Given the following code (that doesn't work):
while True:
# Snip: print out current state
while True:
ok = get_input("Is this ok? (y/n)")
if ok.lower() == "y": break 2 # This doesn't work :(
if ok.lower() == "n": break
# Do more processing with menus and stuff
Is there a way to make this work? Or do I have do one check to break out of the input loop, then another, more limited, check in the outside loop to break out all together if the user is satisfied?
My first instinct would be to refactor the nested loop into a function and use return to break out.
islice be relevant here? And why would you be forced to yield in an inner function that doesn't yield (or build and return) anything? And even if you have a different case in mind, where the loop is yielding, you can still factor it into a function; just use yield from (3.3+), a for loop around yield (3.2-), or return the generator iterator instead of yielding it (either), as appropriate. –
Shrieval return is the right approach. And the reasoning is that, according to the Zen of Python, "flat is better than nested". We have three levels of nesting here and if that starts to get in the way, it is time to reduce the nesting or at least extract the whole nesting into a function of its own. –
Novah return is a good idea in general, but I found that list comprehensions can help too. –
Retardant raise to simulate labelled break. Create a new Exception instance for every different label, use raise LabelException() instead of break label and wrap in try: ... except LabelException: the code that needs to jump to that label. If your code is that complex to need labels adding some exceptions classes a raise and an except wont be that bad, the code is ugly anyway. –
Acescent BreakAllLoops option...for now I have a boolean flag that if halt is true then everyone halts but there are lots of if statements...in fact in my case I explicitly don't want to write a separate function. –
Ovenbird Here's another approach that is short. The disadvantage is that you can only break the outer loop, but sometimes it's exactly what you want.
for a in xrange(10):
for b in xrange(20):
if something(a, b):
# Break the inner loop...
break
else:
# Continue if the inner loop wasn't broken.
continue
# Inner loop was broken, break the outer.
break
This uses the for / else construct explained at: Why does python use 'else' after for and while loops?
Key insight: It only seems as if the outer loop always breaks. But if the inner loop doesn't break, the outer loop won't either.
The continue statement is the magic here. It's in the for-else clause. By definition that happens if there's no inner break. In that situation continue neatly circumvents the outer break.
something(a, b) depends on a too. The outer loop may run as long as something(a, b) is not True. –
Corrective break/continue statements, you'd bet I'm throwing this code out. It's extremely hard to read and also confusing for everyone. –
Lafrance PEP 3136 proposes labeled break/continue. Guido rejected it because "code so complicated to require this feature is very rare". The PEP does mention some workarounds, though (such as the exception technique), while Guido feels refactoring to use return will be simpler in most cases.
return is usually the way to go, I've seen quite a few cases where a simple concise ‘break 2’ statement would just make so much sense. Also, refactor/return doesn't work the same for continue. In these cases, numeric break and continue would be easier to follow and less cluttered than refactoring to a tiny function, raising exceptions, or convoluted logic involving setting a flag to break at each nest level. It's a shame Guido rejected it. –
Selfimmolation break; break would be nice. –
Volsci break through all of the loops. Stupid. –
Bouse for; else or if; else constructs. –
Pes break; break is already valid syntax; it's just two consecutive break statements, same as if they were written on consecutive lines. It therefore makes little sense as a candidate syntax for doing this. Rather than introducing some weird syntactic special case where code that would otherwise parse as two statements gets an alternate meaning, it'd be cleaner and less confusing to support numbered break (e.g. break 2) as other languages do and the PEP suggests. –
Jerol break; break—that's perfectly valid code, but the second break doesn't accomplish anything because it will never be reached. –
Decision First, ordinary logic is helpful.
If, for some reason, the terminating conditions can't be worked out, exceptions are a fall-back plan.
class GetOutOfLoop( Exception ):
pass
try:
done= False
while not done:
isok= False
while not (done or isok):
ok = get_input("Is this ok? (y/n)")
if ok in ("y", "Y") or ok in ("n", "N") :
done= True # probably better
raise GetOutOfLoop
# other stuff
except GetOutOfLoop:
pass
For this specific example, an exception may not be necessary.
On other other hand, we often have "Y", "N" and "Q" options in character-mode applications. For the "Q" option, we want an immediate exit. That's more exceptional.
done). (2) raising an exception. Merging them together into a single solution just makes it look complicated. For future readers: EITHER use all the lines involving done, OR define GetOutOfLoop(Exception) and raise/except that. –
Fertility raise Exception('GetOutOfLoop') and except Exception:. –
Elsi Exception constructor (e.g. raise Exception('bla bla bla')) are valid in both Python 2 and Python 3. The former is preferable in this case because we don't want our except block to catch all exceptions, but only the special exception we're using to exit the loop. If we do things the way you suggest, and then a bug in our code causes an unexpected exception to be raised, it'll be wrongly treated the same as deliberately exiting the loop. –
Jerol Introduce a new variable that you'll use as a 'loop breaker'. First assign something to it(False,0, etc.), and then, inside the outer loop, before you break from it, change the value to something else(True,1,...). Once the loop exits make the 'parent' loop check for that value. Let me demonstrate:
breaker = False #our mighty loop exiter!
while True:
while True:
if conditionMet:
#insert code here...
breaker = True
break
if breaker: # the interesting part!
break # <--- !
If you have an infinite loop, this is the only way out; for other loops execution is really a lot faster. This also works if you have many nested loops. You can exit all, or just a few. Endless possibilities! Hope this helped!
I tend to agree that refactoring into a function is usually the best approach for this sort of situation, but for when you really need to break out of nested loops, here's an interesting variant of the exception-raising approach that @S.Lott described. It uses Python's with statement to make the exception raising look a bit nicer. Define a new context manager (you only have to do this once) with:
from contextlib import contextmanager
@contextmanager
def nested_break():
class NestedBreakException(Exception):
pass
try:
yield NestedBreakException
except NestedBreakException:
pass
Now you can use this context manager as follows:
with nested_break() as mylabel:
while True:
print("current state")
while True:
ok = input("Is this ok? (y/n)")
if ok == "y" or ok == "Y": raise mylabel
if ok == "n" or ok == "N": break
print("more processing")
Advantages: (1) it's slightly cleaner (no explicit try-except block), and (2) you get a custom-built Exception subclass for each use of nested_break; no need to declare your own Exception subclass each time.
First, you may also consider making the process of getting and validating the input a function; within that function, you can just return the value if its correct, and keep spinning in the while loop if not. This essentially obviates the problem you solved, and can usually be applied in the more general case (breaking out of multiple loops). If you absolutely must keep this structure in your code, and really don't want to deal with bookkeeping booleans...
You may also use goto in the following way (using an April Fools module from here):
#import the stuff
from goto import goto, label
while True:
#snip: print out current state
while True:
ok = get_input("Is this ok? (y/n)")
if ok == "y" or ok == "Y": goto .breakall
if ok == "n" or ok == "N": break
#do more processing with menus and stuff
label .breakall
I know, I know, "thou shalt not use goto" and all that, but it works well in strange cases like this.
To break out of multiple nested loops, without refactoring into a function, make use of a "simulated goto statement" with the built-in StopIteration exception:
try:
for outer in range(100):
for inner in range(100):
if break_early():
raise StopIteration
except StopIteration: pass
See this discussion on the use of goto statements for breaking out of nested loops.
keeplooping = True
while keeplooping:
# Do stuff
while keeplooping:
# Do some other stuff
if finisheddoingstuff():
keeplooping = False
or something like that.
You could set a variable in the inner loop, and check it in the outer loop immediately after the inner loop exits, breaking if appropriate. I kind of like the GOTO method, provided you don't mind using an April Fool's joke module - it’s not Pythonic, but it does make sense.
This isn't the prettiest way to do it, but in my opinion, it's the best way.
def loop():
while True:
#snip: print out current state
while True:
ok = get_input("Is this ok? (y/n)")
if ok == "y" or ok == "Y": return
if ok == "n" or ok == "N": break
#do more processing with menus and stuff
I'm pretty sure you could work out something using recursion here as well, but I don't know if that's a good option for you.
Keep looping if two conditions are true.
I think this is a more Pythonic way:
dejaVu = True
while dejaVu:
while True:
ok = raw_input("Is this ok? (y/n)")
if ok == "y" or ok == "Y" or ok == "n" or ok == "N":
dejaVu = False
break
while dejaVu:? You set it to true anyway. –
Rejoin There is no way to do this from a language level. Some languages have a goto others have a break that takes an argument, python does not.
The best options are:
Set a flag which is checked by the outer loop, or set the outer loops condition.
Put the loop in a function and use return to break out of all the loops at once.
Reformulate your logic.
Using Function
def doMywork(data):
for i in data:
for e in i:
return
Using flag
is_break = False
for i in data:
if is_break:
break # outer loop break
for e in i:
is_break = True
break # inner loop break
Factor your loop logic into an iterator that yields the loop variables and returns when done -- here is a simple one that lays out images in rows/columns until we're out of images or out of places to put them:
def it(rows, cols, images):
i = 0
for r in xrange(rows):
for c in xrange(cols):
if i >= len(images):
return
yield r, c, images[i]
i += 1
for r, c, image in it(rows=4, cols=4, images=['a.jpg', 'b.jpg', 'c.jpg']):
... do something with r, c, image ...
This has the advantage of splitting up the complicated loop logic and the processing...
An easy way to turn multiple loops into a single, breakable loop is to use numpy.ndindex
for i in range(n):
for j in range(n):
val = x[i, j]
break # still inside the outer loop!
for i, j in np.ndindex(n, n):
val = x[i, j]
break # you left the only loop there was!
You do have to index into your objects, as opposed to being able to iterate through the values explicitly, but at least in simple cases it seems to be approximately 2-20 times simpler than most of the answers suggested.
There is a hidden trick in the Python while ... else structure which can be used to simulate the double break without much code changes/additions. In essence if the while condition is false, the else block is triggered. Neither exceptions, continue or break trigger the else block. For more information see answers to "Else clause on Python while statement", or Python doc on while (v2.7).
while True:
#snip: print out current state
ok = ""
while ok != "y" and ok != "n":
ok = get_input("Is this ok? (y/n)")
if ok == "n" or ok == "N":
break # Breaks out of inner loop, skipping else
else:
break # Breaks out of outer loop
#do more processing with menus and stuff
The only downside is that you need to move the double breaking condition into the while condition (or add a flag variable). Variations of this exists also for the for loop, where the else block is triggered after loop completion.
How to break out of multiple loops in Python? and the answer should have been "It does not work, try something else". I know it fixes the exact given example of the OP, but does not answer their question. –
Womanhater In this case, as pointed out by others as well, functional decomposition is the way to go. Code in Python 3:
def user_confirms():
while True:
answer = input("Is this OK? (y/n) ").strip().lower()
if answer in "yn":
return answer == "y"
def main():
while True:
# do stuff
if user_confirms():
break
By using a function:
def myloop():
for i in range(1,6,1): # 1st loop
print('i:',i)
for j in range(1,11,2): # 2nd loop
print(' i, j:' ,i, j)
for k in range(1,21,4): # 3rd loop
print(' i,j,k:', i,j,k)
if i%3==0 and j%3==0 and k%3==0:
return # getting out of all loops
myloop()
Try running the above codes by commenting out the return as well.
Without using any function:
done = False
for i in range(1,6,1): # 1st loop
print('i:', i)
for j in range(1,11,2): # 2nd loop
print(' i, j:' ,i, j)
for k in range(1,21,4): # 3rd loop
print(' i,j,k:', i,j,k)
if i%3==0 and j%3==0 and k%3==0:
done = True
break # breaking from 3rd loop
if done: break # breaking from 2nd loop
if done: break # breaking from 1st loop
Now, run the above codes as is first and then try running by commenting out each line containing break one at a time from the bottom.
Another way of reducing your iteration to a single-level loop would be via the use of generators as also specified in the python reference
for i, j in ((i, j) for i in A for j in B):
print(i , j)
if (some_condition):
break
You could scale it up to any number of levels for the loop
The downside is that you can no longer break only a single level. It's all or nothing.
Another downside is that it doesn't work with a while loop. I originally wanted to post this answer on Python - `break` out of all loops but unfortunately that's closed as a duplicate of this one
I'd like to remind you that functions in Python can be created right in the middle of the code and can access the surrounding variables transparently for reading and with nonlocal or global declaration for writing.
So you can use a function as a "breakable control structure", defining a place you want to return to:
def is_prime(number):
foo = bar = number
def return_here():
nonlocal foo, bar
init_bar = bar
while foo > 0:
bar = init_bar
while bar >= foo:
if foo*bar == number:
return
bar -= 1
foo -= 1
return_here()
if foo == 1:
print(number, 'is prime')
else:
print(number, '=', bar, '*', foo)
>>> is_prime(67)
67 is prime
>>> is_prime(117)
117 = 13 * 9
>>> is_prime(16)
16 = 4 * 4
Try using an infinite generator.
from itertools import repeat
inputs = (get_input("Is this ok? (y/n)") for _ in repeat(None))
response = (i.lower()=="y" for i in inputs if i.lower() in ("y", "n"))
while True:
#snip: print out current state
if next(response):
break
#do more processing with menus and stuff
# this version uses a level counter to choose how far to break out
break_levels = 0
while True:
# snip: print out current state
while True:
ok = get_input("Is this ok? (y/n)")
if ok == "y" or ok == "Y":
break_levels = 1 # how far nested, excluding this break
break
if ok == "n" or ok == "N":
break # normal break
if break_levels:
break_levels -= 1
break # pop another level
if break_levels:
break_levels -= 1
break
# ...and so on
# this version breaks up to a certain label
break_label = None
while True:
# snip: print out current state
while True:
ok = get_input("Is this ok? (y/n)")
if ok == "y" or ok == "Y":
break_label = "outer" # specify label to break to
break
if ok == "n" or ok == "N":
break
if break_label:
if break_label != "inner":
break # propagate up
break_label = None # we have arrived!
if break_label:
if break_label != "outer":
break # propagate up
break_label = None # we have arrived!
#do more processing with menus and stuff
With an example: Are these two matrices equal/same?
matrix1 and matrix2 are the same size, n, two-dimensional matrices.
First solution, without a function
same_matrices = True
inner_loop_broken_once = False
n = len(matrix1)
for i in range(n):
for j in range(n):
if matrix1[i][j] != matrix2[i][j]:
same_matrices = False
inner_loop_broken_once = True
break
if inner_loop_broken_once:
break
Second solution, with a function
This is the final solution for my case.
def are_two_matrices_the_same (matrix1, matrix2):
n = len(matrix1)
for i in range(n):
for j in range(n):
if matrix1[i][j] != matrix2[i][j]:
return False
return True
Here's an implementation that seems to work:
break_ = False
for i in range(10):
if break_:
break
for j in range(10):
if j == 3:
break_ = True
break
else:
print(i, j)
The only draw back is that you have to define break_ before the loops.
What I would personally do is use a boolean that toggles when I am ready to break out the outer loop. For example
while True:
#snip: print out current state
quit = False
while True:
ok = input("Is this ok? (y/n)")
if ok.lower() == "y":
quit = True
break # this should work now :-)
if ok.lower() == "n":
quit = True
break # This should work too :-)
if quit:
break
#do more processing with menus and stuff
My reason for coming here is that i had an outer loop and an inner loop like so:
for x in array:
for y in dont_use_these_values:
if x.value==y:
array.remove(x) # fixed, was array.pop(x) in my original answer
continue
do some other stuff with x
As you can see, it won't actually go to the next x, but will go to the next y instead.
what i found to solve this simply was to run through the array twice instead:
for x in array:
for y in dont_use_these_values:
if x.value==y:
array.remove(x) # fixed, was array.pop(x) in my original answer
continue
for x in array:
do some other stuff with x
I know this was a specific case of OP's question, but I am posting it in the hope that it will help someone think about their problem differently while keeping things simple.
break instead of continue would do what you want, wouldn't it? :-) –
Pontormo probably little trick like below will do if not prefer to refactorial into function
added 1 break_level variable to control the while loop condition
break_level = 0
# while break_level < 3: # if we have another level of nested loop here
while break_level < 2:
#snip: print out current state
while break_level < 1:
ok = get_input("Is this ok? (y/n)")
if ok == "y" or ok == "Y": break_level = 2 # break 2 level
if ok == "n" or ok == "N": break_level = 1 # break 1 level
You can define a variable( for example break_statement ), then change it to a different value when two-break condition occurs and use it in if statement to break from second loop also.
while True:
break_statement=0
while True:
ok = raw_input("Is this ok? (y/n)")
if ok == "n" or ok == "N":
break
if ok == "y" or ok == "Y":
break_statement=1
break
if break_statement==1:
break
Hopefully this helps:
x = True
y = True
while x == True:
while y == True:
ok = get_input("Is this ok? (y/n)")
if ok == "y" or ok == "Y":
x,y = False,False #breaks from both loops
if ok == "n" or ok == "N":
break #breaks from just one
I came across this recently and, wanting to avoid a duplicate return statement, which can conceal logical errors, looked at @yak's idea. This works well within nested for loops but is not very elegant. An alternative is to check for the condition before the next loop:
b = None
for a in range(10):
if something(a, b): # should never = True if b is None
break
for b in range(20):
pass
This might not work everywhere but is adaptable and, if required, has the advantage of allowing the condition to be duplicated rather than a potential result.
Trying to minimal changes to the OP's question, I just added a flag before breaking the 1st for loop and check that flag on the outer loop to see if we need to brake once again.
break_2 = False
while True:
# Snip: print out current state
if break_2: break
while True:
ok = get_input("Is this ok? (y/n)")
if ok.lower() == "y": break_2 = True
if break_2: break
if ok.lower() == "n": break
# Do more processing with menus and stuff
while True:
# Snip: print out current state
while True:
ok = get_input("Is this ok? (y/n)")
if ok.lower() == "y":
break_2 = True
if ok.lower() == "n":
break
if break_2:
break
If you just need to test an edge case inside a complex nest of for loops, you can throw in a 1/0 to raise an exception. I promise I won't tell anyone. This comes in handy when you quickly want to test a single iteration of a deeply nested for loop, and you don't want to track down a large amount of break statements or comment out a significant amount of code.
Yes, you could wrap it in a function and use return, but in some contexts that can be infeasibly cumbersome.
example for entry level programmers:
for i in first_iter:
for j in second_iter:
for k in third_iter:
print(i_want_to_run_this_once_and_stop_executing(i,j,k))
1/0
code_that_takes_a_long_time()
expensive_code()
in big jupyter notebook scripts that do some heavy data pre-processing, this is especially handy.
FLAG You can use a flag to break out of the loop:
if found:
break
Here, 'found' is the flag, you initially set it to False, then in the loop you use this code.
found = False
for table_height in range(500):
if found:
break
Here is the full code with three for loops:
found = False
for table_height in range(500):
if found:
break
for cat_height in range(500):
if found:
break
for tort_height in range(500):
equation1 = table_height + cat_height == tort_height + 170
equation2 = table_height + tort_height == cat_height + 130
if equation1 and equation2:
print('table', table_height, ' cat', cat_height, ' tortoise', tort_height)
found = True
break
In this code if equation1 and equation2 are True, it will set the 'found' flag True, and break out of the innermost for loop, and it will also break out of the other two for loops since 'found' is True.
Here is a very short version: Create a file with the name break_out_nested.py
import itertools
import sys
it = sys.modules[__name__] # this allows us to share variables with break_out_nested.py when we import it
def bol(*args):
condi = args[-1] # the condition function
i = args[:-1] # all iterables
for p in itertools.product(*i): # itertools.product creates the nested loop
if condi(): # if the condition is True, we return
return
yield p # if not, we yield the result
Now you need only a few lines to break out of nested loops (data from Rafiq's example)
from break_out_nested import it, bol # import what we have just created
# you need to create new variables as attributes of it,
# because break_out_nested has only access to these variables
it.i, it.j, it.k = 1, 1, 1
# the break condition
def cond(): return it.i % 3 == 0 and it.j % 3 == 0 and it.k % 3 == 0
# The condition will be checked in each loop
for it.i, it.j, it.k in bol(range(1, 6, 1), range(1, 11, 2, ), range(1, 21, 4), cond):
print(it.i, it.j, it.k)
More examples:
def cond(): return it.i + it.j + it.k == 777
it.i, it.j, it.k = 0, 0, 0
for it.i, it.j, it.k in bol(range(100), range(1000), range(10000), cond):
print(it.i, it.j, it.k)
def cond(): return it.i + it.j + it.k >= 100000
it.i, it.j, it.k = 0, 0, 0
# you dont have to use it.i, it.j, it.k as the loop variables, you can
# use anything you want, but you have to update the variables somewhere
for i, j, k in bol(range(100), range(1000), range(10000), cond):
it.i, it.j, it.k = i * 10, j * 100, k * 100
print(it.i, it.j, it.k)
This is my elaboration on the answer by Mark Dickinson:
from contextlib import contextmanager
@contextmanager
def nested_break_on(exception_cls):
try:
yield
except exception_cls:
pass
@contextmanager
def nested_break():
class NestedBreakException(Exception):
__slots__ = ()
def break_out():
raise NestedBreakException
try:
yield break_out
except NestedBreakException:
pass
# Demo of `nested_break_on` CM
class NestedBreakException(Exception):
__slots__ = ()
def break_out():
raise NestedBreakException()
with nested_break_on(NestedBreakException):
while True:
while True:
print("Got in.")
break_out()
print("Broke out.")
# Demo of `nested_break` CM
with nested_break() as break_out:
while True:
while True:
print("Got in.")
break_out()
print("Broke out.")
break for outer and inner while loops:
while True:
while True:
print('Breaks inner "while" loop')
break # Here
print('Breaks outer "while" loop')
break # Here
Or, break for outer and inner while loops with if statement:
while True:
while True:
if True:
print('Breaks inner "while" loop')
break # Here
print('Breaks outer "while" loop')
break # Here
Output:
Breaks inner "while" loop
Breaks outer "while" loop
break for outer and inner for loops:
for _ in iter(int, 1):
for _ in iter(int, 1):
print('Breaks inner "for" loop')
break # Here
print('Breaks outer "for" loop')
break # Here
Or, break for outer and inner for loops with if statement:
for _ in iter(int, 1):
for _ in iter(int, 1):
if True:
print('Breaks inner "for" loop')
break # Here
print('Breaks outer "for" loop')
break # Here
Output:
Breaks inner "for" loop
Breaks outer "for" loop
Similar like the one before, but more compact. (Booleans are just numbers)
breaker = False #our mighty loop exiter!
while True:
while True:
ok = get_input("Is this ok? (y/n)")
breaker+= (ok.lower() == "y")
break
if breaker: # the interesting part!
break # <--- !
Since this question has become a standard question for breaking into a particular loop, I would like to give my answer with example using Exception.
Although there exists no label named breaking of loop in multipally looped construct, we can make use of User-defined Exceptions to break into a particular loop of our choice. Consider the following example where let us print all numbers upto 4 digits in base-6 numbering system:
class BreakLoop(Exception):
def __init__(self, counter):
Exception.__init__(self, 'Exception 1')
self.counter = counter
for counter1 in range(6): # Make it 1000
try:
thousand = counter1 * 1000
for counter2 in range(6): # Make it 100
try:
hundred = counter2 * 100
for counter3 in range(6): # Make it 10
try:
ten = counter3 * 10
for counter4 in range(6):
try:
unit = counter4
value = thousand + hundred + ten + unit
if unit == 4 :
raise BreakLoop(4) # Don't break from loop
if ten == 30:
raise BreakLoop(3) # Break into loop 3
if hundred == 500:
raise BreakLoop(2) # Break into loop 2
if thousand == 2000:
raise BreakLoop(1) # Break into loop 1
print('{:04d}'.format(value))
except BreakLoop as bl:
if bl.counter != 4:
raise bl
except BreakLoop as bl:
if bl.counter != 3:
raise bl
except BreakLoop as bl:
if bl.counter != 2:
raise bl
except BreakLoop as bl:
pass
When we print the output, we will never get any value whose unit place is with 4. In that case, we don't break from any loop as BreakLoop(4) is raised and caught in same loop. Similarly, whenever ten place is having 3, we break into third loop using BreakLoop(3). Whenever hundred place is having 5, we break into second loop using BreakLoop(2) and whenver the thousand place is having 2, we break into first loop using BreakLoop(1).
In short, raise your Exception (in-built or user defined) in the inner loops, and catch it in the loop from where you want to resume your control to. If you want to break from all loops, catch the Exception outside all the loops. (I have not shown this case in example).
The way I solve this is by defining a variable that is referenced to determine if you break to the next level or not. In this example, this variable is called 'shouldbreak'.
Variable_That_Counts_To_Three=1
while 1==1:
shouldbreak='no'
Variable_That_Counts_To_Five=0
while 2==2:
Variable_That_Counts_To_Five+=1
print(Variable_That_Counts_To_Five)
if Variable_That_Counts_To_Five == 5:
if Variable_That_Counts_To_Three == 3:
shouldbreak='yes'
break
print('Three Counter = ' + str(Variable_That_Counts_To_Three))
Variable_That_Counts_To_Three+=1
if shouldbreak == 'yes':
break
print('''
This breaks out of two loops!''')
This gives a lot of control over how exactly you want the program to break, allowing you to choose when you want to break and how many levels to go down.
True or False would at least turn this from horrible to merely unattractive. –
Bevan © 2022 - 2024 — McMap. All rights reserved.
tryexceptblock and escape by raising an exception. It's the most readable approach IMO – Windflower