在C＃中将整数转换为二进制

如何将整数转换为二进制表示形式？

我正在使用此代码：

String input = "8";
String output = Convert.ToInt32(input, 2).ToString();

但这会引发异常：

找不到任何可解析的数字

您的示例具有一个表示为字符串的整数。假设您的整数实际上是一个整数，并且您想获取该整数并将其转换为二进制字符串。

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

返回1000。

从任何经典库转换为C＃中的任何库

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

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

// result == "256"

支持的基数是2、8、10和16

非常简单，没有额外的代码，只需输入，转换和输出。

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;
    }

原始方式：

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)不会将基准转换为基准。假定该字符串在指示的基数中包含有效数字，然后转换为基数10。

因此您会收到错误消息，因为“ 8”不是以2为底的有效数字。

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

将显示15（1111以2为基数= 15以10为基数）

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

将显示61440。

我知道这个答案看起来与这里已经存在的大多数答案相似，但是我注意到几乎没有一个使用for循环。在没有任何特殊功能（例如带有参数的ToString（））的情况下，此代码可以正常工作，并且可以认为很简单，而且时间也不长。也许有些人更喜欢for循环而不是while循环，这可能适合他们。

public static string ByteConvert (int num)
{
    int[] p = new int[8];
    string pa = "";
    for (int ii = 0; ii<= 7;ii = ii +1)
    {
        p[7-ii] = num%2;
        num = num/2;
    }
    for (int ii = 0;ii <= 7; ii = ii + 1)
    {
        pa += p[ii].ToString();
    }
    return pa;
}

using System;

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

        try {

            int i = (int) Convert.ToInt64(args[0]);
            Console.WriteLine("\n{0} converted to Binary is {1}\n", i, ToBinary(i));

        } catch(Exception e) {
            Console.WriteLine("\n{0}\n", e.Message);
        }
    }

    public static string ToBinary(Int64 Decimal) {
        // Declare a few variables we're going to need
        Int64 BinaryHolder;
        char[] BinaryArray;
        string BinaryResult = "";

        while (Decimal > 0) {
            BinaryHolder = Decimal % 2;
            BinaryResult += BinaryHolder;
            Decimal = Decimal / 2;
        }

        BinaryArray = BinaryResult.ToCharArray();
        Array.Reverse(BinaryArray);
        BinaryResult = new string(BinaryArray);

        return BinaryResult;
    }
}

使用Enumerableand的另一种方法也是内联解决方案，它LINQ是：

int number = 25;

string binary = Enumerable.Range(0, (int) Math.Log(number, 2) + 1).Aggregate(string.Empty, (collected, bitshifts) => ((number >> bitshifts) & 1 )+ collected);

此函数将在C＃中将整数转换为二进制：

public static string ToBinary(int N)
{
    int d = N;
    int q = -1;
    int r = -1;

    string binNumber = string.Empty;
    while (q != 1)
    {
        r = d % 2;
        q = d / 2;
        d = q;
        binNumber = r.ToString() + binNumber;
    }
    binNumber = q.ToString() + binNumber;
    return binNumber;
}

class Program
{
    static void Main(string[] args)
    {
        var @decimal = 42;
        var binaryVal = ToBinary(@decimal, 2);

        var binary = "101010";
        var decimalVal = ToDecimal(binary, 2);

        Console.WriteLine("Binary value of decimal {0} is '{1}'", @decimal, binaryVal);
        Console.WriteLine("Decimal value of binary '{0}' is {1}", binary, decimalVal);
        Console.WriteLine();

        @decimal = 6;
        binaryVal = ToBinary(@decimal, 3);

        binary = "20";
        decimalVal = ToDecimal(binary, 3);

        Console.WriteLine("Base3 value of decimal {0} is '{1}'", @decimal, binaryVal);
        Console.WriteLine("Decimal value of base3 '{0}' is {1}", binary, decimalVal);
        Console.WriteLine();


        @decimal = 47;
        binaryVal = ToBinary(@decimal, 4);

        binary = "233";
        decimalVal = ToDecimal(binary, 4);

        Console.WriteLine("Base4 value of decimal {0} is '{1}'", @decimal, binaryVal);
        Console.WriteLine("Decimal value of base4 '{0}' is {1}", binary, decimalVal);
        Console.WriteLine();

        @decimal = 99;
        binaryVal = ToBinary(@decimal, 5);

        binary = "344";
        decimalVal = ToDecimal(binary, 5);

        Console.WriteLine("Base5 value of decimal {0} is '{1}'", @decimal, binaryVal);
        Console.WriteLine("Decimal value of base5 '{0}' is {1}", binary, decimalVal);
        Console.WriteLine();

        Console.WriteLine("And so forth.. excluding after base 10 (decimal) though :)");
        Console.WriteLine();


        @decimal = 16;
        binaryVal = ToBinary(@decimal, 11);

        binary = "b";
        decimalVal = ToDecimal(binary, 11);

        Console.WriteLine("Hexidecimal value of decimal {0} is '{1}'", @decimal, binaryVal);
        Console.WriteLine("Decimal value of Hexidecimal '{0}' is {1}", binary, decimalVal);
        Console.WriteLine();
        Console.WriteLine("Uh oh.. this aint right :( ... but let's cheat :P");
        Console.WriteLine();

        @decimal = 11;
        binaryVal = Convert.ToString(@decimal, 16);

        binary = "b";
        decimalVal = Convert.ToInt32(binary, 16);

        Console.WriteLine("Hexidecimal value of decimal {0} is '{1}'", @decimal, binaryVal);
        Console.WriteLine("Decimal value of Hexidecimal '{0}' is {1}", binary, decimalVal);

        Console.ReadLine();
    }


    static string ToBinary(decimal number, int @base)
    {
        var round = 0;
        var reverseBinary = string.Empty;

        while (number > 0)
        {
            var remainder = number % @base;
            reverseBinary += remainder;

            round = (int)(number / @base);
            number = round;
        }

        var binaryArray = reverseBinary.ToCharArray();
        Array.Reverse(binaryArray);

        var binary = new string(binaryArray);
        return binary;
    }

    static double ToDecimal(string binary, int @base)
    {
        var val = 0d;

        if (!binary.All(char.IsNumber))
            return 0d;

        for (int i = 0; i < binary.Length; i++)
        {
            var @char = Convert.ToDouble(binary[i].ToString());

            var pow = (binary.Length - 1) - i;
            val += Math.Pow(@base, pow) * @char;
        }

        return val;
    }
}

学习资源：

您需要了解的有关二进制的所有信息

包括将十进制转换为二进制的算法

    static void convertToBinary(int n)
    {
        Stack<int> stack = new Stack<int>();
        stack.Push(n);
        // step 1 : Push the element on the stack
        while (n > 1)
        {
            n = n / 2;
            stack.Push(n);
        }

        // step 2 : Pop the element and print the value
        foreach(var val in stack)
        {
            Console.Write(val % 2);
        }
     }

class Program{

   static void Main(string[] args){

      try{

     int i = (int)Convert.ToInt64(args[0]);
         Console.WriteLine("\n{0} converted to Binary is {1}\n",i,ToBinary(i));

      }catch(Exception e){

         Console.WriteLine("\n{0}\n",e.Message);

      }

   }//end Main


        public static string ToBinary(Int64 Decimal)
        {
            // Declare a few variables we're going to need
            Int64 BinaryHolder;
            char[] BinaryArray;
            string BinaryResult = "";

            while (Decimal > 0)
            {
                BinaryHolder = Decimal % 2;
                BinaryResult += BinaryHolder;
                Decimal = Decimal / 2;
            }

            // The algoritm gives us the binary number in reverse order (mirrored)
            // We store it in an array so that we can reverse it back to normal
            BinaryArray = BinaryResult.ToCharArray();
            Array.Reverse(BinaryArray);
            BinaryResult = new string(BinaryArray);

            return BinaryResult;
        }


}//end class Program

提供BCL Convert.ToString(n, 2)很好，但是如果您需要另一种实现，该实现比提供的BCL快几个壁虱。

以下自定义实现适用于所有整数（-ve和+ ve）。原始来源取自https://davidsekar.com/algorithms/csharp-program-to-convert-decimal-to-binary

static string ToBinary(int n)
{
    int j = 0;
    char[] output = new char[32];

    if (n == 0)
        output[j++] = '0';
    else
    {
        int checkBit = 1 << 30;
        bool skipInitialZeros = true;
        // Check the sign bit separately, as 1<<31 will cause
        // +ve integer overflow
        if ((n & int.MinValue) == int.MinValue)
        {
            output[j++] = '1';
            skipInitialZeros = false;
        }

        for (int i = 0; i < 31; i++, checkBit >>= 1)
        {
            if ((n & checkBit) == 0)
            {
                if (skipInitialZeros)
                    continue;
                else
                    output[j++] = '0';
            }
            else
            {
                skipInitialZeros = false;
                output[j++] = '1';
            }
        }
    }

    return new string(output, 0, j);
}

上面的代码是我的实现。所以，我很想听听任何反馈意见:)

    // I use this function
    public static string ToBinary(long number)
    {
        string digit = Convert.ToString(number % 2);
        if (number >= 2)
        {
            long remaining = number / 2;
            string remainingString = ToBinary(remaining);
            return remainingString + digit;
        }
        return digit;
     }

        static void Main(string[] args) 
        {
        Console.WriteLine("Enter number for converting to binary numerical system!");
        int num = Convert.ToInt32(Console.ReadLine());
        int[] arr = new int[16];

        //for positive integers
        if (num > 0)
        {

            for (int i = 0; i < 16; i++)
            {
                if (num > 0)
                {
                    if ((num % 2) == 0)
                    {
                        num = num / 2;
                        arr[16 - (i + 1)] = 0;
                    }
                    else if ((num % 2) != 0)
                    {
                        num = num / 2;
                        arr[16 - (i + 1)] = 1;
                    }
                }
            }
            for (int y = 0; y < 16; y++)
            {
                Console.Write(arr[y]);
            }
            Console.ReadLine();
        }

        //for negative integers
        else if (num < 0)
        {
            num = (num + 1) * -1;

            for (int i = 0; i < 16; i++)
            {
                if (num > 0)
                {
                    if ((num % 2) == 0)
                    {
                        num = num / 2;
                        arr[16 - (i + 1)] = 0;
                    }
                    else if ((num % 2) != 0)
                    {
                        num = num / 2;
                        arr[16 - (i + 1)] = 1;
                    }
                }
            }

            for (int y = 0; y < 16; y++)
            {
                if (arr[y] != 0)
                {
                    arr[y] = 0;
                }
                else
                {
                    arr[y] = 1;
                }
                Console.Write(arr[y]);
            }
            Console.ReadLine();
        }           
    }

如果您想要一个简洁的函数，可以从类内部的main方法中调用，这可能会有所帮助。int.Parse(toBinary(someint))如果您需要数字而不是字符串，则可能仍需要调用，但是我发现此方法效果很好。此外，可以将其调整为使用for循环而不是do-while如果愿意。

    public static string toBinary(int base10)
    {
        string binary = "";
        do {
            binary = (base10 % 2) + binary;
            base10 /= 2;
        }
        while (base10 > 0);

        return binary;
    }

toBinary(10)返回字符串"1010"。

我在编码挑战中遇到了这个问题，您必须将32位十进制转换为二进制并找到子字符串的可能组合。

using System;
using System.Collections.Generic;
using System.Globalization;
using System.Numerics;
using System.IO;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace ConsoleApp2
{
    class Program
    {

        public static void Main()
        {
            int numberofinputs = int.Parse(Console.ReadLine());
            List<BigInteger> inputdecimal = new List<BigInteger>();
            List<string> outputBinary = new List<string>();


            for (int i = 0; i < numberofinputs; i++)
            {
                inputdecimal.Add(BigInteger.Parse(Console.ReadLine(), CultureInfo.InvariantCulture));
            }
            //processing begins 

            foreach (var n in inputdecimal)
            {
                string binary = (binaryconveter(n));
                subString(binary, binary.Length);
            }

            foreach (var item in outputBinary)
            {
                Console.WriteLine(item);
            }

            string binaryconveter(BigInteger n)
            {
                int i;
                StringBuilder output = new StringBuilder();

                for (i = 0; n > 0; i++)
                {
                    output = output.Append(n % 2);
                    n = n / 2;
                }

                return output.ToString();
            }

            void subString(string str, int n)
            {
                int zeroodds = 0;
                int oneodds = 0;

                for (int len = 1; len <= n; len++)
                {

                    for (int i = 0; i <= n - len; i++)
                    {
                        int j = i + len - 1;

                        string substring = "";
                        for (int k = i; k <= j; k++)
                        {
                            substring = String.Concat(substring, str[k]);

                        }
                        var resultofstringanalysis = stringanalysis(substring);
                        if (resultofstringanalysis.Equals("both are odd"))
                        {
                            ++zeroodds;
                            ++oneodds;
                        }
                        else if (resultofstringanalysis.Equals("zeroes are odd"))
                        {
                            ++zeroodds;
                        }
                        else if (resultofstringanalysis.Equals("ones are odd"))
                        {
                            ++oneodds;
                        }

                    }
                }
                string outputtest = String.Concat(zeroodds.ToString(), ' ', oneodds.ToString());
                outputBinary.Add(outputtest);
            }

            string stringanalysis(string str)
            {
                int n = str.Length;

                int nofZeros = 0;
                int nofOnes = 0;

                for (int i = 0; i < n; i++)
                {
                    if (str[i] == '0')
                    {
                        ++nofZeros;
                    }
                    if (str[i] == '1')
                    {
                        ++nofOnes;
                    }

                }
                if ((nofZeros != 0 && nofZeros % 2 != 0) && (nofOnes != 0 && nofOnes % 2 != 0))
                {
                    return "both are odd";
                }
                else if (nofZeros != 0 && nofZeros % 2 != 0)
                {
                    return "zeroes are odd";
                }
                else if (nofOnes != 0 && nofOnes % 2 != 0)
                {
                    return "ones are odd";
                }
                else
                {
                    return "nothing";
                }

            }
            Console.ReadKey();
        }

    }
}

    int x=550;
    string s=" ";
    string y=" ";

    while (x>0)
    {

        s += x%2;
        x=x/2;
    }


    Console.WriteLine(Reverse(s));
}

public static string Reverse( string s )
{
    char[] charArray = s.ToCharArray();
    Array.Reverse( charArray );
    return new string( charArray );
}