What will be returned by this code?

Consider:

>>> squares = []
>>> for x in range(5):
...     squares.append(lambda: x**2)
>>> squares[2]()
>>> squares[4]()

It will return:

>>> squares[2]()
16
>>> squares[4]()
16

This happens because x 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 x is 4, so all the functions now return 4**2, i.e. 16.