我有一个字节数组。 我希望将该数组的每个字节String转换为相应的十六进制值。

Java中是否有将字节数组转换为十六进制的函数?


当前回答

你可以使用Bouncy Castle Provider库中的方法:

org.bouncycastle.util.encoders.Hex.toHexString(byteArray);

Bouncy Castle Crypto包是一个Java实现 加密算法。这个jar包含JCE提供程序和 用于JDK 1.5的Bouncy Castle Cryptography API的轻量级API JDK 1.8。

Maven的依赖:

<dependency>
    <groupId>org.bouncycastle</groupId>
    <artifactId>bcprov-jdk15on</artifactId>
    <version>1.60</version>
</dependency>

或来自Apache Commons Codec:

org.apache.commons.codec.binary.Hex.encodeHexString(byteArray);

Apache Commons Codec包包含简单的编码器和解码器 用于各种格式,如Base64和十六进制。除了 这些编码器和解码器被广泛使用,编解码器包也有 维护语音编码实用程序的集合。

Maven的依赖:

<dependency>
    <groupId>commons-codec</groupId>
    <artifactId>commons-codec</artifactId>
    <version>1.11</version>
</dependency>

其他回答

Use

Integer.toHexString((int)b);

这是你的快速方法:

    private static final String[] hexes = new String[]{
        "00","01","02","03","04","05","06","07","08","09","0A","0B","0C","0D","0E","0F",
        "10","11","12","13","14","15","16","17","18","19","1A","1B","1C","1D","1E","1F",
        "20","21","22","23","24","25","26","27","28","29","2A","2B","2C","2D","2E","2F",
        "30","31","32","33","34","35","36","37","38","39","3A","3B","3C","3D","3E","3F",
        "40","41","42","43","44","45","46","47","48","49","4A","4B","4C","4D","4E","4F",
        "50","51","52","53","54","55","56","57","58","59","5A","5B","5C","5D","5E","5F",
        "60","61","62","63","64","65","66","67","68","69","6A","6B","6C","6D","6E","6F",
        "70","71","72","73","74","75","76","77","78","79","7A","7B","7C","7D","7E","7F",
        "80","81","82","83","84","85","86","87","88","89","8A","8B","8C","8D","8E","8F",
        "90","91","92","93","94","95","96","97","98","99","9A","9B","9C","9D","9E","9F",
        "A0","A1","A2","A3","A4","A5","A6","A7","A8","A9","AA","AB","AC","AD","AE","AF",
        "B0","B1","B2","B3","B4","B5","B6","B7","B8","B9","BA","BB","BC","BD","BE","BF",
        "C0","C1","C2","C3","C4","C5","C6","C7","C8","C9","CA","CB","CC","CD","CE","CF",
        "D0","D1","D2","D3","D4","D5","D6","D7","D8","D9","DA","DB","DC","DD","DE","DF",
        "E0","E1","E2","E3","E4","E5","E6","E7","E8","E9","EA","EB","EC","ED","EE","EF",
        "F0","F1","F2","F3","F4","F5","F6","F7","F8","F9","FA","FB","FC","FD","FE","FF"
    };

    public static String byteToHex(byte b) {
        return hexes[b&0xFF];
    }

如果你使用奇妙仙子,那么有:

package com.google.crypto.tink.subtle;

public final class Hex {
  public static String encode(final byte[] bytes) { ... }
  public static byte[] decode(String hex) { ... }
}

所以像这样的东西应该是有用的:

import com.google.crypto.tink.subtle.Hex;

byte[] bytes = {-1, 0, 1, 2, 3 };
String enc = Hex.encode(bytes);
byte[] dec = Hex.decode(enc)
    byte[] bytes = {-1, 0, 1, 2, 3 };
    StringBuilder sb = new StringBuilder();
    for (byte b : bytes) {
        sb.append(String.format("%02X ", b));
    }
    System.out.println(sb.toString());
    // prints "FF 00 01 02 03 "

另请参阅

java.util.Formatter语法 %(旗帜)(宽度)转换 标记'0' -结果将被填充为零 宽度2 转换'X' -结果被格式化为十六进制整数,大写


看看问题的文本,也有可能是这样要求的:

    String[] arr = {"-1", "0", "10", "20" };
    for (int i = 0; i < arr.length; i++) {
        arr[i] = String.format("%02x", Byte.parseByte(arr[i]));
    }
    System.out.println(java.util.Arrays.toString(arr));
    // prints "[ff, 00, 0a, 14]"

这里的几个答案使用Integer.toHexString(int);这是可行的,但有一些注意事项。由于形参是int型,因此对byte参数执行扩大原语转换,这涉及到符号扩展。

    byte b = -1;
    System.out.println(Integer.toHexString(b));
    // prints "ffffffff"

在Java中有符号的8位字节被符号扩展为32位整型。为了有效地撤销这个符号扩展,可以使用0xFF来屏蔽字节。

    byte b = -1;
    System.out.println(Integer.toHexString(b & 0xFF));
    // prints "ff"

使用toHexString的另一个问题是它不会用零填充:

    byte b = 10;
    System.out.println(Integer.toHexString(b & 0xFF));
    // prints "a"

这两个因素结合起来应该形成字符串。格式解决方案更佳。

参考文献

整型类型和值 对于字节,从-128到127,包括 JLS 5.1.2扩大原语转换

这是一条非常快的路。不需要外部库。

final protected static char[] HEXARRAY = "0123456789abcdef".toCharArray();

    public static String encodeHexString( 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] = HEXARRAY[v >>> 4];
            hexChars[j * 2 + 1] = HEXARRAY[v & 0x0F];
        }
        return new String(hexChars);
    }