I need help converting binary to decimal in my code without using bin

Question

I have to make a menu in my coding class. The user is able to choose a number on the menu and the menu is able to close. I have everything down except for binary to decimal. My code so far is shown below.

base10 is what I am working on. I just don't know the code to get to it since I had to miss class and he went over it in class.

#Name: Menu
#Input: 
#Output:
#Description:
def menu():
  while(True):
    print("Welcome The Menu")
    print("\t 1. Convert Binary to Decimal")
    print("\t 2. Convert Decimal to Binary")
    opt=input("Input a number or 9 to Quit. ")
    if(int(opt)==1):
      base10()
    elif(int(opt)==2):
      base2()   
    elif(int(opt)=="9"):
      break;
  else:
    print("Invalid Input")

#Name: Reverse
#Input: String s
#Output: String r
#Description: This function takes a string as input and reverses that string
#returning it as output. 

def reverse():
  r=""
  for i in s:
    r=i + r
  return r

#Name: Base 10
#Input:
#Output:
#Description

def base10():
  num=int(input("Enter a binary number:"))

  sum=0
  for i in range(0,len()):
      sum+=(int([i])*(2**i))
  print(sum)


#Name: Base 2
#Input: num
#Output: b
#Description: It takes the number from the user and divides it by two. B is the binary number and num is the decimal number.

def base2():
  b=""
  num=int(input("Enter a decimal number: "))
  while(num != 0):
    b=str(num % 2) + b
    num=num // 2
    int(num == 10)
    print(b)

menu()

The expected result is for me to enter a number in my menu to choose either binary to decimal or decimal to binary. Then I should be able to put in the base2 and base10 numbers and get the opposite back. (I cannot use bin!)

Answer 1

A binary representation of a (positive) integer can be viewed as a weighted sum.

Consider 1011 = 11.

1011
abcd

In other words: a = c = d = 1
           and: b = 0

Can be represented as a weighted sum:

8*a + 4*b + 2*c + 1*d

By plugging in a,b,c,d as defined you'll get 11, but another way to look at the weighted sum is:

(2**3)*a  +  (2**2)*b  +  (2**1)*c  +  (2**0)*d

or

 (2**3)*a  +  (2**2)*b  +  (2**1)*c  +  (2**0)*d
     └ exponent   └ exponent   └ exponent   └ exponent

Here, 2 raised to a decreasing exponent, times the binary digit.

In other words:

(2**3)*1 + (2**2)*0 + (2**1)*1 + (2**0)*1 == 11

It might be easier to reverse the string first, then the exponent counts upwards from zero:

(2**0)*1 + (2**1)*1 + (2**2)*0 + (2**3)*1 == 11

Without giving too much away, assuming you had the binary string in a variable binstr , I'd take a look at the output of:

print(list(enumerate(reversed(binstr))))

or

for (i, x) in enumerate(reversed(binstr)):
    print(i, x)

One way might end up looking like:

intval = sum(<something> for (i,x) in enumerate(reversed(binstr)))

But that's not the only way. There are other ways, including using reduce (or a similar approach).

Of course, there's always int(binstr, 2) but I'm guessing that's not allowed either.

Answer 2

The int constructor accepts an optional base. You can rewrite base10 as

def base10():
  num = int(input("Enter a binary number:"), 2)
  print(num )

Converting the other way would normally be done with bin . Another way to do it is to check the bits of a number one by one. Your base2 implementation is close, but you shouldn't print in the loop.

Something like:

output = []
while num:
    output.append(num % 2)
    num //= 2
print(''.join(reversed(output)))

An alternative is to use the bit_length method of int along with a simple but twiddle to create the number in the correct order:

b = ''.join(str(int(bool(num & (1 << n)))) for n in range(num.bit_length() - 1, -1, -1))

The monstrosity str(int(bool(num & (1 << n)))) converts a bit flag that is a power of 2 into an boolean, and then uses the integer representation of the boolean to generate a string that is either 0 or 1 . It's equivalent to doing '1' if num & (1 << n) else '0' .