benoitc/gunicorn

sendfile example does not use sendfile

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#2,279 opened on Feb 27, 2020

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 (4 comments) (0 reactions) (1 assignee)Python (9,334 stars) (1,707 forks)batch import
To Schedulegood first issue

Description

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']):

Contributor guide

Tech stack
python
Domain
backend
Issue type
bug
DifficultyEstimated implementation difficulty for a new contributor, from 1 for very small changes to 5 for expert-level work.
3
Estimated timeA rough time range for an experienced contributor to investigate, implement, test, and prepare a pull request.
1-3 hours
Activity statusHow available the issue appears right now: fresh, active, stale, blocked, or waiting on maintainer input.
stale
ClarityHow clearly the issue explains the expected change, acceptance criteria, and next step.
clear
Prerequisites
understanding of Python decoratorsbasic WSGI knowledge
Newbie friendlinessA 1-100 score estimating how approachable this issue is for first-time contributors.
60
Research direction
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.
sendfile example does not use sendfile · benoitc/gunicorn#2279 | Good First Issue