Conversion of denary number to binary number problem (Python)-CodePudding

Im stuck on a problem where I have to write a function that converts a denary number into a binary number using the repeated division by two algorithm. Steps Include:

The number to be converted is divided by two.
The remainder from the division is the next binary digit. Digits are added to the front of the sequence.
The result is truncated so that the input to the next division by two is always an integer.
The algorithm continues until the result is 0.

Please click the link below to see what the output should be like: https://i.stack.imgur.com/pifUO.png

def dentobi(user):
  denary = user
  divide = user / 2
  remainder = user % 2
  binary = remainder
  
  if user != 0:
    print("Denary:", denary)
    print("Divide by 2:", divide)
    print("Remainder:", remainder)
    print("Binary:", binary)
  
user = int(input("Please enter a number: "))
dentobi(user)

This is what I have done so far but Im not getting anywhere.

Can someone explain how I would do this?

CodePudding user response：

One way, using divmod to divide by 2 and get the remainder in one step:

def binary(num):
    b = ""
    while num:
        num, digit = divmod(num, 2)
        b = f"{digit}{b}"
    return b

binary(26)
'11010'

This assumes a positive number but can easily be extended to work for 0 and negatives.

CodePudding user response：

The Answer provided by @user2390182 is functionally correct except that it returns an empty string when num is zero. However, I have noted on several occasions that divmod() is rather slow. This function is more complex but is faster:

def binary(n):
    r = []
    while n > 0:
        r.append('1' if n & 1 else '0')
        n //= 2
    return '0' if len(r) == 0 else ''.join(r[::-1])