Cell variable defined in loop PYL-W0640
Bug risk
Major
3 years ago
—
3 years old
Cell variable listener defined in loop
326 interesting = True
327
328 def anonymous():
329 result = listener.on_request(mid, pid, sender, command) 330 if self.__connection is not None:
331 self.__connection.send_reply(key, result)
332 self.logger.warning(Cell variable listener defined in loop
297 interesting = True
298
299 def anonymous():
300 listener.on_message(uri, headers, decoded_payloads) 301
302 self.__worker.submit(anonymous)
303 dispatched = True
Description
A variable used in a closure is defined in a loop. This will result in all closures using the same value for the closed-over variable, which can pave way for a hideous bug.
Consider the following code snippet:
# Motivation: Create 3 functions that return `x**2`, `x**3`, and `x**4`.
# Note: This will be using the loop variable to show the bug.
powers = [lambda x: x**i for i in range(2,5)]
So, powers is supposed to contain 3 functions to return 2nd, 3rd and 4th powers of a given number.
On execution, these are the results:
In [1]: powers = [lambda x: x**i for i in range(2,5)]
In [2]: powers[0](2) # Expected result: 4 (2**2)
Out[2]: 16
In [3]: powers[1](2) # Expected result: 8 (2**3)
Out[3]: 16
In [4]: powers[2](2) # Expected result: 16 (2**4)
Out[4]: 16
This happens because i is not local to the lambdas, but is defined in the outer scope, and it is accessed when the lambda is called — not when it is defined. At the end of the loop, the value of i is 4, so all the functions now return x**4.
Recommended
In order to avoid this, you need to save the values in variables local to the lambdas, so that they don’t rely on the value of the global i:
In [1]: powers = [lambda x, n =i: x**n for i in range(2,5)]
In [2]: powers[0](2)
Out[2]: 4
In [3]: powers[1](2)
Out[3]: 8
In [4]: powers[2](2)
Out[4]: 16