python / python/cpython

Protocol named "Protocol" or "Generic" has no members.

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stdlib topic-typing type-bug
主要語言
Python
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描述

Bug report

Bug description:

If a protocol is named Protocol or Generic then it will ignore any defined members.

when you get lazy with naming in your unit tests this makes for quite a head scratcher...

Minimal Reproduction
import typing


class Protocol(typing.Protocol):
    a: int


members = typing.get_protocol_members(Protocol)

assert members == {"a"}, f"Expected members to be {{'a'}}, got {members}"
Output:
Traceback (most recent call last):
  File ".../test_protocol.py", line 10, in <module>
    assert members == {"a"}, f"Expected members to be {{'a'}}, got {members}"
           ^^^^^^^^^^^^^^^^
AssertionError: Expected members to be {'a'}, got frozenset()
Expected Output:

Nothing, script should succeed without error.

Investigation

The problem appears to stem from the _get_protocol_attrs function used by _ProtocolMeta which is defined as follows:

https://github.com/python/cpython/blob/171133aa84cd2fa8738bdbb0c76435645810e8d3/Lib/typing.py#L1879-L1899

The if statement on line 1887 is too loose in what it matches.

Possibile Fix

The check for base needs to be more specific, either using the instance itself:

if base in {Protocol, Generic}:
    continue

or checking the module too, in order to ensure its the correct class being excluded:

if base.__name__ in {'Protocol', 'Generic'} and base.__module__ == 'typing':
    continue
CPython versions tested on:

3.14

Operating systems tested on:

macOS

Linked PRs
  • gh-145724

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研究方向

從報告中所識別的函式 _get_protocol_attrs 所在的 Lib/typing.py 開始,並使用 Python 3.14 執行提供的最小重現。完成的條件是:名為 Protocol 或 Generic 的使用者定義類別能透過 typing.get_protocol_members 公開其宣告的協定成員;開始前請檢查 gh-145724,因為該 issue 連結了一個現有的 pull request。

由索引模型根據 Issue 內容生成。

評估

技術堆疊
python
領域
backend
Issue 類型
缺陷
難度
2/5
預估耗時
1-3 小時
活躍度
停滯
描述清晰度
描述清楚
新手友好度
35/100

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