python - 被调用者是否有可能强制其调用者在 python 中返回？

Question

被调用者是否可以强制其调用者在 python 中返回？如果是这样，这是一个好方法吗？这不违反Zen of Python的显式优于隐式句吗？ ？

例子:

import inspect

class C(object):
    def callee(self):
        print 'in callee'
        caller_name = inspect.stack()[1][3]
        caller = getattr(self, caller_name)
        # force caller to return
        # so that "in caller after callee" gets never printed

        caller.return() # ???

    def caller(self):
        print 'in caller before calle'
        self.callee()
        print 'in caller after callee'

c = C()

c.caller()

print 'resume'

输出:

in caller before callee
in callee
resume

最后，感谢@Andrew Jaffe 关于上下文管理器的建议，我用一个简单的装饰器解决了这个问题。

# In my real code this is not a global variable
REPORT_ERRORS = True

def error_decorator(func):
    """
    Returns Event instance with result of the
    decorated function, or caught exception.
    Or reraises that exception.
    """

    def wrap():
        error = None
        user = None

        try:
            user = func()
        except Exception as e:
            error = e
        finally:
            if REPORT_ERRORS:
                return Event(user, error)
            else:
                raise error 
    return wrap


@error_decorator
def login():

    response = fetch_some_service()

    if response.errors:
        # flow ends here
        raise BadResponseError

    user = parse_response(response)

    return user

Answer 1

从被调用方返回一个值以供调用方读取并据此执行相应操作有什么问题？

代替

caller.return() # ???

写

return False

在

def caller(self):
        print 'in caller before calle'

        rc = self.callee()
        if not rc:
             return
        print 'in caller after callee'

当然你可以引发异常并在被调用者中捕获它并相应地表现或者简单地减少它失败

mgilson 的副本

我主张基于返回值的检查的原因

1. 显式优于隐式
2. 被调用者不应控制调用者行为。这是一种糟糕的编程习惯。相反，调用者应该根据被调用者的行为改变其行为。