Preventing python definition from execution - python

I want to know what is the best way of checking an condition in Python definition and prevent it from further execution if condition is not satisfied. Right now i am following the below mentioned scheme but it actually prints the whole trace stack. I want it to print only an error message and do not execute the rest of code. Is there any other cleaner solution for doing it.
def Mydef(n1,n2):
if (n1>n2):
raise ValueError("Arg1 should be less than Arg2)
# Some Code
Mydef(2,1)

That is what exceptions are created for. Your scheme of raising exception is good in general; you just need to add some code to catch it and process it
try:
Mydef(2,1)
except ValueError, e:
# Do some stuff when exception is raised, e.message will contain your message
In this case, execution of Mydef stops when it encounters raise ValueError line of code, and goes to the code block under except.
You can read more about exceptions processing in the documentation.
If you don't want to deal with exceptions processing, you can gracefully stop function to execute further code with return statement.
def Mydef(n1,n2):
if (n1>n2):
return

def Mydef(n1,n2):
if (n1>n2):
print "Arg1 should be less than Arg2"
return None
# Some Code
Mydef(2,1)
Functions stop executing when they reach to return statement or they run the until the end of definition. You should read about flow control in general (not specifically to python)

Related

Execute when error occurs python

I guess I'm not the first asking this question, but I haven't found a solution that I could use/understand yet. And the issue is probably not as simple as i first expected.
I think it can be boiled down to two general questions:
1) Is there a way to avoid Python to stop when an error occur and just jump on to the next line of code in the script?
2) Is there a way to make Python execute a line of code if an error occurs? Like, if error then...
My concrete problem:
I have a very large program with a lot of functions and other stuff, which would take forever to adjust individually by using "try" for example (if i understand it correctly)
My program run as a large loop that gather information and keeps running. This means that it does not really matter to me, that my program fails multiple time as long as it keeps running. I can easily handle that some of the information is with error and would just like my program to take a note of it and keep going.
Is there a solution to this?
As you rightly pointed out, the try/catch block in Python is by far your best ally:
for i in range(N):
try: do_foo() ; except: do_other_foo()
try: do_bar() ; except: do_other_bar()
Alternatively, you could also use, in case you didn't need the Exception:
from contextlib import suppress
for i in range(N):
with suppress(Exception):
do_foo()
with suppress(Exception):
do_bar()
Your only possibility is to rely on the try/except clause. Keep in mind that the try/except may use also finally and else (see documentation:
try:
print("problematic code - error NOT raised")
except:
print("code that gets executed only if an error occurs")
else:
print("code that gets executed only if an error does not occur")
finally:
print("code that gets ALWAYS executed")
# OUTPUT:
# problematic code - error NOT raised
# code that gets executed only if an error does not occur
# code that gets ALWAYS executed
or, when an error is raised:
try:
print("problematic code - error raised!")
raise "Terrible, terrible error"
except:
print("code that gets executed only if an error occurs")
else:
print("code that gets executed only if an error does not occur")
finally:
print("code that gets ALWAYS executed")
# OUTPUT:
# problematic code - error raised!
# code that gets executed only if an error occurs
# code that gets ALWAYS executed
I urge to point out, by the way, that ignoring everything makes me shiver:
you really should (at least, more or less) identify which exception can be raised, catch them (except ArithmeticError: ..., check built-in exceptions) and handle them individually. What you're trying to do will probably snowball into an endless chain of problems, and ignoring them will probably create more problems!
I think that this question helps to understand what a robust software is, meanwhile on this one you can see how SO community thinks python exceptions should be handled

Does 'finally' always execute in Python?

For any possible try-finally block in Python, is it guaranteed that the finally block will always be executed?
For example, let’s say I return while in an except block:
try:
1/0
except ZeroDivisionError:
return
finally:
print("Does this code run?")
Or maybe I re-raise an Exception:
try:
1/0
except ZeroDivisionError:
raise
finally:
print("What about this code?")
Testing shows that finally does get executed for the above examples, but I imagine there are other scenarios I haven't thought of.
Are there any scenarios in which a finally block can fail to execute in Python?
"Guaranteed" is a much stronger word than any implementation of finally deserves. What is guaranteed is that if execution flows out of the whole try-finally construct, it will pass through the finally to do so. What is not guaranteed is that execution will flow out of the try-finally.
A finally in a generator or async coroutine might never run, if the object never executes to conclusion. There are a lot of ways that could happen; here's one:
def gen(text):
try:
for line in text:
try:
yield int(line)
except:
# Ignore blank lines - but catch too much!
pass
finally:
print('Doing important cleanup')
text = ['1', '', '2', '', '3']
if any(n > 1 for n in gen(text)):
print('Found a number')
print('Oops, no cleanup.')
Note that this example is a bit tricky: when the generator is garbage collected, Python attempts to run the finally block by throwing in a GeneratorExit exception, but here we catch that exception and then yield again, at which point Python prints a warning ("generator ignored GeneratorExit") and gives up. See PEP 342 (Coroutines via Enhanced Generators) for details.
Other ways a generator or coroutine might not execute to conclusion include if the object is just never GC'ed (yes, that's possible, even in CPython), or if an async with awaits in __aexit__, or if the object awaits or yields in a finally block. This list is not intended to be exhaustive.
A finally in a daemon thread might never execute if all non-daemon threads exit first.
os._exit will halt the process immediately without executing finally blocks.
os.fork may cause finally blocks to execute twice. As well as just the normal problems you'd expect from things happening twice, this could cause concurrent access conflicts (crashes, stalls, ...) if access to shared resources is not correctly synchronized.
Since multiprocessing uses fork-without-exec to create worker processes when using the fork start method (the default on Unix), and then calls os._exit in the worker once the worker's job is done, finally and multiprocessing interaction can be problematic (example).
A C-level segmentation fault will prevent finally blocks from running.
kill -SIGKILL will prevent finally blocks from running. SIGTERM and SIGHUP will also prevent finally blocks from running unless you install a handler to control the shutdown yourself; by default, Python does not handle SIGTERM or SIGHUP.
An exception in finally can prevent cleanup from completing. One particularly noteworthy case is if the user hits control-C just as we're starting to execute the finally block. Python will raise a KeyboardInterrupt and skip every line of the finally block's contents. (KeyboardInterrupt-safe code is very hard to write).
If the computer loses power, or if it hibernates and doesn't wake up, finally blocks won't run.
The finally block is not a transaction system; it doesn't provide atomicity guarantees or anything of the sort. Some of these examples might seem obvious, but it's easy to forget such things can happen and rely on finally for too much.
Yes. Finally always wins.
The only way to defeat it is to halt execution before finally: gets a chance to execute (e.g. crash the interpreter, turn off your computer, suspend a generator forever).
I imagine there are other scenarios I haven't thought of.
Here are a couple more you may not have thought about:
def foo():
# finally always wins
try:
return 1
finally:
return 2
def bar():
# even if he has to eat an unhandled exception, finally wins
try:
raise Exception('boom')
finally:
return 'no boom'
Depending on how you quit the interpreter, sometimes you can "cancel" finally, but not like this:
>>> import sys
>>> try:
... sys.exit()
... finally:
... print('finally wins!')
...
finally wins!
$
Using the precarious os._exit (this falls under "crash the interpreter" in my opinion):
>>> import os
>>> try:
... os._exit(1)
... finally:
... print('finally!')
...
$
I'm currently running this code, to test if finally will still execute after the heat death of the universe:
try:
while True:
sleep(1)
finally:
print('done')
However, I'm still waiting on the result, so check back here later.
According to the Python documentation:
No matter what happened previously, the final-block is executed once the code block is complete and any raised exceptions handled. Even if there's an error in an exception handler or the else-block and a new exception is raised, the code in the final-block is still run.
It should also be noted that if there are multiple return statements, including one in the finally block, then the finally block return is the only one that will execute.
Well, yes and no.
What is guaranteed is that Python will always try to execute the finally block. In the case where you return from the block or raise an uncaught exception, the finally block is executed just before actually returning or raising the exception.
(what you could have controlled yourself by simply running the code in your question)
The only case I can imagine where the finally block will not be executed is when the Python interpretor itself crashes for example inside C code or because of power outage.
I found this one without using a generator function:
import multiprocessing
import time
def fun(arg):
try:
print("tried " + str(arg))
time.sleep(arg)
finally:
print("finally cleaned up " + str(arg))
return foo
list = [1, 2, 3]
multiprocessing.Pool().map(fun, list)
The sleep can be any code that might run for inconsistent amounts of time.
What appears to be happening here is that the first parallel process to finish leaves the try block successfully, but then attempts to return from the function a value (foo) that hasn't been defined anywhere, which causes an exception. That exception kills the map without allowing the other processes to reach their finally blocks.
Also, if you add the line bar = bazz just after the sleep() call in the try block. Then the first process to reach that line throws an exception (because bazz isn't defined), which causes its own finally block to be run, but then kills the map, causing the other try blocks to disappear without reaching their finally blocks, and the first process not to reach its return statement, either.
What this means for Python multiprocessing is that you can't trust the exception-handling mechanism to clean up resources in all processes if even one of the processes can have an exception. Additional signal handling or managing the resources outside the multiprocessing map call would be necessary.
You can use a finally with an if statement, below example is checking for network connection and if its connected it will run the finally block
try:
reader1, writer1 = loop.run_until_complete(self.init_socket(loop))
x = 'connected'
except:
print("cant connect server transfer") #open popup
x = 'failed'
finally :
if x == 'connected':
with open('text_file1.txt', "r") as f:
file_lines = eval(str(f.read()))
else:
print("not connected")

Python: Make exceptions 'exiting'

In Python, is there any (proper) way to change the the default exception handling behaviour so that any uncaught exception will terminate/exit the program?
I don't want to wrap the entire program in a generic try-except block:
try:
// write code here
except Exception:
sys.exit(1)
For those asking for more specificity and/or claiming this is already the case, it's my understanding that not all Python exceptions are system-exiting: docs
Edit: It looks like I have forked processes complicating matters so won't be posting any specific details about my own mess.
If you're looking for an answer to the original question, Dmitry's comment is interesting and useful, references the 2nd answer to this question
You can use Specific exception instead of Exception because Exception is a Base class for all exceptions. For more details refer Exception tutorial
You can write your script like this-
try:
# write code here
except OverflowError:
raise SystemExit
except ArithmeticError:
sys.exit()
except IOError:
quit()
Try this different approaches to find what is exactly you are missing.
Edit 1 - Maintain Program Execution
In order to maintain your program execution try this one-
consider error_handler function is raising SystemExit exception then In your main method you need to add below code so you can maintain your program execution.
try:
error_handler()
except SystemExit:
print "sys.exit was called but I'm proceeding anyway (so there!-)."

When is KeyboardInterrupt raised in Python?

All the docs tell us is,
Raised when the user hits the interrupt key (normally Control-C or Delete). During execution, a check for interrupts is made regularly.
But from the point of the code, when can I see this exception? Does it occur during statement execution? Only between statements? Can it happen in the middle of an expression?
For example:
file_ = open('foo')
# <-- can a KeyboardInterrupt be raised here, after the successful
# completion of open but prior to the try? -->
try:
# try some things with file_
finally:
# cleanup
Will this code leak during a well-timed KeyboardInterrupt? Or is it raised during the execution of some statements or expressions?
According to a note in the unrelated PEP 343:
Even if you write bug-free code, a KeyboardInterrupt exception can still cause it to exit between any two virtual machine opcodes.
So it can occur essentially anywhere. It can indeed occur during evaluation of a single expression. (This shouldn't be surprising, since an expression can include function calls, and pretty much anything can happen inside a function call.)
Yes, a KeyboardInterrupt can occur in the place you marked.
To deal with this, you should use a with block:
with open('foo') as file_:
# do some things
raise KeyboardInterrupt
# file resource is closed no matter what, even if a KeyboardInterrupt is raised
However, the exception could occur even between the open() call and the assignment to file_. It's probably not worth worrying about this, because usually a ctrl-c will mean your program is about to end, so the "leaked" file handle will be cleaned up by the OS. But if you know that it is important, you can use a signal handler to catch the signal that raises KeyboardInterrupt (SIGINT).

try-except-raise clause, good behaviour?

I have noticed me writing try-except clauses like the following very much in the past. The main reason for this is to write less code.
class Synchronizer(object):
# ...
def _assert_dir(self, dirname, argname, argnum):
""" *Private*. Raises OSError if the passed string does not point
to an existing directory on the file-system. """
if not os.path.isdir(dirname):
message = 'passed `%s` argument (%d) does not point to a ' \
'directory on the file-system.'
raise OSError(message % (argname, argnum))
def synchronize(self, source_dir, dest_dir, database):
# Ensure the passed directories do exist.
try:
self._assert_dir(source_dir, 'source_dir', 2)
self._assert_dir(dest_dir, 'dest_dir', 3)
except OSError:
raise
# ...
I was doing it this way, because otherwise I would've needed to write
class Synchronizer(object):
# ...
def synchronize(self, source_dir, dest_dir, database):
# Ensure the passed directories do exist.
if not os.path.isdir(source_dir):
message = 'passed `source_dir` argument (2) does not point to a ' \
'directory on the file-system.'
raise OSError(message)
if not os.path.isdir(dest_dir):
message = 'passed `dest_dir` argument (3) does not point to a ' \
'directory on the file-system.'
raise OSError(message)
# ...
I actually like the idea of writing methods doing check-and-raise operations, but I see one big disadvantage: Readability. Especially for editors that do code-folding, the try statement is not very much telling the reader what happens inside of it, while if not os.path.isdir(source_dir) is quite a good hint.
IMHO the try-except clause is required because it would confuse the catcher of the exception (reader of the traceback) where the exception comes from.
What do you think about this design? Is it awful, great or confusing to you? Or do you have any ideas on how to improve the situation?
There are two questions that I ask myself before using try for handling exceptional conditions and if the answer is YES to both, only then I will try to handle the exception.
Q1. Is this truly an exception scenario? I do not want to execute try blocks if the condition occurs 90% of the time. It is better to use if - else in such a case.
Q2. Can I recover from the error? It makes little sense to handle the exception if I cannot recover from it. It's better to propagate it to a higher level which happens automatically without me having to write extra code.
The code posted by you does not do anything to recover if the directory does not exist and it does not appear that you can do much about it. Why not let the error propagate to a higher level? Why do you even need a try block there?
This depends upon your requirement..
If you want to catch some exception, and continue with the code in your method, then you should use the 2nd scenario. Have yout try-except block inside your method.
def function():
try:
raise IOError
except IOError e:
// Handle
//continue with reset of the function
print "This will get printed"
function()
But if you want to handle all the exception at one place, with specific action for specific type, or you just want to halt your function, if one exception is raised, you can better handle them outside your function: -
def function():
raise IOError
// subsequent code Will not execute
print "This will not get printed"
try:
function()
except IOError e:
// Handle IOError
except EOFError e1:
// Handle EOF Error
By using the 2nd way, you are actually increasing the chance of some of your codes not getting executed. In general, your try-except block should be small. They should be separated for handling exception at different points and not all the exceptions should be handled at one place.
As far as I'm concerned, I generally like to minimize my try-except block as much as possible. That way I know where exactly my exception was raised.

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