python 使用`inspect`或`__source__`属性获取python中函数的源代码

Posted 2021-05-09

def function_code(fn):
    """ returns the source code of the given function 
        by: Cody Kochmann 
      """
    from inspect import getsource
    assert callable(fn), 'function_source needs its argument to be callable'
    if '__source__' in dir(fn):
        return fn.__source__
    else:
        try:
            return getsource(fn)
        except OSError:
            return None