Break and continue

Question

Why this code counts 3 and then prints "continue"
Shouldn't the output be like this?
1
2
Skipping
4
5
Python for begginers(26.1lesson)
https://code.sololearn.com/cPhq9Gd845R3/?ref=app

Answer 1

Mohammad Hasan

Do this

i = 0
while i < 5:
i += 1

if i == 3:
print("Skipping 3")
continue
print(i)

https://code.sololearn.com/cYxjZn39Wx2O/?ref=app

Answer 2

You are first printing the value of i
1st iteration

Program print 1 than update the value . Value of i(2) is not equals to 3 so again checks the while condition and prints 2 then update the value now value of i is 3 if condition becomes true and skipping 3 is printed then continue(which has no use here)
Then the while condition is checked 3<=5 which is true so 3 get printed .

If you want the desired output -
i = 0
while i<=4:
i = i+1
if i == 3:
print("Skipping 3")
continue
print(i)

Answer 3

This is because when i is equal to 3, "skipping 3" is printed and 3 is also printed. You haven't coded something that will not display 3 when i=3. I mean code something inside the if statement that will not display 3 and instead of that "skipping 3" will be printed.

I didn't know much about python, that why I am unable to help!

Answer 4

A͢J - S͟o͟l͟o͟H͟e͟l͟p͟e͟r͟
Thanks but what's the difference?
Yours worked but sololearn's didn't:/

Answer 5

Mohammad Hasan

That's not a problem of Sololearn compiler, that's problem in your code.

You are 1st printing value then incrementing so after print 2 i will be 3 and now according to condition printing "Skipping 3" but i still hold 3 so 3 will be print then increment by 1

Answer 6

Mohammad Hasan

I have already shared code in my first reply, you can check that.

Answer 7

A͢J - S͟o͟l͟o͟H͟e͟l͟p͟e͟r͟
No I didn't wrote this code it's in lessons
What do I need to do to fix that problem?