在python中测试两个函数的相等性

Question

I want to make two functions equal to each other, like this:

def fn_maker(fn_signature):
  def _fn():
    pass
  _fn.signature = fn_signature
  return _fn

# test equality of two function instances based on the equality of their signature values
>>> fa = fn_maker(1)
>>> fb = fn_maker(1)
>>> fc = fn_maker(2)
>>> fa == fb # should be True, same signature values
True
>>> fa == fc # should be False, different signature values
False

How should I do it? I know I could probably override eq and ne if fa, fb, fc are instances of some class. But here eq is not in dir(fa) and adding it the list doesnt work. I figured out some workaround like using a cache, eg,

def fn_maker(fn_signature):
  if fn_signature in fn_maker.cache:
    return fn_maker.cache[fn_signature]
  def _fn():
    pass
  _fn.signature = fn_signature
  fn_maker.cache[fn_signature] = _fn
  return _fn
fn_maker.cache = {}

By this way there is a guarantee that there is only one function for the same signature value (kinda like a singleton). But I am really looking for some neater solutions.

Answer 1

如果将函数转换为覆盖__call__()以及比较运算符的某些类的实例，则可以非常轻松地实现所需的语义。

Answer 2

It is not possible to override the __eq__ implementation for functions (tested with Python 2.7)

>>> def f():
...   pass
...
>>> class A(object):
...   pass
...
>>> a = A()
>>> a == f
False
>>> setattr(A, '__eq__', lambda x,y: True)
>>> a == f
True
>>> setattr(f.__class__, '__eq__', lambda x,y: True)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: can't set attributes of built-in/extension type 'function'

Answer 3

I don't think it's possible.

But overriding __call__ seems a nice solution to me.