How to fix “SyntaxError: ‘break’ outside loop” in Python
Python raises “SyntaxError: ‘break’ outside loop” whenever it encounters a break
statement outside a loop. The most common cases are using break
within an if
block (that’s not part of a loop) or when you accidentally use it instead of return
to return from a function.
Here’s what the error looks like:
File /dwd/sandbox/test.py, line 2
break
^^^^^
SyntaxError: 'break' outside loop
The break
statement is a control flow feature used to break out of the innermost loop. For instance, when you reach a specific value. That's pretty much like the C language.
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Based on Python syntax, the break
keyword is only valid inside loops - for and while.
Here's an example:
values = [7, 8, 9.5, 12]
for i in values:
# Break out of the loop if i > 10
if (i > 10):
break
print(i)
The above code iterates over a list and prints out the values less than 10. Once it reaches a value greater than 10, it breaks out of the loop.
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How to fix SyntaxError: 'break' outside loop
This syntax error occurs under two scenarios:
- When using break inside an if block that's not part of a loop
- When using break (instead of return) to return from a function
Let's see some examples with their solutions.
When using break inside an if block that's not part of a loop:
if item > 100
break # π« SyntaxError: 'break' outside loop
# some code here
There's no point in breaking out of an if
block. If the condition isn't met, the code isn't executed anyway. The above code only would make sense if it's inside a loop:
values = [7, 8, 9.5, 12]
for item in values:
if (item > 10):
break
print(i)
Otherwise, it'll be useless while being a SyntaxError too! However, if you want to keep the if
block for syntactical reasons, you can replace the break
keyword with the pass
keyword. A pass statement does nothing in Python. However, you can always use it when a statement is required syntactically, but no action is needed.
When using break (instead of return) to return from a function: Another reason behind this error is to accidentally use the break
keyword (instead of return
) to return from a function:
def checkAge(age):
if (age < 12):
break # π« SyntaxError: 'break' outside loop
# some code here
To return from a function, you should always use return
(with or without a value):
def checkAge(age):
if (age <= 12):
return
# some code here
Alright, I think it does it. I hope this quick guide helped you solve your problem.
Thanks for reading.
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