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);
}

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 ) );

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);
}

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();
}

private static String bytesToHexString(byte[] bytes, int length) {
        if (bytes == null || length == 0) return null;

        StringBuilder ret = new StringBuilder(2*length);

        for (int i = 0 ; i < length ; i++) {
            int b;

            b = 0x0f & (bytes[i] >> 4);
            ret.append("0123456789abcdef".charAt(b));

            b = 0x0f & bytes[i];
            ret.append("0123456789abcdef".charAt(b));
        }

        return ret.toString();
    }

这个简单的联机程序适合我 String result = new BigInteger(1, inputBytes).toString(16); EDIT -使用此命令将删除前导0，但在我的用例中它们是有效的。感谢@Voicu指出这一点