Python Development Mode¶
New in version 3.7.
The Python Development Mode introduces additional runtime checks that are too expensive to be enabled by default. It should not be more verbose than the default if the code is correct; new warnings are only emitted when an issue is detected.
See also Python debug build.
Effects of the Python Development Mode¶
Enabling the Python Development Mode is similar to the following command, but with additional effects described below:
PYTHONMALLOC=debug PYTHONASYNCIODEBUG=1 python3 -W default -X faulthandler
Effects of the Python Development Mode:
defaultwarning filter. The following warnings are shown:
Normally, the above warnings are filtered by the default warning filters.
It behaves as if the
-W defaultcommand line option is used.
Install debug hooks on memory allocators to check for:
Memory allocator API violation
Unsafe usage of the GIL
It behaves as if the
PYTHONMALLOCenvironment variable is set to
To enable the Python Development Mode without installing debug hooks on memory allocators, set the
PYTHONMALLOCenvironment variable to
faulthandler.enable()at Python startup to install handlers for the
SIGILLsignals to dump the Python traceback on a crash.
It behaves as if the
PYTHONASYNCIODEBUGenvironment variable is set to
By default, for best performance, the errors argument is only checked at the first encoding/decoding error and the encoding argument is sometimes ignored for empty strings.
The Python Development Mode does not enable the
tracemalloc module by
default, because the overhead cost (to performance and memory) would be too
large. Enabling the
tracemalloc module provides additional information
on the origin of some errors. For example,
ResourceWarning logs the
traceback where the resource was allocated, and a buffer overflow error logs
the traceback where the memory block was allocated.
The Python Development Mode can only be enabled at the Python startup. Its
value can be read from
Changed in version 3.8: The
io.IOBase destructor now logs
Changed in version 3.9: The encoding and errors arguments are now checked for string encoding and decoding operations.
Example of a script counting the number of lines of the text file specified in the command line:
import sys def main(): fp = open(sys.argv) nlines = len(fp.readlines()) print(nlines) # The file is closed implicitly if __name__ == "__main__": main()
The script does not close the file explicitly. By default, Python does not emit any warning. Example using README.txt, which has 269 lines:
$ python3 script.py README.txt 269
Enabling the Python Development Mode displays a
$ python3 -X dev script.py README.txt 269 script.py:10: ResourceWarning: unclosed file <_io.TextIOWrapper name='README.rst' mode='r' encoding='UTF-8'> main() ResourceWarning: Enable tracemalloc to get the object allocation traceback
In addition, enabling
tracemalloc shows the line where the file was
$ python3 -X dev -X tracemalloc=5 script.py README.rst 269 script.py:10: ResourceWarning: unclosed file <_io.TextIOWrapper name='README.rst' mode='r' encoding='UTF-8'> main() Object allocated at (most recent call last): File "script.py", lineno 10 main() File "script.py", lineno 4 fp = open(sys.argv)
The fix is to close explicitly the file. Example using a context manager:
def main(): # Close the file explicitly when exiting the with block with open(sys.argv) as fp: nlines = len(fp.readlines()) print(nlines)
Not closing a resource explicitly can leave a resource open for way longer than expected; it can cause severe issues upon exiting Python. It is bad in CPython, but it is even worse in PyPy. Closing resources explicitly makes an application more deterministic and more reliable.
Bad file descriptor error example¶
Script displaying the first line of itself:
import os def main(): fp = open(__file__) firstline = fp.readline() print(firstline.rstrip()) os.close(fp.fileno()) # The file is closed implicitly main()
By default, Python does not emit any warning:
$ python3 script.py import os
The Python Development Mode shows a
ResourceWarning and logs a “Bad file
descriptor” error when finalizing the file object:
$ python3 script.py import os script.py:10: ResourceWarning: unclosed file <_io.TextIOWrapper name='script.py' mode='r' encoding='UTF-8'> main() ResourceWarning: Enable tracemalloc to get the object allocation traceback Exception ignored in: <_io.TextIOWrapper name='script.py' mode='r' encoding='UTF-8'> Traceback (most recent call last): File "script.py", line 10, in <module> main() OSError: [Errno 9] Bad file descriptor
os.close(fp.fileno()) closes the file descriptor. When the file object
finalizer tries to close the file descriptor again, it fails with the
file descriptor error. A file descriptor must be closed only once. In the
worst case scenario, closing it twice can lead to a crash (see bpo-18748
for an example).
The fix is to remove the
os.close(fp.fileno()) line, or open the file with