如何在使用朴素递归将十进制数字转换为二进制数字时捕获第一位数字？

Question

I am trying to convert a decimal number to a binary digit in the below way using recursion.

def getbin(x: int, s: str):
    if int(x/2) > 0:
        y = int(x/2)
        s += str(y % 2)
        print(f'int(x/2): {int(x/2)}, y: {y}, st: {s}')
        getbin(y, s)
    elif int(x/2) == 1:
        s += '11'
        return s


if __name__ == '__main__':
    print(getbin(28, ''))

But when I call this, I can see in the output that the first digit of the binary number is not getting captured. I ran two test cases:

For the number 28, the expected output should be 00111 but the output is 0111 :

For the number 5, the output should be 101 but the output is 01

Could anyone let me know what is the mistake I am making here and how can I correct it?

Answer 1

Your problem is that you are testing against x/2 instead of testing against x . Thus you lose the most significant bit of the result. Try something like this:

def getbin(x: int, s: str):
    s += str(x % 2)
    y = x // 2
    if y > 0:
        return getbin(y, s)
    return s

Note also that you need to reverse the result of getbin to get the correct binary string.