🔒 Dealing with iterations of a while loop (Very simple ,IT WILL TAKE YOU A MINUTE)

[valid tapir]

x = 23
while x != 0:
g = x % 2
x = x //2

basically this small program converts a number to binary right . I want to store the value printed out by g every new iterations value of g to some variable . when I try to do some like this : o = g + 1 .it adds 1 to every iteration of the value of one . please and let me know if I need to clarify

[civic notch]

@valid tapir

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[bronze marlin]

You want to represent binary as a string.

[valid tapir]

not exactly I want to just add the values of binary together basically sum them within a while loop

[bronze marlin]

Otherwise you're just copying the number in a roundabout way

[valid tapir]

yeah it just keep adding the new iteration and not storing the value and constantly updating it

I hope im making sense

[bronze marlin]

Try adding each digit to a list instead.

g.append(x % 2)

or g += x % 2 to skip the sum logic later

!e ```py
x = 23
print(bin(x))
g = 0
while x != 0:
g += x % 2
x = x //2
print(g)

[civic notch]

:white_check_mark: Your 3.13 eval job has completed with return code 0.

001 | 0b10111
002 | 4

[valid tapir]

okay let me try that

it worked thank you so much

[scarlet flare]

FYI :- @valid tapir if this is not an assignment or a task by university, there is a python built-in function which does exactly this.

[bronze marlin]

f"{x:b}".count("1")

[runic agate]

!d int.bit_count

[civic notch]

int.bit\_count

int.bit_count()```
Return the number of ones in the binary representation of the absolute value of the integer. This is also known as the population count. Example:

```py
>>> n = 19
>>> bin(n)
'0b10011'
>>> n.bit_count()
3
>>> (-n).bit_count()
3
```...

[bronze marlin]

Why does this exist?

[marsh gyro]

popcount is a common-ish operation, so it makes sense that it's exposed.

[valid tapir]

this was a university thing but im just redoing it for practice

[civic notch]

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