Python - Can I access the object who call me?

问题:

If I have this:

class A:
    def callFunction(self, obj):
        obj.otherFunction()

class B:
    def callFunction(self, obj):
        obj.otherFunction()

class C:
    def otherFunction(self):
        # here I wan't to have acces to the instance of A or B who call me.

...

# in main or other object (not matter where)
a = A()
b = B()
c = C()
a.callFunction(c) # How 'c' know that is called by an instance of A...
b.callFunction(c) # ... or B

Despite the design or other issues, this is only a question of an inquiring mind.

Note: This has to be done without changing otherFunction signature

回答1:

If this is for debugging purposes you can use inspect.currentframe():

import inspect

class C:
    def otherFunction(self):
        print inspect.currentframe().f_back.f_locals

Here is the output:

>>> A().callFunction(C())
{'self': <__main__.A instance at 0x96b4fec>, 'obj': <__main__.C instance at 0x951ef2c>}

回答2:

Here is a quick hack, get the stack and from last frame get locals to access self

class A:
    def callFunction(self, obj):
        obj.otherFunction()

class B:
    def callFunction(self, obj):
        obj.otherFunction()

import inspect

class C:
    def otherFunction(self):
        lastFrame = inspect.stack()[1][0]
        print lastFrame.f_locals['self'], "called me :)"

c = C()

A().callFunction(c)
B().callFunction(c)

output:

<__main__.A instance at 0x00C1CAA8> called me :)
<__main__.B instance at 0x00C1CAA8> called me :)

回答3:

Examine the stack with the inspect module with inspect.stack(). You can then get the instance from each element in the list with f_locals['self']