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Pyflame fails on multithreaded applications with setuid #140

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torkve opened this issue Mar 7, 2018 · 3 comments
Open

Pyflame fails on multithreaded applications with setuid #140

torkve opened this issue Mar 7, 2018 · 3 comments

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@torkve
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torkve commented Mar 7, 2018

Reproducible on:

  • Pyflame 1.6.3 (commit 4a9c3de) linux-gnu x86_64 (ABI list: 26)
  • Python 2.7.8 built with GCC 4.4.6 on Linux

Details:

glibc with pthreads uses RT signal to interrupt syscalls and notify all threads when any thread changes uid of the process. Pyflame is unaware of this signal and fails on it.

How to reproduce:

test_setuid.py:

import threading
import time
import random
import os

def task():
    time.sleep(random.randint(1, 6))
    os.setresuid(random.randint(1000, 1005), 0, 0)
    time.sleep(100)


threads = []
for _ in xrange(10):
    t = threading.Thread(target=task)
    t.daemon = True
    t.start()
    threads.append(t)

for t in threads:
    t.join()

run:

 sudo ./pyflame --threads -t python test_setuid.py > /dev/null
Unexpected ptrace(2) exception: waitpid() indicated a WIFSTOPPED process, but got unexpected signal 33
@eklitzke
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Thanks, this is a very good bug report. I have been working on other projects recently, but I will fix this next time I have some cycles for Pyflame. I imagine this would be a pretty small change if anyone else wants to get a PR in.

@fillest
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fillest commented Dec 1, 2018

I get the same error without using setuid in the following case:

$ cat test.py
import time
import os

print os.getpid()
while True:
	time.sleep(1)

In one terminal: python test.py. In another sudo src/pyflame --threads -p 9699 -s 10. In the third kill 9699 (alternatively, just ctrl+c the script).

Unexpected ptrace(2) exception: waitpid() indicated a WIFSTOPPED process, but got unexpected signal 15

("signal 2" in case with ctrl+c)
Ubuntu 18, latest pyflame, Python 2.7.15rc1

@fillest
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fillest commented Dec 1, 2018

And looks like the signals get caught only by pyflame - the script continues to run.

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3 participants