Get class of caller's method (via inspect) in Python; or: super(Class,self).method() replacement wit
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by Slava Vishnyakov
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Published on 2010-04-24T23:48:00Z
Indexed on
2010/04/25
0:53 UTC
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python
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
?
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