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

Java 17最终包含HexFormat类,所以你可以简单地做:

HexFormat.of().formatHex(bytes);

它支持配置为小写/大写,分隔符,前缀,后缀等。

Apache Commons Codec库有一个Hex类用于完成这种类型的工作。

import org.apache.commons.codec.binary.Hex;

String foo = "I am a string";
byte[] bytes = foo.getBytes();
System.out.println( Hex.encodeHexString( bytes ) );

我们不需要使用任何外部库,也不需要基于循环和常量编写代码。 这样就够了:

byte[] theValue = .....
String hexaString = new BigInteger(1, theValue).toString(16);

//移位字节更有效 //你也可以用这个

public static String getHexString (String s) 
{
    byte[] buf = s.getBytes();

    StringBuffer sb = new StringBuffer();

    for (byte b:buf)
    {
        sb.append(String.format("%x", b));
    }


        return sb.toString();
}
public static String toHexString(byte[] bytes) {

    StringBuilder sb = new StringBuilder();

    if (bytes != null) 
        for (byte b:bytes) {

            final String hexString = Integer.toHexString(b & 0xff);

            if(hexString.length()==1)
                sb.append('0');

            sb.append(hexString);//.append(' ');
        }

      return sb.toString();//.toUpperCase();
}

使用DatatypeConverter:

public String toHexString(byte... bytes) {

    return Optional.ofNullable(bytes)
            .filter(bs->bs.length>0)
            .map(DatatypeConverter::printHexBinary)
            .map(str->IntStream.range(0, str.length())
                    .filter(i->(i%2)==0)        // take every second index
                    .mapToObj(i->"0x" + str.substring(i, i+2))
                    .collect(Collectors.joining(" ")))
            .orElse("");
}