CODEWITHSUNDEEP
c#binarydecimalconsole-application
user2892096
user2892096

Reputation: 3

trouble converting binary into its decimal equivalent in c#

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);

        }     
    }
}

Upvotes: 0

Views: 348

Answers (3)

Cameron Tinker
Cameron Tinker

Reputation: 9789

If you need to go from base 2 (binary) to base 10 (decimal), you can use this approach:
1112 = 1*2^2 + 1*2^1 + 1*2^0
...... = 4 + 2 + 1 = 710

11012 = 1*2^3 + 1*2^2 + 0*2^1 + 1*2^0
........ = 8 + 4 + 0 + 1 = 1310

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:
12110 = 1*10^2 + 2*10^1 + 1*10^0
......... = 100 + 20 + 1 = 12110

254310 = 2*10^3 + 5*10^2 + 4*10^1 + 3*10^0
........... = 2000 + 500 + 40 + 3 = 254310

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

1A3C16 = 1*16^3 + 10*16^2 + 3*16^1 + 12*16^0
............ = 4096 + 2560 + 48 + 12 = 671610

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.
7228 = 7*8^2 + 2*8^1 + 2*8^0
........ = 448 + 16 + 2 = 46610

Upvotes: 1

Iłya Bursov
Iłya Bursov

Reputation: 24145

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;

        }

Upvotes: 0

zimdanen
zimdanen

Reputation: 5626

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.

Upvotes: 2

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