Re: How to retry something with a timeout in Python?

2009-04-29 Thread tinnews 
Scott David Daniels scott.dani...@acm.org wrote:
 tinn...@isbd.co.uk wrote:
  This feels like it should be simple but I can't see a clean way of
  doing it at the moment.
  
  I want to retry locking a file for a number of times and then give up,
  in pseudo-code it would be something like:-
  
  
  for N times
  try to lock file
  if successful break out of for loop
  if we don't have a lock then give up and exit
 
  for attempt in range(N):
   try:
   lock_file_with_timeout(per_try)  # change to what you mean
   except LockAttemptFailure: # or however the failure is shown
   pass  # here the attempt+1th try failed.
   else:
   break # success -- have the lock
  else:
  raise ImTiredError  # however you handle N attempts w/o success
  rest_of_code
 
Ah, yes, it's the 'else:' with the 'for' that makes it easier, I've
come from languages which don't have that, thank you!  :-)

 Often it is easiest to stick it in a function:
 
  def retry_lock(tries=3, wait_per_attempt=.5):
  for attempt in range(tries):
  try:
  # change following to whatever you do to attempt a lock.
  lock_file_with_timeout(wait_per_attempt)
  except LockAttemptFailure: # or however the failure is shown
  pass  # here the attempt+1th try failed.
  else:
  return # success -- have the lock
  raise ImTiredError
 
 --Scott David Daniels
 scott.dani...@acm.org

-- 
Chris Green

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Re: How to retry something with a timeout in Python?

2009-04-28 Thread Scott David Daniels 

tinn...@isbd.co.uk wrote:

This feels like it should be simple but I can't see a clean way of
doing it at the moment.

I want to retry locking a file for a number of times and then give up,
in pseudo-code it would be something like:-


for N times
try to lock file
if successful break out of for loop
if we don't have a lock then give up and exit


for attempt in range(N):
 try:
 lock_file_with_timeout(per_try)  # change to what you mean
 except LockAttemptFailure: # or however the failure is shown
 pass  # here the attempt+1th try failed.
 else:
 break # success -- have the lock
else:
raise ImTiredError  # however you handle N attempts w/o success
rest_of_code

Often it is easiest to stick it in a function:

def retry_lock(tries=3, wait_per_attempt=.5):
for attempt in range(tries):
try:
# change following to whatever you do to attempt a lock.
lock_file_with_timeout(wait_per_attempt)
except LockAttemptFailure: # or however the failure is shown
pass  # here the attempt+1th try failed.
else:
return # success -- have the lock
raise ImTiredError

--Scott David Daniels
scott.dani...@acm.org
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http://mail.python.org/mailman/listinfo/python-list

Re: How to retry something with a timeout in Python?

2009-04-28 Thread Gabriel Genellina 

En Tue, 28 Apr 2009 17:56:13 -0300, tinn...@isbd.co.uk escribió:


I want to retry locking a file for a number of times and then give up,
in pseudo-code it would be something like:-


for N times
try to lock file
if successful break out of for loop
if we don't have a lock then give up and exit


How does one do this tidily in python?


for i in range(N):
  if try_to_lock_file():
break
else:
  raise RuntimeError, Could not lock file
do_real_work()

Note the indentation of the `else` clause, it belongs to the `for` loop  
and is executed only when the iteration is completely exhausted.


--
Gabriel Genellina

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