How to get the caller class name inside a function of another class in python?

Kaushik picture Kaushik · Jun 12, 2013 · Viewed 27.9k times · Source

My objective is to stimulate a sequence diagram of an application for this I need the information about a caller and callee class names at runtime. I can successfully retrieve the caller function but not able to get a caller class name?

#Scenario caller.py:

import inspect

class A:

    def Apple(self):
        print "Hello"
        b=B()
        b.Bad()



class B:

    def Bad(self):
        print"dude"
        print inspect.stack()


a=A()
a.Apple()

When I printed the stack there was no information about the caller class. So is it possible to retrieve the caller class during runtime ?

Answer

brice picture brice · Jun 12, 2013

Well, after some digging at the prompt, here's what I get:

stack = inspect.stack()
the_class = stack[1][0].f_locals["self"].__class__.__name__
the_method = stack[1][0].f_code.co_name

print("I was called by {}.{}()".format(the_class, the_method))
# => I was called by A.a()

When invoked:

➤ python test.py
A.a()
B.b()
  I was called by A.a()

given the file test.py:

import inspect

class A:
  def a(self):
    print("A.a()")
    B().b()

class B:
  def b(self):
    print("B.b()")
    stack = inspect.stack()
    the_class = stack[1][0].f_locals["self"].__class__.__name__
    the_method = stack[1][0].f_code.co_name
    print("  I was called by {}.{}()".format(the_class, the_method))

A().a()

Not sure how it will behave when called from something other than an object.