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Ricardo Busquet
Ricardo Busquet

Posted on • Updated on • Originally published at

The dynamic nature of Python's MRO

Imagine the following class relationships:

class Base:
    def chain(self):
        return 'Base'

class A(Base):

class B(Base):
    def chain(self):
        return f"{super().chain()} <- B"

class C(A, B):

class D(C):
    def chain(self):
        return f"{super().chain()} <- D"

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Calling chain in an instance of D will result in the following string:

In [1]: d.chain()
Out[1]: 'Base <- B <- D'
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What would happen if the following code runs?

In [2]: A.chain = A.chain
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Inoffensive, right? Now try calling d.chain() again...

In [3]: d.chain()
Out[3]: 'Base <- D'
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Let's observe D's Method Resolution Order (MRO, the order of classes where Python will look for when resolving methods and attributes):

In [4]: D.mro()
Out[4]: [__main__.D, __main__.C, __main__.A, __main__.B, __main__.Base, object]
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When calling D.chain(), Python will look in this list and return the first instance where chain is present as a member of the class. In our example, D implements chain.

D.chain in turn will call super().chain(). What will Python do? it will grab the next class from D and try to find chain in this new list. C doesn't implement it; neither does A. B does! The result will be whatever B.chain returns, plus the bit " <- D".

B.chain does the super() call again... we know what we're doing. Base implements it and there are no more super() calls. So we have Base.chain returning "Base", B.chain() returning "Base <- B", and D.chain() finally returning "Base <- B <- D".

So what’s going on after we do A.chain = A.chain? Why is B.chain() ignored? Let’s dissect what’s going on with it. What’s A's MRO?

In [5]: A.mro()
Out[5]: [__main__.A, __main__.Base, object]
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Pretty simple. What happens when you do A.chain? Python will look at A, which does not implement it. But Base implements it, so that's the implementation it's gonna use. But what's going on when we do the assignment?

Python evaluates this assignment right to left: A.chain = A.chain then means "find chain for A", which returns Base.chain. Then, python will effectively create a new member in the class A named chain!

Let's go back to D's MRO:

[__main__.D, __main__.C, __main__.A, __main__.B, __main__.Base, object]
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The way these classes were constructed, A's members will be tested before B! What does it mean for the resolution order? When D.chain() calls super().chain(), it will now grab the newly added member of A and call it. And from A's point of view, the implementation is the same as Base.chain, which doesn't have any super() call! 🤯

This happens due to the dynamic nature of Python. The entry for super() in Python's documentation has an amazing description of this issue, and how it's a unique use case due to Python's nature.

Python's multiple inheritance feature, plus its dynamic nature, make it so the actual method resolution order and class hierarchy are only known at runtime and can change at any time during the application lifetime.

Classes designs MUST be collaborative. The example above has many issues, but there is a simple set of rules that help designing collaborative classes:

  • methods called by super() need to exist
  • methods called and their implementations need to have a matching argument signature
  • every occurrence of the method needs to use super()

The point here is that the actual implementation of the method called by super() is only known at runtime and cannot be easily defined statically.

Design your classes collaboratively, and watch for mutations in classes in your runtime.

Want to know more about Python's MRO? This amazing article in the Python docs introduces how the algorithm works after version 2.3. Check it out!

This is my first post. Please let me know if there's anything I can do to make this better!

Photo by Adam Śmigielski on Unsplash

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