有没有办法可以看到加载的cloudpickle对象的python源码？ 例如，下面我正在加载对象
In [17]: new_squared = pickle.loads (b'\x80\x02ccloudpickle.cloudpickle\n_fill_function\nq\x00(ccloudpickle.cloudpickle\n_make_skel_func\nq\x01ccloudpickle.cloudpickle\n_builtin_type\nq\x02X\x08\x00\x00\x00CodeTypeq\x03\x85q\x04Rq\x05(K\x01
...: K\x00K\x01K\x02KCc_codecs\nencode\nq\x06X\x08\x00\x00\x00|\x00\x00d\x01\x00\x13Sq\x07X\x06\x00\x00\x00latin1q\x08\x86q\tRq\nNK\x02\x86q\x0b)X\x01\x00\x00\x00xq\x0c\x85q\rX\x1e\x00\x00\x00<ipython-input-3-fefd3c33c8fb>q\x0eX\x08\x00
...: \x00\x00<lambda>q\x0fK\x01c__builtin__\nbytes\nq\x10)Rq\x11))tq\x12Rq\x13]q\x14}q\x15\x87q\x16Rq\x17}q\x18N}q\x19tR.')
...:
...:
接下来，我将验证它是否有效：
In [18]: new_squared(9)
Out[18]: 81
..确实如此。
如何查看实际的lambda表达式？
如果我尝试使用inspect，我会收到以下错误：
In [19]: import inspect
In [20]: h = inspect.getsource(new_squared)
OSError
Traceback (most recent call last)
<ipython-input-20-62e15704e160> in <module>()
----> 1 h = inspect.getsource(new_squared)
/usr/bin/anaconda/envs/py35/lib/python3.5/inspect.py in getsource(object)
942 or code object. The source code is returned as a single string. An
943 OSError is raised if the source code cannot be retrieved."""
--> 944 lines, lnum = getsourcelines(object)
945 return ''.join(lines)
946
/usr/bin/anaconda/envs/py35/lib/python3.5/inspect.py in getsourcelines(object)
929 raised if the source code cannot be retrieved."""
930 object = unwrap(object)
--> 931 lines, lnum = findsource(object)
932
933 if ismodule(object):
/usr/bin/anaconda/envs/py35/lib/python3.5/inspect.py in findsource(object)
760 lines = linecache.getlines(file)
761 if not lines:
--> 762 raise OSError('could not get source code')
763
764 if ismodule(object):
OSError: could not get source code