Re: how to get an object instance from its class name?

From:

red floyd <no.spam@here.dude>

Newsgroups:

comp.lang.c++

Date:

Fri, 14 Mar 2008 09:29:30 -0700

Message-ID:

<cKxCj.1102$qT6.135@nlpi070.nbdc.sbc.com>

dotNeter wrote:

On Mar 14, 4:51 pm, Michael DOUBEZ <michael.dou...@free.fr> wrote:

dotNeter a ?crit :

Is there any way for this?
If I get a string representing a class name, how do I new an instance
upon this name?

There is no native way. You have to use a factory i.e. an object that
knows how to build an object from a parameter. That also means that all
such object have a common ancestor.

There are many way to implement such a factory (builder, prototype ...).
But you have to know the types in advance or have a way to register
them into the factory.

An alternative non-portable way is to put those classes in a dynamic
library and define for each a factory function (make_<Class>() by
example) and then use the dlopen(), dlsym() calls to open the library
and call the factory function.

Michael

This is very fascinating. I've spent some time on factory pattern, but
I don't like to write so many switch-case in my system.
One further question is that if it truly fit my future requirement
since the classes may get enriched someday.

Don't use a switch-case. Use a std::map.

E.g.:

class base { };
class derived1 : public base { };
class derived2 : public base { };
....
class derivedn : public base { };

typedef base (*createfunc )(/*params if needed*/);
std::map<std::string, createfunc> factory_map;

then look up in the map instead of using a case statement.