trouble converting binary into its decimal equivalent in c#

Question

I'm currently experiencing a major difficulty trying to get my program to convert binary numbers into their proper decimal equivalent. With what I have it gives me the proper amount for some numbers like 111=7, but it gives me the wrong amount like 1101 = 11 when it is supposed to be 13.

namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args)
        {

            int num;
            //int n;



            Console.WriteLine("Enter a number");
             num = Convert.ToInt16(Console.ReadLine());

            //n = num;

            bintonum(num);
        }


        public static void bintonum (int num)
        {
            int dig;
            double sum = 0;


                while (num > 0)
                {

                dig = num % 10; //takes the number and breaks it down into each digit
                sum = dig + (sum * 2); //reverses the number and adds the digit aquired from the previous line
                num = num / 10; // reduces the number by one digit to get to zero

                }

                Console.WriteLine("{0}", sum);

        }     
    }
}

Answer 1

This functionality is provided by the .NET framework:

Convert.ToInt32(num, 2)

Here, num is the string, and 2 is the base in which the string is represented.

More details here .

Answer 2

If you need to go from base 2 (binary) to base 10 (decimal), you can use this approach:
111 ₂ = 1 *2^2 + 1 *2^1 + 1 *2^0
...... = 4 + 2 + 1 = 7 ₁₀

1101 ₂ = 1 *2^3 + 1 *2^2 + 0 *2^1 + 1 *2^0
........ = 8 + 4 + 0 + 1 = 13 ₁₀

The basic premise is that there are two digits for the binary system and every place holder in a binary number can be multiplied by a power of two to get its decimal equivalent. The same goes for places in a decimal number:
121 ₁₀ = 1 *10^2 + 2 *10^1 + 1 *10^0
......... = 100 + 20 + 1 = 121 ₁₀

2543 ₁₀ = 2 *10^3 + 5 *10^2 + 4 *10^1 + 3 *10^0
........... = 2000 + 500 + 40 + 3 = 2543 ₁₀

Also, you can use this approach for any base to base 10 (decimal). Hexadecimal looks like this:
0 = 0, 1 = 1, 2 = 2, 3 = 3, 4 = 4, 5 = 5, 6 = 6, 7 = 7, 8 = 8,
9 = 9, A = 10, B = 11, C = 12, D = 13, E = 14, F = 15

1A3C ₁₆ = 1 *16^3 + 10 *16^2 + 3 *16^1 + 12 *16^0
............ = 4096 + 2560 + 48 + 12 = 6716 ₁₀

Here is an octal example:
The octal number system ranges from 0-7 and is often used for representing groups of three of binary numbers.
722 ₈ = 7 *8^2 + 2 *8^1 + 2 *8^0
........ = 448 + 16 + 2 = 466 ₁₀

Answer 3

as you're checking right 'bit' (or 0th bit), you need another loop, check:

        int num = 1101;
        int sum = 0;
        int n2 = 1;
        while (num > 0)
        {

            int dig = num % 10;
            sum = dig*n2 + sum;
            num = num / 10;
            n2 = n2 * 2;

        }