[Solved] SyntaxError: ‘return’ outside function in Python

🚫 SyntaxError: ‘return’ outside function

This post was originally published on decodingweb.dev as part of the Python Syntax Errors series.

Python raises the error “SyntaxError: ‘return’ outside function” once it encounters a return statement outside a function.

Here’s what the error looks like:

File /dwd/sandbox/test.py, line 4
  return True
  ^^^^^^^^^^^
SyntaxError: 'return' outside function

Based on Python's syntax & semantics a return statement may only be used in a function - to return a value to the caller.

However, if - for some reason - a return statement isn't nested in a function, Python's interpreter raises the "SyntaxError: 'return' outside function" error.

Using the return statement outside a function isn't something you'd do on purpose, though; This error usually happens when the indentation-level of a return statement isn't consistent with the rest of the function.

Additionally, it can occur when you accidentally use a return statement to break out of a loop (rather than using the break statement)

How to fix SyntaxError: 'return' outside function

This SyntaxError happens under various scenarios:

1. Inconsistent indentation
2. Using the return statement to break out of a loop
3. Let's explore each scenario with some examples.

1. Inconsistent indentation: A common cause of this syntax error is an inconsistent indentation, meaning Python doesn't consider the return statement a part of a function because its indentation level is different.

In the following example, we have a function that accepts a number and checks if it's an even number:

def isEven(value):
  remainder = value % 2

  # if the remainder of the division is zero, it's even
return remainder == 0

As you probably noticed, we hadn't indented the return statement relative to the isEven() function.

To fix it, we correct the indentation like so:

# ✅ Correct
def isEven(value):
  remainder = value % 2

  # if the remainder of the division is zero, it's even
  return remainder == 0

Problem solved!

Let's see another example:

def check_age(age):
  print('checking the rating...')

# if the user is under 12, don't play the movie
if (age < 12):
  print('The movie can\'t be played!')
  return

In the above code, the if block has the same indentation level as the top-level code. As a result, the return statement is considered outside the function.

To fix the error, we bring the whole if block to the same indentation level as the function.

# ✅ Correct
def check_age(age):
  print('checking the rating...')

  # if the user is under 12, don't play the movie
  if (age < 12):
    print('The movie can\'t be played!')
    return

  print('Playing the movie')

check_age(25)
# output: Playing the movie

Using the return statement to break out of a loop: Another reason for this error is using a return statement to stop a for loop located in the top-level code.

The following code is supposed to print the first fifteen items of a range object:

items = range(1, 100)

# print the first 15 items
for i in items:
  if i > 15:
    return
  print(i)

However, based on Python's semantics, the return statement isn't used to break out of functions - You should use the break statement instead:

# ✅ Correct
items = range(1, 100)

# print the first 15 items
for i in items:
  if i > 15:
    break
  print(i)

And that's how you'd fix the error "SyntaxError: 'return' outside function" in Python.

In conclusion, always make sure the return statement is indented relative to its surrounding function. Or if you're using it to break out of a loop, replace it with a break statement.

Alright, I think it does it. I hope this quick guide helped you solve your problem.

Thanks for reading.