I want to print odd numbers within a range using Python with the code below, but I don't understand what is wrong with it and it only prints 1 and stops.
a=1
while a<11:
if a%2==0:
continue
print(a)
a+=1
-
5Follow what your code does step by step: pythontutor.comdeceze– deceze ♦2022-03-08 12:56:40 +00:00Commented Mar 8, 2022 at 12:56
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4 Answers
The problem is that you don't increment a when it's not odd in the if clause.
a=1
while a<11:
if a%2==0:
a+=1
continue
print(a)
a+=1
You must increment a also when it's even, else it will be stuck in an infinite loop at a=2.
A better variation would be to avoid continue and do as follows:
a=1
while a<11:
if a%2!=0:
print(a)
a+=1
answered Mar 8, 2022 at 12:58
Muhammad Mohsin Khan
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5 Comments
Josip Juros
Using continue as a whole is a bad practice IMO and should be avoided, which here it can be easily avoided.
Muhammad Mohsin Khan
Agreed. I was explaining to them the problem in their code. But I have taken your suggestion to improve my answer.
Josip Juros
Providing both options is 100% best here. Good answer.
leoOrion
While in this case
continue is not needed, saying using continue is a bad practice is not true at all. There is nothing wrong with using continue.Josip Juros
@leoOrion Agree to disagree.
x = 1
while x < 15:
if x % 2 != 0:
print(x)
x = x + 1
This is how you do it. Main issue with your code is that continue skips everything, including the increment. Meaning your code causes an infinite loop. Continue skips all other lines in that iteration of a loop and goes to a new one, meaning a never increments meaning a is always less than 11.
Comments
Her you go mate:
a=1
while a<11:
if a%2!=0:
print(a)
a+=1
Comments
you need to add a+=1 before you continue, just know that you can write something more simple then that, and more pretty ;)
a=1
while a<11:
if a%2==0:
a+=1 ///add this
continue
print(a)
a+=1
