Why the output of this code not include print(2)

Question

try:
... print(1)
... print(20/0)
... print(2)
... except ZeroDivisionError:
... print(3)
... finally:
... print(4)

1
3
4

Answer 1

print(20/0)
print(2)

You are using 'print(20/0)' above 'print(2)'. So when 20 is devided by 0, 'except' catches it and hence there is no scope left for 2 to be printed. So, it prints whatever inside the ZeroDivisionError or it just prints 3. Then finally it prints 4!

So the output you get is :

1
3
4

However, if you write it like this :

print(2)
print(20/0)

Then you'll get this :))

1
2
3
4

https://code.sololearn.com/cb24nsAbAhUm/?ref=app

Answer 2

Execution of statements stopped after there was error for 20/0

Answer 3

Because you can't divide by zero.

So at that point, execution jumps into the except block.

Answer 4

Because after line 3 (where a error occur) I program stop and go to see exception which is ZeroDivisionError of line 3
And print code under it which print 3
And code under finally execute at end of program always which print 4