Convert integer to binary in C#

Question

Your example has an integer expressed as a string. Let's say your integer was actually an integer, and you want to take the integer and convert it to a binary string.

int value = 8;
string binary = Convert.ToString(value, 2);

Which returns 1000.

Convert from any classic base to any base in C#

string number = "100";
int fromBase = 16;
int toBase = 10;

string result = Convert.ToString(Convert.ToInt32(number, fromBase), toBase);

// result == "256"

Supported bases are 2, 8, 10 and 16

Very Simple with no extra code, just input, conversion and output.

using System;

namespace _01.Decimal_to_Binary
{
    class DecimalToBinary
    {
        static void Main(string[] args)
        {
            Console.Write("Decimal: ");
            int decimalNumber = int.Parse(Console.ReadLine());

            int remainder;
            string result = string.Empty;
            while (decimalNumber > 0)
            {
                remainder = decimalNumber % 2;
                decimalNumber /= 2;
                result = remainder.ToString() + result;
            }
            Console.WriteLine("Binary:  {0}",result);
        }
    }
}

http://zamirsblog.blogspot.com/2011/10/convert-decimal-to-binary-in-c.html

    public string DecimalToBinary(string data)
    {
        string result = string.Empty;
        int rem = 0;
        try
        {
            if (!IsNumeric(data))
                error = "Invalid Value - This is not a numeric value";
            else
            {
                int num = int.Parse(data);
                while (num > 0)
                {
                    rem = num % 2;
                    num = num / 2;
                    result = rem.ToString() + result;
                }
            }
        }
        catch (Exception ex)
        {
            error = ex.Message;
        }
        return result;
    }

primitive way:

public string ToBinary(int n)
{
    if (n < 2) return n.ToString();

    var divisor = n / 2;
    var remainder = n % 2;

    return ToBinary(divisor) + remainder;
}

Convert.ToInt32(string, base) does not do base conversion into your base. It assumes that the string contains a valid number in the indicated base, and converts to base 10.

So you're getting an error because "8" is not a valid digit in base 2.

String str = "1111";
String Ans = Convert.ToInt32(str, 2).ToString();

Will show 15 (1111 base 2 = 15 base 10)

String str = "f000";
String Ans = Convert.ToInt32(str, 16).ToString();

Will show 61440.