tags:

views:

260

answers:

4

Hi,

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?

Thanks

+9  A: 

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?

Mark Byers
and how could I do so the scripts stops here?
Werner
`sys.exit()` terminates the program immediately.
Mark Byers
`raise SystemExit` terminates the program; `sys.exit()` raises it, if you prefer a function call.
ΤΖΩΤΖΙΟΥ
+2  A: 

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.

Steve314
+2  A: 

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

TheCandyMan666
A: 

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 finished the if block rather than breaking the loop. Consider that I cannot think of a case that you'd use break any way but inside an if statement to break the loop the if statement is in.

Mike Graham