如何在中声明@staticmethod佐佩·伊恩

2024-06-16 11:26:04 发布

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我尝试用@staticmethod和@classmethod创建接口。声明类方法很简单。但是我找不到声明静态方法的正确方法。你知道吗

考虑类接口及其实现:

#!/usr/bin/python3
from zope.interface import Interface, implementer, verify


class ISerializable(Interface):

    def from_dump(slice_id, intex_list, input_stream):
        '''Loads from dump.'''

    def dump(out_stream):
        '''Writes dump.'''

    def load_index_list(input_stream):
        '''staticmethod'''


@implementer(ISerializable)
class MyObject(object):

    def dump(self, out_stream):
        pass

    @classmethod
    def from_dump(cls, slice_id, intex_list, input_stream):
        return cls()

    @staticmethod
    def load_index_list(stream):
        pass

verify.verifyClass(ISerializable, MyObject)
verify.verifyObject(ISerializable, MyObject())
verify.verifyObject(ISerializable, MyObject.from_dump(0, [], 'stream'))

输出:

Traceback (most recent call last):
  File "./test-interface.py", line 31, in <module>
    verify.verifyClass(ISerializable, MyObject)
  File "/usr/local/lib/python3.4/dist-packages/zope/interface/verify.py", line 102, in verifyClass
    return _verify(iface, candidate, tentative, vtype='c')
  File "/usr/local/lib/python3.4/dist-packages/zope/interface/verify.py", line 97, in _verify
    raise BrokenMethodImplementation(name, mess)
zope.interface.exceptions.BrokenMethodImplementation: The implementation of load_index_list violates its contract
        because implementation doesn't allow enough arguments.

如何在此接口中正确声明静态方法?你知道吗


Tags: fromzope声明inputstreamusrdefload
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1楼 · 发布于 2024-06-16 11:26:04

显然verifyClass既不能正确理解classmethod也不能正确理解staticmethod。问题是,在python3中,如果在python3中执行getattr(MyObject, 'load_index_list'),则会得到一个裸函数,并且verifyClass认为它是另一个未绑定方法,然后期望隐式self是第一个参数。你知道吗

最简单的解决方法是在那里使用classmethod而不是staticmethod。你知道吗

我想有人也可以做一个错误报告。你知道吗

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