Missing create python bounded method from closure

Thanks for this great package! 👏

I'm implementing dynamic generated heap class with pyo3 in rust for an ORM.

Assign a PythonCFunction to a class won't make it behave like a method. This is different from Python methods which has a __get__ method to bound.

This will not work

let my_method = PyCFunction::new_method_closure(py, cls, Some(function_name), Some("Represent."), |args, kwargs| {
    ...
})?;
my_class.setattr(py, "my_method", my_method)?;

When accessing, the my_method is not bounded. Self is not passed in as the first argument.

There are no args passed

let my_method = PyCFunction::new_method_closure(py, cls, Some(function_name), Some("Represent."), |args, kwargs| {
    println!("see args: {:?} {:?}", args, kwargs) // (,) and None
})?;
my_class.setattr(py, "my_method", my_method)?;

After digging into some details of the implementation, I found that something like this with trampolines need to be implemented

    pub const fn cmethod_function_with_keywords(
        name: &'static str,
        cfunction: PyCMethod,
        doc: &'static str,
    ) -> Self {
        Self {
            ml_name: name,
            ml_meth: PyMethodType::PyCMethod(cfunction),
            ml_flags: ffi::METH_VARARGS | ffi::METH_KEYWORDS | ffi::METH_METHOD,
            ml_doc: doc,
        }
    }

However, I'm not skilled and able to do this. I need your help. 😃

Start by reading PyCFunction::new_method_closure and the cmethod_function_with_keywords example in the issue, including PyMethodType and the METH_VARARGS, METH_KEYWORDS, and METH_METHOD flags. Trace how the closure is installed on a class and confirm that accessing it passes the instance as the first argument, including keyword arguments.