NAME::reg(#NAME)

I'm sure there are better names for those two though. Another thing which probably makes sense to use here is shared_ptr.

If you have a set of unrelated types that have no common base-class, you can give the function pointer a return type of boost::variant<A, B, C, D, ...> instead. Like if you have a class Foo, Bar and Baz, it looks like this:

typedef boost::variant<Foo, Bar, Baz> variant_type;
template<typename T> variant_type createInstance() { 
    return variant_type(T()); 
}

typedef std::map<std::string, variant_type (*)()> map_type;

A boost::variant is like an union. It knows which type is stored in it by looking what object was used for initializing or assigning to it. Have a look at its documentation here. Finally, the use of a raw function pointer is also a bit oldish. Modern C++ code should be decoupled from specific functions / types. You may want to look into Boost.Function to look for a better way. It would look like this then (the map):

typedef std::map<std::string, boost::function<variant_type()> > map_type;

std::function will be available in the next version of C++ too, including std::shared_ptr.