Python: rasing from except causes finally to execute? - python

Say I have the following code:
try:
[...]
except:
raise Exception([...])
finally:
[code]
My question is: if the code in the try block raises an exception which is caught in except, is [code] from the finally clause executed since a new exception is raised in the except clause? And if so, when is it executed? Before the new exception is raised or after the new exception is propagated through the method stack?

an example is worth a 1000 words, why didn't you just try what you've written?
>>> def foo():
>>> try:
>>> print "2 try block"
>>> raise Exception("1")
>>> print "never printed"
>>> except:
>>> print "3 first except block"
>>> raise Exception("2")
>>> finally:
>>> print "4 finally block"
>>> print "end of function"
>>>
>>> try:
>>> print "1 before foo"
>>> foo()
>>> print "never printed too"
>>> except:
>>> print "5 outter except clause"
1 before foo
2 try block
3 first except block
4 finally block
5 outter except clause
Before the new exception is raised or after the new exception is propagated through the method stack?
so as you can tell from the example, the finally block is called after the except block it has been defined in (i.e. after leaving the try/except/finally block), but before getting to the outer try/except block.
Which is logical, you want the finally to always be triggered when you exit the try block however you exit it, so you can be sure that your code environment is coherent when executing code outside of the try statement (whether you're releasing resources, or resetting values, or rolling back/committing modifications...).

finally is executed no matter the try block succeeds or the except block is run due to exceptions!
even if your except block raises an exception the new exception will be handled by another try catch handler but after executing the finally block, instead of forming a recursive loop:
try:
try:
[...]
except:
raise Exception([...]) #this is line number xyz
finally:
[code]
except:
[...] #this code will be running after line number xyz

Related

Python debugging in eclipse stop on outer except block

I have nested exceptions handling in my code, when inner blocks do some stuff before re-raising the exception to upper layers.
Traceback always reports the line number that started the exception handling.
However, when running in UI debugger (Pydev/Eclipse) it stops on the outer exception block in some cases.
Consider the following code for example:
import sys
def f(a):
c=5/a
return c
def sub(d):
print("Entered sub(%d)" % d)
try:
print("Entered try #sub")
e = f(d)/(d-1)
return e
except:
print("Do some staff before re-raising the exception upwards")
raise
def main():
try:
print("Entered try #main")
d = int(sys.argv[1])
sub(d)
except:
print("Reached except block #main")
raise
if __name__ == '__main__':
main()
When running with argument = 0, the exception is caused at line#4 and the debugger stops on that line:
However, when running with argument = 1, the exception is caused at line#11 (as reported in the printed traceback) but the debugger stops at line#15.
Once the debugger stops at the incorrect location it is very difficult to watch internal variables and handle the error, especially when the code in the try block uses loops.
In Pydev->Manage Exceptions, I checked only the "suspend on uncaught exceptions".
There is a checkbox "Skip exceptions caught in same function" that seem related (because it seems like the debugger skipped the first exception in sub and stopped on "raise" statement which can be considered another exception in same function, although the documentation says that it should re-raise the same exception).
This checkbox is grayed out unless I first check "Suspend on caught exceptions*", but once enabling it the debugger gets stuck and does not stop anywhere...
Will appreciate your help.
-Moshe

Python: pseudo try before actually try and finding exception

I know I'm not clear on this thing but I couldn't be more specific in the title. Consider the following code:
try:
print "Try this out"
a = int("Blah blah")
except:
print "I got the exception"
The output of this code is-
Try this out
I got the exception
What I want the python to do is to check if it might raise an exception in the try: block first and then execute it. Else, just execute the except: block. Is it possible to do such a thing without nesting the multiple try-except blocks.
No, it's not possible, because you get an exception during execution. But you can do like this:
try:
a = int("Blah blah")
print "Try this out"
except:
print "I got the exception"
Your try statement prints up as no error happens in printing it ,but as soon as it tries to execute a, gets an error so except executes.
Try this way to catch the exception properly:
try:
a = int("Blah blah")
print ("Try this out")
except:
print ("I got the exception")

Break statement in finally block swallows exception

Consider:
def raiseMe( text="Test error" ):
raise Exception( text )
def break_in_finally_test():
for i in range(5):
if i==2:
try:
raiseMe()
except:
raise
else:
print "succeeded!"
finally:
print "testing this!"
break
if __name__=='__main__':
break_in_finally_test()
I expected to see Exception( "Test error" ) to be raised, but instead only "testing this" is printed. The intention, of course, was to call raiseMe() only once, no matter if we succeed or not - but if it raises an exception, I would have wanted to see that!
Why does break swallow the exception that I explicitly raise?
From https://docs.python.org/2.7/reference/compound_stmts.html#finally:
If finally is present, it specifies a ‘cleanup’ handler. The try clause is executed, including any except and else clauses. If an exception occurs in any of the clauses and is not handled, the exception is temporarily saved. The finally clause is executed. If there is a saved exception, it is re-raised at the end of the finally clause. If the finally clause raises another exception or executes a return or break statement, the saved exception is discarded
This also reflects the behaviour expected from the try...finally statement before PEP341:
This is how a try except finally block looked like pre PEP341:
try:
try:
raiseMe()
except:
raise
finally:
#here is where cleanup is supposed to happen before raising error
break
#after finally code: raise error
As the raising of errors never happens in the finally block it is never actually raised.
To maintain backwards compatibility with Python<=2.4, it had to be done in this way.
From the docs Error Handling Docs:
A finally clause is always executed before leaving the try statement, whether an exception has occurred or not.
Your exception never gets raised because you break before the try statement gets fully evaluated.
I think upon reflection that it is due to the fact that break actually raises a StopIteration to "break" out of the for-loop. This is really not very intuitive and not particularly well documented (not mentioned on 1, for example). Could maybe someone confirm/explain it better?
Have the following code structure:
def func():
try:
driver.do("something")
except TimeoutException:
pass
finally:
result = driver.do("something else")
return result
Got the exception by pylint:
return statement in finally block may swallow exception (lost-exception)
Solution was to put return out of the finally statement:
def func():
try:
driver.do("something")
except TimeoutException:
pass
finally:
result = driver.do("something else")
return result # <<

Exit python properly

I knew that sys.exit() raises an Exit exception, so when I run this I knew it wouldn't exit:
In [25]: try:
....: sys.exit()
....: except:
....: print "oops"
....:
oops
But I thought that os._exit() was meant to exit using a C call, but it's also causing an exception:
In [28]: try:
....: os._exit()
....: except:
....: print "oops"
....:
oops
Is there a way of doing this without killing the PID?
I'm not sure I understand your question, but os._exit isn't raising an exception -- you calling os._exit with the wrong number of arguments is:
try:
os._exit()
except Exception as e:
print e
#output: _exit() takes exactly 1 argument (0 given)
Don't use except without an Exception class, so sys.exit will just work fine without triggering the exception handling:
>>> import sys
>>> try:
... sys.exit()
... except Exception:
... print 'oops'
...
$
There are other exceptions which are triggered with a plain except clause (and in general shouldn't), especially KeyboardInterrupt.

In Python try until no error

I have a piece of code in Python that seems to cause an error probabilistically because it is accessing a server and sometimes that server has a 500 internal server error. I want to keep trying until I do not get the error. My solution was:
while True:
try:
#code with possible error
except:
continue
else:
#the rest of the code
break
This seems like a hack to me. Is there a more Pythonic way to do this?
It won't get much cleaner. This is not a very clean thing to do. At best (which would be more readable anyway, since the condition for the break is up there with the while), you could create a variable result = None and loop while it is None. You should also adjust the variables and you can replace continue with the semantically perhaps correct pass (you don't care if an error occurs, you just want to ignore it) and drop the break - this also gets the rest of the code, which only executes once, out of the loop. Also note that bare except: clauses are evil for reasons given in the documentation.
Example incorporating all of the above:
result = None
while result is None:
try:
# connect
result = get_data(...)
except:
pass
# other code that uses result but is not involved in getting it
Here is one that hard fails after 4 attempts, and waits 2 seconds between attempts. Change as you wish to get what you want form this one:
from time import sleep
for x in range(0, 4): # try 4 times
try:
# msg.send()
# put your logic here
str_error = None
except Exception as str_error:
pass
if str_error:
sleep(2) # wait for 2 seconds before trying to fetch the data again
else:
break
Here is an example with backoff:
from time import sleep
sleep_time = 2
num_retries = 4
for x in range(0, num_retries):
try:
# put your logic here
str_error = None
except Exception as e:
str_error = str(e)
if str_error:
sleep(sleep_time) # wait before trying to fetch the data again
sleep_time *= 2 # Implement your backoff algorithm here i.e. exponential backoff
else:
break
Maybe something like this:
connected = False
while not connected:
try:
try_connect()
connected = True
except ...:
pass
When retrying due to error, you should always:
implement a retry limit, or you may get blocked on an infinite loop
implement a delay, or you'll hammer resources too hard, such as your CPU or the already distressed remote server
A simple generic way to solve this problem while covering those concerns would be to use the backoff library. A basic example:
import backoff
#backoff.on_exception(
backoff.expo,
MyException,
max_tries=5
)
def make_request(self, data):
# do the request
This code wraps make_request with a decorator which implements the retry logic. We retry whenever our specific error MyException occurs, with a limit of 5 retries. Exponential backoff is a good idea in this context to help minimize the additional burden our retries place on the remote server.
The itertools.iter_except recipes encapsulates this idea of "calling a function repeatedly until an exception is raised". It is similar to the accepted answer, but the recipe gives an iterator instead.
From the recipes:
def iter_except(func, exception, first=None):
""" Call a function repeatedly until an exception is raised."""
try:
if first is not None:
yield first() # For database APIs needing an initial cast to db.first()
while True:
yield func()
except exception:
pass
You can certainly implement the latter code directly. For convenience, I use a separate library, more_itertools, that implements this recipe for us (optional).
Code
import more_itertools as mit
list(mit.iter_except([0, 1, 2].pop, IndexError))
# [2, 1, 0]
Details
Here the pop method (or given function) is called for every iteration of the list object until an IndexError is raised.
For your case, given some connect_function and expected error, you can make an iterator that calls the function repeatedly until an exception is raised, e.g.
mit.iter_except(connect_function, ConnectionError)
At this point, treat it as any other iterator by looping over it or calling next().
Here's an utility function that I wrote to wrap the retry until success into a neater package. It uses the same basic structure, but prevents repetition. It could be modified to catch and rethrow the exception on the final try relatively easily.
def try_until(func, max_tries, sleep_time):
for _ in range(0,max_tries):
try:
return func()
except:
sleep(sleep_time)
raise WellNamedException()
#could be 'return sensibleDefaultValue'
Can then be called like this
result = try_until(my_function, 100, 1000)
If you need to pass arguments to my_function, you can either do this by having try_until forward the arguments, or by wrapping it in a no argument lambda:
result = try_until(lambda : my_function(x,y,z), 100, 1000)
Maybe decorator based?
You can pass as decorator arguments list of exceptions on which we want to retry and/or number of tries.
def retry(exceptions=None, tries=None):
if exceptions:
exceptions = tuple(exceptions)
def wrapper(fun):
def retry_calls(*args, **kwargs):
if tries:
for _ in xrange(tries):
try:
fun(*args, **kwargs)
except exceptions:
pass
else:
break
else:
while True:
try:
fun(*args, **kwargs)
except exceptions:
pass
else:
break
return retry_calls
return wrapper
from random import randint
#retry([NameError, ValueError])
def foo():
if randint(0, 1):
raise NameError('FAIL!')
print 'Success'
#retry([ValueError], 2)
def bar():
if randint(0, 1):
raise ValueError('FAIL!')
print 'Success'
#retry([ValueError], 2)
def baz():
while True:
raise ValueError('FAIL!')
foo()
bar()
baz()
of course the 'try' part should be moved to another funcion becouse we using it in both loops but it's just example;)
Like most of the others, I'd recommend trying a finite number of times and sleeping between attempts. This way, you don't find yourself in an infinite loop in case something were to actually happen to the remote server.
I'd also recommend continuing only when you get the specific exception you're expecting. This way, you can still handle exceptions you might not expect.
from urllib.error import HTTPError
import traceback
from time import sleep
attempts = 10
while attempts > 0:
try:
#code with possible error
except HTTPError:
attempts -= 1
sleep(1)
continue
except:
print(traceback.format_exc())
#the rest of the code
break
Also, you don't need an else block. Because of the continue in the except block, you skip the rest of the loop until the try block works, the while condition gets satisfied, or an exception other than HTTPError comes up.
what about the retrying library on pypi?
I have been using it for a while and it does exactly what I want and more (retry on error, retry when None, retry with timeout). Below is example from their website:
import random
from retrying import retry
#retry
def do_something_unreliable():
if random.randint(0, 10) > 1:
raise IOError("Broken sauce, everything is hosed!!!111one")
else:
return "Awesome sauce!"
print do_something_unreliable()
e = ''
while e == '':
try:
response = ur.urlopen('https://https://raw.githubusercontent.com/MrMe42/Joe-Bot-Home-Assistant/mac/Joe.py')
e = ' '
except:
print('Connection refused. Retrying...')
time.sleep(1)
This should work. It sets e to '' and the while loop checks to see if it is still ''. If there is an error caught be the try statement, it prints that the connection was refused, waits 1 second and then starts over. It will keep going until there is no error in try, which then sets e to ' ', which kills the while loop.
Im attempting this now, this is what i came up with;
placeholder = 1
while placeholder is not None:
try:
#Code
placeholder = None
except Exception as e:
print(str(datetime.time(datetime.now()))[:8] + str(e)) #To log the errors
placeholder = e
time.sleep(0.5)
continue
Here is a short piece of code I use to capture the error as a string. Will retry till it succeeds. This catches all exceptions but you can change this as you wish.
start = 0
str_error = "Not executed yet."
while str_error:
try:
# replace line below with your logic , i.e. time out, max attempts
start = raw_input("enter a number, 0 for fail, last was {0}: ".format(start))
new_val = 5/int(start)
str_error=None
except Exception as str_error:
pass
WARNING: This code will be stuck in a forever loop until no exception occurs. This is just a simple example and MIGHT require you to break out of the loop sooner or sleep between retries.

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