Python: 'break' outside loop

Question

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?

Asked By: Open the way

||

Source

Answer 1

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?

Answered By: Mark Byers

Answer 2

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.

Answered By: user180247

Answer 3

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

Answered By: enricog

Answer 4

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.

Answered By: Mike Graham

Answer 5

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

while 1:
    if blah:
        break

Answered By: Joshua Stafford

Python: 'break' outside loop

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