Python: 'Break' Outside Loop
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...
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?
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?
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.
Because break can only be used inside a loop. It is used to break out of a loop (stop the loop).
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.
This is an old question, but if you wanted to break out of an if statement, you could do:
while 1:
if blah:
break
James Sterling reports on renewable energy developments, climate policy, ecological conservation, and green tech innovations around the globe.
