Get class of caller's method (via inspect) in Python; or: super(Class,self).method() replacement wit
- by Slava Vishnyakov
Is it possible to get reference to class B in this example?
class A(object): pass
class B(A):
def test(self):
test2()
class C(B): pass
import inspect
def test2():
frame = inspect.currentframe().f_back
cls = frame.[?something here?]
# cls here should == B (class)
c = C()
c.test()
Basically, C is child of B, B is child of A. Then we create c of type C. Then the call to c.test() actually calls B.test() (via inheritance), which calls to test2().
test2() can get the parent frame frame; code reference to method via frame.f_code;
self via frame.f_locals['self']; but type(frame.f_locals['self']) is C (of course), but not B, where method is defined.
Any way to get B?
