bin to dec

written on June 5th, 2023 by alex

A quick snippet to convert binary numbers to their decimal values



In in the decimal number system each digit has a 10^n value. In the number 123 the 3 is in the 10^0 place, which gives it a value of 3 * 10^0 = 3 * 1 = 3. Or said another way, there are 3 ones.

Similarly, there are 2 tens, or 2 * 10^1 = 2 * 10 = 20. And 1 one-hundreds, or 1 * 10^2 = 1 * 100 = 100. The sum of these is of course 123

Binary is analogous, but it uses powers of 2 instead. So reading from right to left the place values of each digit are 2^0, 2^1, 2^2, …, 2^n



So here’s a quick way to convert a string representation of a binary number into its decimal value:

def convert_binary_to_decimal(binary_str: str) -> int:
    decimal = 0
    for char_digit in binary_str:
        one_or_zero = int(char_digit)
        # neat trick on below line
        decimal = decimal * 2 + one_or_zero

    return decimal

The reason it works is that each iteration of the loop retroactively accounts for the “true” value of of the previously processed binary digit.

Because we’re parsing from left to right, the first digit is the most significant. But we assume it’s in the 2^0 place (least significant) and only change that assumption if and when we process additional digits.

For example, if the input were 1

decimal = decimal * 2 + one_or_zero
decimal = 0 * 2 + 1

Since the current value of decimal is 0 and the digit being processed is 1.

If the input were 11 then we’d move to the next digit, the second 1. At the start of this iteration decimal is equal to 1 (from previous iteration)

decimal = decimal * 2 + one_or_zero
decimal = 1 * 2 + 1

Which equals 3, which is the correct decimal value of input binary value 11.

Here’s the trick - multiplying decimal by 2 each additional iteration is like increasing the power of 2 that we multiplied the previous binary digit by in the previous iteration.

That is,

\[(x * 2^3) * 2 = x * 2^4\]



In the last iteration, one_or_zero will be in the 2^0 place, which always evaluates to 1. So the last one_or_zero will always have a decimal value of 1 for binary digit 1 or a decimal value of 0 for binary digit 0. Meaning that we can just add it to decimal.



The advantages of this method is that we can parse the input in left-to-right order without needing to first find the place-values (2^n) of each digit in each unique input.



Of course, you could use the built-in int() function instead,

# second parameter indicates which base, 2 for binary/base-2
decimal = int(binary_num, 2)

Also, note that I used a string representation of a binary sequence but most languages have binary types or some better way to represent them.