Finally always runs just before the return in try block, then why update in finally block not affect value of variable returned by try block?

Question

Finally block runs just before the return statement in the try block, as shown in the below example - returns False instead of True:

>>> def bool_return(): ...  try: ...    return True ...  finally: ...    return False ... >>> bool_return() False 

Similarly, the following code returns value set in the Finally block:

>>> def num_return(): ...  try: ...    x=100 ...    return x ...  finally: ...    x=90 ...    return x ... >>> num_return() 90 

However, for variable assignment without return statement in the finally block, why does value of variable updated by the finally block not get returned by the try block? Is the variable from finally block scoped locally in the finally block? Or is the return value from the try block held in memory buffer and unaffected by assignment in finally block? In the below example, why is the output 100 instead of 90?

>>> def num_return(): ...  try: ...    x=100 ...    return x ...  finally: ...    x=90 ... >>> num_return() 100 

Similarly the following example:

In [1]: def num_return():    ...:   try:    ...:     x=[100]    ...:     return x    ...:   finally:    ...:     x[0] = 90    ...:  In [2]: num_return() Out[2]: [90]  In [3]: def num_return():    ...:   try:    ...:     x=[100]    ...:     return x[0]    ...:   finally:    ...:     x[0] = 90    ...:  In [4]: num_return() Out[4]: 100 

Answer 1

A little experiment to help confirm what others have answered, is to replace x with a single-value list, like this:

def num_return():   try:     x=[100]     return x   finally:     x[0] = 90 

now the value returned is [90], so the list is indeed modified in the finally block.

BUT if you return x[0], you get 100 (even though we just based the fact that the list itself does change in the finally block).

 In [1]: def num_return():    ...:   try:    ...:     x=[100]    ...:     return x    ...:   finally:    ...:     x[0] = 90    ...:  In [2]: num_return() Out[2]: [90]  In [3]: def num_return():    ...:   try:    ...:     x=[100]    ...:     return x[0]    ...:   finally:    ...:     x[0] = 90    ...:  In [4]: num_return() Out[4]: 100 

Answer 2

When you say return x, Python saves the value of the variable x at that point as the return value. Updating x later doesn't affect this.