Add optional param to base class function, ignore in subclass

Question

If I have a class hierarchy and use it like follows, including a third-party subclass that I can't change

class Base:
  def doit(self):
    pass

class Sub(Base):
  def doit(self):
    print('hi')

class SubInThirdPartyLibrary(Base):
  def doit(self):
    print('hi from library')

def doit(x: Base):
  x.doit()

doit(Sub())
doit(SubInThirdPartyLibrary())

Is there any way I can add an optional parameter to the base class, use that in the subclass, without breaking the third-party class?

class Base:
  def doit(self, msg: str = 'hi'):
    pass

class Sub(Base):
  def doit(self, msg: str = 'hi'):
    print(msg)

class SubInThirdPartyLibrary(Base):
  def doit(self):  # This isn't mine, I can't add the parameter here but want it to keep working.
    print('hi from library')

def doit(x: Base):
  x.doit(hi='hello')

doit(Sub())
doit(SubInThirdPartyLibrary())  # TypeError: doit() takes 1 positional argument but 2 were given

No; this is a breaking change that invalidates the previous public interface that SubInThirdPartyLibrary relies on. — chepner
– chepner, Commented Dec 31, 2021 at 19:10
You can overwrite SubInThirdPartyLibrary.doit from anywhere after the import of the module. The function has to accept a self argument then, just like a class's method. — TheEagle
– TheEagle, Commented Dec 31, 2021 at 19:29

Francisco · Accepted Answer · 2021-12-31 23:36:43Z

2

You can use inspect.signature:

import inspect

def doit(x: Base):
    sig = inspect.signature(x.doit)
    if 'msg' in sig.parameters:
        x.doit(msg='hello')
    else:
        x.doit()

I would recommend against doing this, and instead try another solution.

answered Dec 31, 2021 at 23:36

Francisco

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