how can I use a function to exit for loop

Question

I want to write a function to exit a for loop. Typically, we can

for i in range(10):
    if i==8:
       break

I hope to write a function like,

def interrupt(i):
   if i == val:
      ...


for i in range(10):
   interrupt(i)

Edit

I understand the mechanism and why I can't do this now. Just out of curiosity, does python offer something like 'macro expansion', so that I don't actually create a function, but a name to repalce the code block?

You can have interrupt throw an exception of some kind and catch it in your for loop, breaking, or doing whatever you please after excepting it. — Brian
– Brian, Commented Sep 9, 2019 at 20:59
@CeliusStingher, break and continue do different things. — Brian
– Brian, Commented Sep 9, 2019 at 20:59
@CeliusStingher And neither one can cross function-call boundaries; they only apply to the innermost loop they appear in lexically. — chepner
– chepner, Commented Sep 9, 2019 at 21:00
break exits the loop; continue skips the rest of the body and gets the next value from the iterator. — chepner
– chepner, Commented Sep 9, 2019 at 21:01
No, Python does not offer any kind of macro expansion. Such constructs usually cause more problems than they solve. — chepner
– chepner, Commented Sep 9, 2019 at 21:09

chepner · Accepted Answer · 2019-09-09 20:59:31Z

3

You can't. The for loop continues until an explicit break statement is encountered directly in the loop, or until the iterator raises StopIteration.

The best you can do is examine the return value of interrupt or catch an exception raised by interrupt, and use that information to use break or not.

answered Sep 9, 2019 at 20:59

chepner

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Christian Sloper · Accepted Answer · 2019-09-09 21:03:36Z

3

I don't think it is possible to throw a break, you can fake it a bit like this for example:

def interrupt(i):
    if i > 2:
        return True
    return False

for i in range(10):
    if interrupt(i):
        break

    #.. do loop stuff

answered Sep 9, 2019 at 21:03

Christian Sloper

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Massifox · Accepted Answer · 2019-09-09 21:31:38Z

2

The following code will solve your problem:

def interrupt(i, pred=lambda x:x==8):
    if pred(i):
        raise StopIteration(i)

try:
    for i in range(10):
        interrupt(i)
except StopIteration as e:
    print('Exit for loop with i=%s' % e)

# output: Exit for loop with i=8

The scheme is simple, and works in general with any kind of exception:

Write a function foo() that raises one or more exceptions when one or more events occur
Insert the call to the function foo() in the try-except construct
Manage exceptions in except blocks

edited Sep 9, 2019 at 21:31

answered Sep 9, 2019 at 21:17

Massifox

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