[C++-sig] Executing scripts in other namespace than __main__

Fabzter faboster at gmail.com
Thu Mar 18 05:42:35 CET 2010

Hi everybody, it's my annoying again.
I have a pure virtual class in C++ wich I have wrapped and exposed
(call it BaseClass), and I let the users inherit from it to make their
own class in their script. It's worth noting that I do not know the
name of such class they are creating. When I load only one script,
there's no problem in doing this:

object module_main = import("__main__");
object namespace_module_main = module_main.attr("__dict__");
exec("import sys\n", namespace_module_main,

object ignored = exec_file(boost::python::str("path/file1.py"),

//and as i don't know how the class they defined was named...
object ignored_ =
        exec("l = dir()\n" //get everything defined in current dictionary
             "class_ = None\n"
             "for i in l:\n"
             "    res = eval(i)\n"
             "    if type(res) == type(BaseClass): class_ = res\n"
//if we've got a class derived from BaseClass, we save it to "class_"
             "if class_ is None: raise Error()\n"
                , this->namespace_modulo_local, this->namespace_modulo_local);

object class_ = namespace_module_main["class_"]; //get the class object
object instance_ = class_();
BaseClass *bc = extract<BaseClass*>(instance_);

This works quite well when I only load one module. But if I load 2 or
more modules, the method I use to extract a class derived from
BaseClass is useless, since there are two or more classes derived from
BaseClass in __main__ and I don't know which one I will get.

So I thought I could run each script in it's own namespace, and then
using the same method for retrieving the class, but in every
individual namespace, ensuring I will get the derived class from the
specified script. How can I acommplish this?


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