<<< Binary to Decimal | Logical Flow of Code (LFoC)

The given code is the implementation of a binary to decimal conversion algorithm. It takes a binary number as input and converts it into its equivalent decimal representation.

Here's a step-by-step description of the code:

1. Initialize variables:

   - `n`: This is the input binary number.

   - `rem`: This variable will store the remainder after dividing `n` by 10.

   - `decimal`: This variable will store the final decimal representation of the binary number.

   - `i`: This is a loop counter that will keep track of the position of the digits in the binary number.

2. Loop to convert binary to decimal:

   The code uses a `for` loop to iterate through the binary digits. The loop continues as long as the value of `n` is not equal to 0. In each iteration, it performs the following steps:

   a. Calculate the remainder:

      The `rem` variable is assigned the value of the last digit of the binary number `n` by calculating `n % 10`. This step isolates the rightmost digit of the binary number.

   b. Update `n`:

      The binary number `n` is then updated by performing an integer division `n /= 10`, which effectively removes the last digit from `n`. This prepares `n` for the next iteration.

   c. Convert binary to decimal:

      The decimal equivalent of the current binary digit is calculated and added to the `decimal` variable. The formula used is `decimal += rem * pow(2, i)`. This is done by multiplying the remainder (`rem`) by 2 raised to the power of `i`. The `i` variable keeps track of the position of the digit in the binary number, starting from the rightmost position with `i = 0`.

3. Result:

   After the loop finishes, the `decimal` variable will hold the decimal representation of the input binary number.

Example:

Input binary number: 101011

- In the first iteration: `rem = 1`, `n = 10101`, `decimal = 1 * pow(2, 0) = 1`.

- In the second iteration: `rem = 1`, `n = 1010`, `decimal = 1 + 1 * pow(2, 1) = 3`.

- In the third iteration: `rem = 0`, `n = 101`, `decimal = 3 + 0 * pow(2, 2) = 3`.

- In the fourth iteration: `rem = 1`, `n = 10`, `decimal = 3 + 1 * pow(2, 3) = 11`.

- In the fifth iteration: `rem = 0`, `n = 1`, `decimal = 11 + 0 * pow(2, 4) = 11`.

- In the sixth iteration: `rem = 1`, `n = 0`, `decimal = 11 + 1 * pow(2, 5) = 43`.

Therefore, the output of the code for the input binary number 101011 will be the decimal number 43.

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