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
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);
}
其他回答
我发现了三种不同的方法: http://www.rgagnon.com/javadetails/java-0596.html
他还提到,我认为最优雅的一个是:
static final String HEXES = "0123456789ABCDEF";
public static String getHex( byte [] raw ) {
if ( raw == null ) {
return null;
}
final StringBuilder hex = new StringBuilder( 2 * raw.length );
for ( final byte b : raw ) {
hex.append(HEXES.charAt((b & 0xF0) >> 4))
.append(HEXES.charAt((b & 0x0F)));
}
return hex.toString();
}
@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
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();
}
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);
}
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;
}
}
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