Python – Printing Daemonic Thread Exceptions in Python

multithreadingpython

Python does not print traceback messages from exceptions raised in daemon threads.

For example, this code creates a daemonic thread and raises an exception in the new thread:

def error_raiser():
    raise Exception

import threading
thread = threading.Thread(target=error_raiser)
thread.daemon = True
thread.start()

but does not print a traceback. (It gives no output).

However, if the thread is not set as a daemon thread, Python will print the traceback. Here is the same code with one line commented out:

def error_raiser():
    raise Exception

import threading
thread = threading.Thread(target=error_raiser)
# thread.daemon = True
thread.start()

and the output:

Exception in Thread-1:
Traceback (most recent call last):
  File "C:\Python26\lib\threading.py", line 525, in __bootstrap_inner
    self.run()
  File "C:\Python26\lib\threading.py", line 477, in run
    self.__target(*self.__args, **self.__kwargs)
  File "test.py", line 2, in error_raiser
    raise Exception
Exception

Executing this code in both Python 2.6.2 and Python 3.0.1 and gives the same results. Interestingly, however, if I execute the code by importing it in the IPython shell, the exception is displayed whether the thread is daemonic or not.

According to the documentation, the only significance of the 'daemon' flag is that "the entire Python program exits when only daemon threads are left." This would make me believe that not printing a traceback after an exception is a bug in Python, unless I missed something in the documentation.

Is this a bug, or did I miss something in the documentation and this behavior is intentional? If it is intentional, how can I force Python to print the traceback in daemon threads without using IPython?

Best Answer

According to Wikipedia, by definition a daemon should detach itself from the controlling tty, so I think it's correct that no exception is shown (and after all, a daemon should keep working even if you close the shell that launched it)..
See here.

As for how to print the traceback, I think that a simple try/except_then_log_to_file would do the trick :)

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Python – How to check whether a file exists without exceptions

If the reason you're checking is so you can do something like if file_exists: open_it(), it's safer to use a try around the attempt to open it. Checking and then opening risks the file being deleted or moved or something between when you check and when you try to open it.

If you're not planning to open the file immediately, you can use os.path.isfile

Return True if path is an existing regular file. This follows symbolic links, so both islink() and isfile() can be true for the same path.

import os.path
os.path.isfile(fname)

if you need to be sure it's a file.

Starting with Python 3.4, the pathlib module offers an object-oriented approach (backported to pathlib2 in Python 2.7):

from pathlib import Path

my_file = Path("/path/to/file")
if my_file.is_file():
    # file exists

To check a directory, do:

if my_file.is_dir():
    # directory exists

To check whether a Path object exists independently of whether is it a file or directory, use exists():

if my_file.exists():
    # path exists

You can also use resolve(strict=True) in a try block:

try:
    my_abs_path = my_file.resolve(strict=True)
except FileNotFoundError:
    # doesn't exist
else:
    # exists

Python – How to execute a program or call a system command

Use the subprocess module in the standard library:

import subprocess
subprocess.run(["ls", "-l"])

The advantage of subprocess.run over os.system is that it is more flexible (you can get the stdout, stderr, the "real" status code, better error handling, etc...).

Even the documentation for os.system recommends using subprocess instead:

The subprocess module provides more powerful facilities for spawning new processes and retrieving their results; using that module is preferable to using this function. See the Replacing Older Functions with the subprocess Module section in the subprocess documentation for some helpful recipes.

On Python 3.4 and earlier, use subprocess.call instead of .run:

subprocess.call(["ls", "-l"])

Best Answer

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Python – How to execute a program or call a system command

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