我有一个字节数组充满十六进制数字和打印它的简单方式是相当没有意义的,因为有许多不可打印的元素。我需要的是精确的十六进制形式:3a5f771c


当前回答

这是一个java.util。类似base64的实现,是不是很漂亮?

import java.util.Arrays;

public class Base16/* a.k.a. Hex */ {
    public static class Encoder{
        private static char[] toLowerHex={'0','1','2','3','4','5','6','7','8','9','a','b','c','d','e','f'};
        private static char[] toUpperHex={'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};
        private boolean upper;
        public Encoder(boolean upper) {
            this.upper=upper;
        }
        public String encode(byte[] data){
            char[] value=new char[data.length*2];
            char[] toHex=upper?toUpperHex:toLowerHex;
            for(int i=0,j=0; i<data.length; i++){
                int octet=data[i]&0xFF;
                value[j++]=toHex[octet>>4];
                value[j++]=toHex[octet&0xF];
            }
            return new String(value);
        }
        static final Encoder LOWER_CASE=new Encoder(false);
        static final Encoder UPPER_CASE=new Encoder(true);
    }
    public static Encoder getEncoder(){
        return Encoder.LOWER_CASE;
    }
    public static Encoder getUpperEncoder(){
        return Encoder.UPPER_CASE;
    }

    public static class Decoder{
      private static int maxIndex=102;
      private static int[] toIndex;
      static {
        toIndex=new int[maxIndex+1];
        Arrays.fill(toIndex, -1);
        char[] chars={'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F','a','b','c','d','e','f'};
        for(int i=0; i<chars.length; i++) {
          toIndex[(int)chars[i]]=i;
        }
      }
      public Decoder() {
      }
      public byte[] decode(String str) {
          char[] value=str.toCharArray();
          int start=0;
          if(value.length>2 && value[0]=='0' && (value[1]=='x' || value[1]=='X')) {
            start=2;
          }
          int byteLength=(value.length-start)/2; // ignore trailing odd char if exists
          byte[] data=new byte[byteLength];
          for(int i=start,j=0;i<value.length;i+=2,j++){
              int i1;
              int i2;
              char c1=value[i];
              char c2=value[i+1];
              if(c1>maxIndex || (i1=toIndex[(int)c1])<0 || c2>maxIndex || (i2=toIndex[(int)c2])<0) {
                throw new IllegalArgumentException("Invalid character at "+i);
              }
              data[j]=(byte)((i1<<4)+i2);
          }
          return data;
      }
      static final Decoder IGNORE_CASE=new Decoder();
  }
  public static Decoder getDecoder(){
      return Decoder.IGNORE_CASE;
  }
}

其他回答

@maybewecouldstealavan提出的解决方案的一个小变种,它让你在输出十六进制字符串中可视化地捆绑N个字节:

 final static char[] HEX_ARRAY = "0123456789ABCDEF".toCharArray();
 final static char BUNDLE_SEP = ' ';

public static String bytesToHexString(byte[] bytes, int bundleSize /*[bytes]*/]) {
        char[] hexChars = new char[(bytes.length * 2) + (bytes.length / bundleSize)];
        for (int j = 0, k = 1; j < bytes.length; j++, k++) {
                int v = bytes[j] & 0xFF;
                int start = (j * 2) + j/bundleSize;

                hexChars[start] = HEX_ARRAY[v >>> 4];
                hexChars[start + 1] = HEX_ARRAY[v & 0x0F];

                if ((k % bundleSize) == 0) {
                        hexChars[start + 2] = BUNDLE_SEP;
                }   
        }   
        return new String(hexChars).trim();    
}

那就是:

bytesToHexString("..DOOM..".toCharArray().getBytes(), 2);
2E2E 444F 4F4D 2E2E

bytesToHexString("..DOOM..".toCharArray().getBytes(), 4);
2E2E444F 4F4D2E2E

如果你正在使用Spring Security框架,你可以使用:

import org.springframework.security.crypto.codec.Hex

final String testString = "Test String";
final byte[] byteArray = testString.getBytes();
System.out.println(Hex.encode(byteArray));

我更喜欢用这个:

final protected static char[] hexArray = "0123456789ABCDEF".toCharArray();
public static String bytesToHex(byte[] bytes, int offset, int count) {
    char[] hexChars = new char[count * 2];
    for ( int j = 0; j < count; j++ ) {
        int v = bytes[j+offset] & 0xFF;
        hexChars[j * 2] = hexArray[v >>> 4];
        hexChars[j * 2 + 1] = hexArray[v & 0x0F];
    }
    return new String(hexChars);
}

它是对公认答案的稍微灵活的改编。 就我个人而言,我既保留了公认的答案,也保留了这个重载,以便在更多的环境中使用。

在这一页上找不到任何解决方案吗

使用循环 使用javax.xml.bind.DatatypeConverter,它编译良好,但经常在运行时抛出java.lang.NoClassDefFoundError。

这里有一个解决方案,它没有上面的缺陷(不保证我的没有其他缺陷)

import java.math.BigInteger;

import static java.lang.System.out;
public final class App2 {
    // | proposed solution.
    public static String encode(byte[] bytes) {          
        final int length = bytes.length;

        // | BigInteger constructor throws if it is given an empty array.
        if (length == 0) {
            return "00";
        }

        final int evenLength = (int)(2 * Math.ceil(length / 2.0));
        final String format = "%0" + evenLength + "x";         
        final String result = String.format (format, new BigInteger(bytes));

        return result;
    }

    public static void main(String[] args) throws Exception {
        // 00
        out.println(encode(new byte[] {})); 

        // 01
        out.println(encode(new byte[] {1})); 

        //203040
        out.println(encode(new byte[] {0x20, 0x30, 0x40})); 

        // 416c6c20796f75722062617365206172652062656c6f6e6720746f2075732e
        out.println(encode("All your base are belong to us.".getBytes()));
    }
}   

我不能在62个操作码下得到这个,但如果你可以在第一个字节小于0x10的情况下没有0填充,那么下面的解决方案只使用23个操作码。真正展示了“容易实现自己”的解决方案,如“如果字符串长度为奇数,则填充为零”,如果本机实现还不可用(或者在本例中,如果BigInteger在toString中有一个以零作为前缀的选项),则可能会非常昂贵。

public static String encode(byte[] bytes) {          
    final int length = bytes.length;

    // | BigInteger constructor throws if it is given an empty array.
    if (length == 0) {
        return "00";
    }

    return new BigInteger(bytes).toString(16);
}

这个怎么样?

    String byteToHex(final byte[] hash)
    {
        Formatter formatter = new Formatter();
        for (byte b : hash)
        {
            formatter.format("%02x", b);
        }
        String result = formatter.toString();
        formatter.close();
        return result;
    }