Boost.Python: how to get super() methods called?

Question

I have a module with two C++ classes exposed which both have a method foo():

struct MyClassA{
    void foo() { std::cout << "MyClassA::foo()" << std::endl; }
};

struct MyClassB{
    void foo() { std::cout << "MyClassB::foo()" << std::endl; }
};

BOOST_PYTHON_MODULE(my_module){
    class_<MyClassA>("MyClassA", init<>()).def("foo", &MyClassA::foo);
    class_<MyClassB>("MyClassB", init<>()).def("foo", &MyClassB::foo);
}

In Python, I create a class that is derived from both classes:

from my_module import MyClassA, MyClassB

class Derived(MyClassA, MyClassB):
    def foo(self):
        super().foo()  # should be unnessessary - doesn't work anyway

a = MyClassA()
a.foo()  # works
b = MyClassB()
b.foo()  # works
d = Derived()
d.foo()  # only prints 'MyClassA::foo()'

Now I'd love to have d.foo() call MyClassA.foo() as well as MyClassB.foo(). But while Derived.mro() looks good:

[<class '__main__.Derived'>, <class 'my_module.MyClassA'>, <class 'my_module.MyClassB'>, <class 'Boost.Python.instance'>, <class 'object'>]

.. only MyClassA.foo() gets called.

How do I make the C++ methods call their super() methods? And does that work for __init__() as well?

score 2 · Answer 1 · answered Feb 21 '18 at 15:53

The C++ layer doesn't have direct access to the MRO information Python uses to automate superclass calls via super. If you polluted the C++ code with explicit knowledge of Python, you could directly create an object of PySuper_Type (must be called with two arguments; the magic that enables no-arg super would not be available) and use that, but mixing the Python code with the C++ code that heavily would get ugly.

In practice, the best advice with Boost.Python is probably the same as for pybind11 (a C++11 specific header-only tool that serves a similar purpose without depending on the whole Boost ecosystem), and the recommended approach there is to explicitly call the C++ level parent class methods, rather than doing so implicitly via super. There is simply no reasonable way to merge the C++ and Python approaches to multiple inheritance that doesn't lead to hopelessly polluting the non-Python code with Python specific code. pybind11's approach would be:

class Derived(MyClassA, MyClassB):
    # Must define __init__ even though you only defer to parent classes,
    # or only MyClassA will be created
    def __init__(self):
        MyClassA.__init__(self)
        MyClassB.__init__(self)
    def foo(self):
        MyClassA.foo(self)
        MyClassB.foo(self)

frans · Accepted Answer · 2018-02-22T13:17:19.247

My current approach (adapted from this answer) needs some sort of wrapper which handles the super()-calls manually. The resulting classes don't need any special handling any more:

C++ code:

struct MyClassA{
    void foo() { std::cout << "MyClassA::foo()" << std::endl; }
};

struct MyClassB{
    void foo() { std::cout << "MyClassB::foo()" << std::endl; }
};

BOOST_PYTHON_MODULE(my_module){
    class_<MyClassA>("MyClassACpp", init<>()).def("foo", &MyClassA::foo);
    class_<MyClassB>("MyClassBCpp", init<>()).def("foo", &MyClassB::foo);
}

wrapper:

from my_module import MyClassACpp, MyClassBCpp

def call_super(cls, instance, method, *args):
    mro = instance.__class__.mro()
    for next_class in mro[mro.index(cls) + 1:]:
        if not hasattr(next_class, method):
            continue
        if next_class.__module__ in {'Boost.Python', 'builtins'}:
            continue
        getattr(next_class, method)(instance, *args)
        if next_class.__module__ != 'my_module':
            break

class MyClassA(MyClassACpp):
    def __init__(self):
        call_super(MyClassA, self, '__init__')
        print('MyClassA.__init__()')

    def foo(self):
        call_super(MyClassA, self, 'foo')

class MyClassB(MyClassBCpp):
    def __init__(self):
        call_super(MyClassB, self, '__init__')
        print('MyClassB.__init__()')

    def foo(self):
        call_super(MyClassB, self, 'foo')

usage:

class Derived(MyClassA, MyClassB):
    def foo(self):
        super().foo()

d = Derived()
d.foo()

output:

MyClassA.__init__()
MyClassB.__init__()
MyClassA::foo()
MyClassB::foo()

Boost.Python: how to get super() methods called?

2 Answers2