Python: 'break' outside loop
python
Solution
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?
Problem
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?