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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?
605
The Python "SyntaxError: 'break' outside loop" occurs when we use the break statement outside of a loop. To solve the error, use a return statement to return a value from a function, or use the sys. exit() method to exit the interpreter.
Sep 25, 2020. A break statement instructs Python to exit a loop. If you use a break statement outside of a loop, for instance, in an if statement without a parent loop, you'll encounter the “SyntaxError: 'break' outside loop” error in your code.
The break statement in Python is used to get out of the current loop. We can't use break statement outside the loop, it will throw an error as “SyntaxError: 'break' outside loop“. We can use break statement with for loop and while loops. If the break statement is present in a nested loop, it terminates the inner loop.
Using break outside a loop or switch will result in a compiler error break cannot be used outside of a loop or a switch. As its name suggests, break statement terminates loop execution. It breaks the loop as soon as it is encountered.
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?
131
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.
43
Because break can only be used inside a loop. It is used to break out of a loop (stop the loop).
5
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