Does 'raise' executes after all the statements? What is the work flow when we use raise?

Question

try:
print(1/0)
except:
print('error')
raise TypeError
finally:
print('last line')
Output:
error
last line
ZeroDivisionError
TypeError
Why did the ZeroDivisionError has occurred here even we have print statement instead of ZeroDivisionError?
Other e.g:
try:
print(4/0)
except:
raise TypeError
print('error')
finally:
print('last line')
Output:
Last line
ZeroDivisionError
TypeError
Why didn't the print statement got executed here?

Answer 1

If you raise an error (by force or unexpected), program execution stops - unless you catch it with an except. Therefore it needs to be in a try block.

If an error gets raised in an except, there is nothing to catch it.

Why is the 'old' exception printed out? When in one error situation other errors occur, they get chained together for complete error information.

In a complex situation that can help you understand the bug.

Answer 2

Okay, in more detail.

try:
print(4/0)

ZeroDivisionError is raised.

except:

This except stops the Error - for now.

raise TypeError

But now you manually raise an Error of your own. Since this error is not in a try block, program stops.

print('error')

So this line is not executed.

You get a complete report of both errors.

finally:
print('last line')

This though is executed, since finally has this job, to execute the code in there no matter what.

Answer 3

Cool. By seeing the output we can understand that finally's statement is executed first. Is that is the work flow? HonFu

Answer 4

The finally block is executed the moment an exception can't be caught anymore. Then the program terminates and you get your full report.

Answer 5

https://code.sololearn.com/cE1IlCogyxRk/?ref=app HonFu please run this code you can understand