Python: 'break' outside loop

47
votes

in the following python code:

narg=len(sys.argv)
print "@length arg= ", narg
if narg == 1:
        print "@Usage: input_filename nelements nintervals"
        break

I get:

SyntaxError: 'break' outside loop

Why?

5
python
What do you think break does here? What do you expect to happen? - S.Lott
The optparse module (or similar) will make parsing argv much simpler. docs.python.org/library/optparse.html - Roger Pate

5 Answers

82
votes

Because break cannot be used to break out of an if - it can only break out of loops. That's the way Python (and most other languages) are specified to behave.

What are you trying to do? Perhaps you should use sys.exit() or return instead?

5
votes

break breaks out of a loop, not an if statement, as others have pointed out. The motivation for this isn't too hard to see; think about code like

for item in some_iterable:
    ...
    if break_condition():
        break

The break would be pretty useless if it terminated the if block rather than terminated the loop -- terminating a loop conditionally is the exact thing break is used for.

4
votes

Because break can only be used inside a loop. It is used to break out of a loop (stop the loop).

3
votes

Because the break statement is intended to break out of loops. You don't need to break out of an if statement - it just ends at the end.

1
votes

This is an old question, but if you wanted to break out of an if statement, you could do:

while 1:
    if blah:
        break