benoitc/gunicorn

sendfile example does not use sendfile

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#2,279 创建于 2020年2月27日

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 (4 评论) (0 反应) (1 负责人)Python (9,334 star) (1,707 fork)batch import
To Schedulegood first issue

描述

The sendfile example application does not actually result in a call to sendfile. I verified this by adding code to http/wsgi.py and also looking at the output of strace.

It works if I remove the @validator decorator on the app method in sendfile.py.

I guess this is because the worker is checking for an instance of the write type, but the type is not the same with the validator wrapper. workers/sync.py handle_request() says

if isinstance(respiter, environ['wsgi.file_wrapper']):

贡献者指南

技术栈
python
领域
backend
议题类型
bug
难度面向新贡献者的预计实现难度，1 表示很小改动，5 表示专家级工作。
3
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clear
前置要求
understanding of Python decoratorsbasic WSGI knowledge
新手友好度1-100 的估计分数，表示该议题对首次贡献者的友好程度。
60
研究方向
Examine the type check in workers/sync.py handle request, specifically the line 'if isinstance(respiter, environ["wsgi.file wrapper"])'. The issue suggests that a decorator changes the type, so the fix may involve stripping the wrapper or adjusting the type check. Review the sendfile.py example and consider using functools.wraps to preserve type information. Test with strace to confirm sendfile syscall is invoked.