20

我有一个家庭作业，我需要在十进制、二进制和十六进制之间进行三向转换。我需要帮助的功能是将十进制转换为十六进制。我几乎不了解十六进制，但是如何将十进制转换为十六进制。我需要一个接受int dec并返回 a的函数String hex。不幸的是，我没有这个功能的任何草稿，我完全迷路了。我只有这个。

  public static String decToHex(int dec)
  {
    String hex = "";


    return hex;
  }

我也不能使用像 Integer.toHexString() 这样的预制函数或任何东西，我需要实际制作算法，否则我不会学到任何东西。

4

13 回答 13

37

一种可能的解决方案：

import java.lang.StringBuilder;

class Test {
  private static final int sizeOfIntInHalfBytes = 8;
  private static final int numberOfBitsInAHalfByte = 4;
  private static final int halfByte = 0x0F;
  private static final char[] hexDigits = { 
    '0', '1', '2', '3', '4', '5', '6', '7', 
    '8', '9', 'A', 'B', 'C', 'D', 'E', 'F'
  };

  public static String decToHex(int dec) {
    StringBuilder hexBuilder = new StringBuilder(sizeOfIntInHalfBytes);
    hexBuilder.setLength(sizeOfIntInHalfBytes);
    for (int i = sizeOfIntInHalfBytes - 1; i >= 0; --i)
    {
      int j = dec & halfByte;
      hexBuilder.setCharAt(i, hexDigits[j]);
      dec >>= numberOfBitsInAHalfByte;
    }
    return hexBuilder.toString(); 
  }

  public static void main(String[] args) {
     int dec = 305445566;
     String hex = decToHex(dec);
     System.out.println(hex);       
  }
}

输出：

1234BABE

无论如何，有一个库方法：

String hex = Integer.toHexString(dec);
于 2012-11-20T01:04:02.953 回答
26

简单的：

  public static String decToHex(int dec)
  {
        return Integer.toHexString(dec);
  }

如此处所述：Java将整数转换为十六进制整数

于 2016-08-16T07:34:13.043 回答
13

我需要一个接受 int dec 并返回 String hex 的函数。

我从http://introcs.cs.princeton.edu/java/31datatype/Hex2Decimal.java.html找到了一个更优雅的解决方案 。我从原来的改变了一点（见编辑）

// precondition:  d is a nonnegative integer
public static String decimal2hex(int d) {
    String digits = "0123456789ABCDEF";
    if (d <= 0) return "0";
    int base = 16;   // flexible to change in any base under 16
    String hex = "";
    while (d > 0) {
        int digit = d % base;              // rightmost digit
        hex = digits.charAt(digit) + hex;  // string concatenation
        d = d / base;
    }
    return hex;
}

免责声明：我在编码面试中问过这个问题。我希望这个解决方案不会太受欢迎:)

2016 年 6 月 17 日编辑：我添加了base变量以灵活地更改为任何基数：二进制、八进制、7 的基数……
根据评论，这个解决方案是最优雅的，所以我删除了Integer.toHexString().

编辑 2015 年 9 月 4 日：我找到了一个更优雅的解决方案http://introcs.cs.princeton.edu/java/31datatype/Hex2Decimal.java.html

于 2014-10-09T08:27:18.987 回答
2

考虑下面的 dec2m 方法，用于从 dec 转换为 hex、oct 或 bin。

样本输出为

28 dec == 11100 bin 28 dec == 34 oct 28 dec == 1C hex

public class Conversion {
    public static void main(String[] argv) {
        int x = 28;                           // sample number
        if (argv.length > 0)
            x = Integer.parseInt(argv[0]);    // number from command line

        System.out.printf("%d dec == %s bin\n", i, dec2m(x, 2));
        System.out.printf("%d dec == %s oct\n", i, dec2m(x, 8));
        System.out.printf("%d dec == %s hex\n", i, dec2m(x, 16));
    }

    static String dec2m(int N, int m) {
        String s = "";
        for (int n = N; n > 0; n /= m) {
            int r = n % m;
            s = r < 10 ? r + s : (char) ('A' - 10 + r) + s;
        }
        return s;
    }
}
于 2014-10-31T01:18:28.920 回答
1

这是任何数字的代码：

import java.math.BigInteger;

public class Testing {

/**
 * @param args
 */
static String arr[] ={"0","1","2","3","4","5","6","7","8","9","A","B","C","D","E","F"}; 
public static void main(String[] args) {
    String value = "214";
    System.out.println(value + " : " + getHex(value));
}


public static String getHex(String value) {
    String output= "";
    try {
        Integer.parseInt(value);
        Integer number = new Integer(value);
        while(number >= 16){
            output = arr[number%16] + output;
            number = number/16;
        }
        output = arr[number]+output;

    } catch (Exception e) {
        BigInteger number = null;
        try{
            number = new BigInteger(value);
        }catch (Exception e1) {
            return "Not a valid numebr";
        }
        BigInteger hex = new BigInteger("16");
        BigInteger[] val = {};

        while(number.compareTo(hex) == 1 || number.compareTo(hex) == 0){
            val = number.divideAndRemainder(hex);
            output = arr[val[1].intValue()] + output;
            number = val[0];
        }
        output = arr[number.intValue()] + output;
    }

    return output;
}

}
于 2013-12-05T06:48:03.473 回答
1

另一种可能的解决方案：

public String DecToHex(int dec){
  char[] hexDigits = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9',
              'A', 'B', 'C', 'D', 'E', 'F'};
  String hex = "";
  while (dec != 0) {
      int rem = dec % 16;
      hex = hexDigits[rem] + hex;
      dec = dec / 16;
  }
  return hex;
}
于 2015-02-10T22:19:02.427 回答
1

最简单的方法是：

String hexadecimalString = String.format("%x", integerValue);
于 2019-02-15T07:46:09.780 回答
0

我会用

Long a = Long.parseLong(cadenaFinal, 16 );

因为有一些十六进制可以比整数大，它会抛出异常

于 2015-02-11T10:01:20.960 回答
0

将 DECIMAL 转换为 -> BINARY、OCTAL、HEXADECIMAL 的代码

public class ConvertBase10ToBaseX {
    enum Base {
        /**
         * Integer is represented in 32 bit in 32/64 bit machine.
         * There we can split this integer no of bits into multiples of 1,2,4,8,16 bits
         */
        BASE2(1,1,32), BASE4(3,2,16), BASE8(7,3,11)/* OCTAL*/, /*BASE10(3,2),*/ 
        BASE16(15, 4, 8){       
            public String getFormattedValue(int val){
                switch(val) {
                case 10:
                    return "A";
                case 11:
                    return "B";
                case 12:
                    return "C";
                case 13:
                    return "D";
                case 14:
                    return "E";
                case 15:
                    return "F";
                default:
                    return "" + val;
                }

            }
        }, /*BASE32(31,5,1),*/ BASE256(255, 8, 4), /*BASE512(511,9),*/ Base65536(65535, 16, 2);

        private int LEVEL_0_MASK;
        private int LEVEL_1_ROTATION;
        private int MAX_ROTATION;

        Base(int levelZeroMask, int levelOneRotation, int maxPossibleRotation) {
            this.LEVEL_0_MASK = levelZeroMask;
            this.LEVEL_1_ROTATION = levelOneRotation;
            this.MAX_ROTATION = maxPossibleRotation;
        }

        int getLevelZeroMask(){
            return LEVEL_0_MASK;
        }
        int getLevelOneRotation(){
            return LEVEL_1_ROTATION;
        }
        int getMaxRotation(){
            return MAX_ROTATION;
        }
        String getFormattedValue(int val){
            return "" + val;
        }
    }

    public void getBaseXValueOn(Base base, int on) {
        forwardPrint(base, on);
    }

    private void forwardPrint(Base base, int on) {

        int rotation = base.getLevelOneRotation();
        int mask = base.getLevelZeroMask();
        int maxRotation = base.getMaxRotation();
        boolean valueFound = false;

        for(int level = maxRotation; level >= 2; level--) {
            int rotation1 = (level-1) * rotation;
            int mask1 = mask << rotation1 ;
            if((on & mask1) > 0 ) {
                valueFound = true;
            }
            if(valueFound)
            System.out.print(base.getFormattedValue((on & mask1) >>> rotation1));
        }
        System.out.println(base.getFormattedValue((on & mask)));
    }

    public int getBaseXValueOnAtLevel(Base base, int on, int level) {
        if(level > base.getMaxRotation() || level < 1) {
            return 0; //INVALID Input
        }
        int rotation = base.getLevelOneRotation();
        int mask = base.getLevelZeroMask();

        if(level > 1) {
            rotation = (level-1) * rotation;
            mask = mask << rotation;
        } else {
            rotation = 0;
        }


        return (on & mask) >>> rotation;
    }

    public static void main(String[] args) {
        ConvertBase10ToBaseX obj = new ConvertBase10ToBaseX();

        obj.getBaseXValueOn(Base.BASE16,12456); 
//      obj.getBaseXValueOn(Base.BASE16,300); 
//      obj.getBaseXValueOn(Base.BASE16,7); 
//      obj.getBaseXValueOn(Base.BASE16,7);

        obj.getBaseXValueOn(Base.BASE2,12456);
        obj.getBaseXValueOn(Base.BASE8,12456);
        obj.getBaseXValueOn(Base.BASE2,8);
        obj.getBaseXValueOn(Base.BASE2,9);
        obj.getBaseXValueOn(Base.BASE2,10);
        obj.getBaseXValueOn(Base.BASE2,11);
        obj.getBaseXValueOn(Base.BASE2,12);
        obj.getBaseXValueOn(Base.BASE2,13);
        obj.getBaseXValueOn(Base.BASE2,14);
        obj.getBaseXValueOn(Base.BASE2,15);
        obj.getBaseXValueOn(Base.BASE2,16);
        obj.getBaseXValueOn(Base.BASE2,17);


        System.out.println(obj.getBaseXValueOnAtLevel(Base.BASE2, 4, 1)); 
        System.out.println(obj.getBaseXValueOnAtLevel(Base.BASE2, 4, 2)); 
        System.out.println(obj.getBaseXValueOnAtLevel(Base.BASE2, 4, 3)); 
        System.out.println(obj.getBaseXValueOnAtLevel(Base.BASE2, 4, 4)); 

        System.out.println(obj.getBaseXValueOnAtLevel(Base.BASE16,15, 1)); 
        System.out.println(obj.getBaseXValueOnAtLevel(Base.BASE16,30, 2)); 
        System.out.println(obj.getBaseXValueOnAtLevel(Base.BASE16,7, 1)); 
        System.out.println(obj.getBaseXValueOnAtLevel(Base.BASE16,7, 2)); 

        System.out.println(obj.getBaseXValueOnAtLevel(Base.BASE256, 511, 1)); 
        System.out.println(obj.getBaseXValueOnAtLevel(Base.BASE256, 511, 2)); 
        System.out.println(obj.getBaseXValueOnAtLevel(Base.BASE256, 512, 1));
        System.out.println(obj.getBaseXValueOnAtLevel(Base.BASE256, 512, 2)); 
        System.out.println(obj.getBaseXValueOnAtLevel(Base.BASE256, 513, 2)); 


    }
}
于 2015-08-27T11:44:06.087 回答
0

这是我的

public static String dec2Hex(int num)
{
    String hex = "";

    while (num != 0)
    {
        if (num % 16 < 10)
            hex = Integer.toString(num % 16) + hex;
        else
            hex = (char)((num % 16)+55) + hex;
        num = num / 16;
    }

    return hex;
}
于 2016-05-28T16:15:47.913 回答
0

将 Decimal 转换为 HexaDecimal 的更好解决方案，这个解决方案不太复杂

import java.util.Scanner;
public class DecimalToHexa
{
    public static void main(String ar[])
    {
        Scanner sc=new Scanner(System.in);
        System.out.println("Enter a Decimal number: ");
        int n=sc.nextInt();
        if(n<0)
        {
            System.out.println("Enter a positive integer");
            return;
        }
        int i=0,d=0;
        String hx="",h="";
        while(n>0)
        {
            d=n%16;`enter code here`
            n/=16;
            if(d==10)h="A";
            else if(d==11)h="B";
            else if(d==12)h="C";
            else if(d==13)h="D";
            else if(d==14)h="E";
            else if(d==15)h="F";
            else h=""+d;            
            hx=""+h+hx;
        }
        System.out.println("Equivalent HEXA: "+hx);
    }
}
于 2017-10-04T12:43:30.353 回答
0

查看下面的代码进行十进制到十六进制的转换，

import java.util.Scanner;

public class DecimalToHexadecimal
{
   public static void main(String[] args)
   {
      int temp, decimalNumber;
      String hexaDecimal = "";
      char hexa[] = {'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};
 
      Scanner sc = new Scanner(System.in);
      System.out.print("Please enter decimal number : ");
      decimalNumber = sc.nextInt();
 
      while(decimalNumber > 0)
      {
         temp = decimalNumber % 16;
         hexaDecimal = hexa[temp] + hexaDecimal;
         decimalNumber = decimalNumber / 16;
      }
 
      System.out.print("The hexadecimal value of " + decimalNumber + " is : " + hexaDecimal);      
      sc.close();
   }
}

您可以在以下链接中了解有关将十进制转换为十六进制的不同方法的更多信息 >> java convert decimal to hexadecimal。

于 2017-11-21T10:30:07.933 回答
0

以下将十进制转换为具有时间复杂度的十六进制十进制： O(n) 线性时间，没有任何 java 内置函数

private static String decimalToHexaDecimal(int N) {
    char hexaDecimals[] = { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F' };
    StringBuilder builder = new StringBuilder();
    int base= 16;
    while (N != 0) {
        int reminder = N % base;
        builder.append(hexaDecimals[reminder]);
        N = N / base;
    }

    return builder.reverse().toString();
}
于 2018-05-17T17:33:03.713 回答