Recursion Python Challenge, please help

Question

Hello guys, following:
The given code defines a recursive function convert(), which needs to convert its argument from decimal to binary.
However, the code has an error.
Fix the code by adding the base case for the recursion, then take a number from user input and call the convert() function, to output the result.
Sample Input:
8
Sample Output:
1000
This is my code so far:
def convert(num):
if num<=1:
return 0
else:
return (num % 2 + 10 * convert(num // 2))
bin=int(input())
convert(bin)
i´ve got a few questions, maybe you can help me solve them:
-Is my Base case correct? I thought of choosing it <1 because otherwise it would keep up the division nearly infinite. But i didnt know what to return after it reaches <1 so i thought about zero what obviously isnt correct.
-Do i have an input problem? i neither get any output.
-Or did i do something wrong with the function.
Would be cool if someone could help me out, really struggling with this one and couldnt find any help in the web.
Thanks and have a good start in the week!

Answer 1

Currently you don't have any output and don't save the conversion result. Replace last lines with
dec = int(input())
bin = convert(dec)
print(bin)
Input is decimal, so I took the freedom to change the variable name

Answer 2

def convert(num):
if num == 0 or num < 2:
return 0
else:
return (num % 2 + 10 * convert(num // 2))
print(convert(int(input()))

Answer 3

Why do you think returning 0 when num<1 isn't correct? Also it doesn't run nearly infinite, because recursion is done with num // 2, which is integer division:
7//2 = 6//2 = 3
5//2 = 4//2 = 2
3//2 = 2//2 = 1
1//2 = 0
So base case can either be
if num==0: return 0
Or
if num<2: return num

Answer 4

Benjamin Jürgens , right, but anyways do i have an output problem, do you have an idea whats wrong?

Answer 5

def convert(num):
if num==0:
return 0
elif num<2:
return num
else:
return (num % 2 + 10 * convert(num // 2))
print(convert(int(input())))

Answer 6

Benjamin Jürgens ahh thank you a lot. that was the problem. case closed, einen schönen tag dir und danke für die hilfe! :)

Answer 7

Lucas Kutman 👍
Dir auch einen schönen Tag!

Answer 8

def convert(num):
if num == 1:
return 1
return (num % 2 + 10 * convert(num // 2))
num = int(input())
print(convert(num))

Answer 9

def convert(num):
if num<=1:
return 1
else:
return (num % 2 + 10 * convert(num // 2))
bin=int(input())
print(convert(bin))