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;
    }

其他回答

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));
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("");
}

我的解决方案是基于maybeWeCouldStealAVan的解决方案,但不依赖于任何额外分配的查找表。它不使用任何“int-to-char”类型强制转换(实际上,Character.forDigit()做到了这一点,执行一些比较来检查数字的真实情况),因此可能会稍慢一些。请随意在任何你想用的地方使用。欢呼。

public static String bytesToHex(final byte[] bytes)
{
    final int numBytes = bytes.length;
    final char[] container = new char[numBytes * 2];

    for (int i = 0; i < numBytes; i++)
    {
        final int b = bytes[i] & 0xFF;

        container[i * 2] = Character.forDigit(b >>> 4, 0x10);
        container[i * 2 + 1] = Character.forDigit(b & 0xF, 0x10);
    }

    return new String(container);
}

从这里的讨论,特别是这个答案,这是我目前使用的函数:

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

My own tiny benchmarks (a million bytes a thousand times, 256 bytes 10 million times) showed it to be much faster than any other alternative, about half the time on long arrays. Compared to the answer I took it from, switching to bitwise ops --- as suggested in the discussion --- cut about 20% off of the time for long arrays. (Edit: When I say it's faster than the alternatives, I mean the alternative code offered in the discussions. Performance is equivalent to Commons Codec, which uses very similar code.)

2k20版本,相对于Java 9的压缩字符串:

private static final byte[] HEX_ARRAY = "0123456789ABCDEF".getBytes(StandardCharsets.US_ASCII);
public static String bytesToHex(byte[] bytes) {
    byte[] hexChars = new byte[bytes.length * 2];
    for (int j = 0; j < bytes.length; j++) {
        int v = bytes[j] & 0xFF;
        hexChars[j * 2] = HEX_ARRAY[v >>> 4];
        hexChars[j * 2 + 1] = HEX_ARRAY[v & 0x0F];
    }
    return new String(hexChars, StandardCharsets.UTF_8);
}

最简单的解决方案,没有外部库,没有数字常量:

public static String byteArrayToHex(byte[] a) {
   StringBuilder sb = new StringBuilder(a.length * 2);
   for(byte b: a)
      sb.append(String.format("%02x", b));
   return sb.toString();
}