Related
I'm trying to implement a string conversion from Base64 to Hex, that must yield the same results as
this website. Which means for example (Base64: bACAAAAAAAA=) is deconverted to (Hex: 6c00800000000000). This implementation in Javascript yields the correct output. So I tried to implement the equivalent in Java:
private static String base64ToHex(String input) {
byte[] raw = Base64.getDecoder().decode(input.getBytes());
String result = "";
for (int i = 0; i < raw.length; i++) {
String hex = Integer.toString(raw[i], 16);
result += (hex.length() == 2 ? hex : '0' + hex);
}
return result.toUpperCase();
}
Unfortunately this does not give the required output. So could you please give me hint about what I am missing?
Try below - You need to decode, not encode. Consider only value from decoded, ignore sign part. i.e take only absolute value, using Math.abs().
private static String base64ToHex(String input) {
byte[] raw = Base64.getDecoder().decode(input.getBytes());
String result = "";
for (int i = 0; i < raw.length; i++) {
String hex = Integer.toString(Math.abs(raw[i]), 16);
result += (hex.length() == 2 ? hex : '0' + hex);
}
return result.toUpperCase();
}
private static String base64ToHex(String input) {
byte[] raw = Base64.getDecoder().decode(
input.getBytes(StandardCharsets.US_ASCII));
StringBuilder result = new StringBuilder(raw.length * 2);
for (byte b : raw) {
result.append(String.format("%02X", b & 0xFF));
}
return result.toString();
}
The problem is that java byte is signed, hence the masking & 0xFF.
I am looking for a way to convert a long string (from a dump), that represents hex values into a byte array.
I couldn't have phrased it better than the person that posted the same question here.
But to keep it original, I'll phrase it my own way: suppose I have a string "00A0BF" that I would like interpreted as the
byte[] {0x00,0xA0,0xBf}
what should I do?
I am a Java novice and ended up using BigInteger and watching out for leading hex zeros. But I think it is ugly and I am sure I am missing something simple.
Update (2021) - Java 17 now includes java.util.HexFormat (only took 25 years):
HexFormat.of().parseHex(s)
For older versions of Java:
Here's a solution that I think is better than any posted so far:
/* s must be an even-length string. */
public static byte[] hexStringToByteArray(String s) {
int len = s.length();
byte[] data = new byte[len / 2];
for (int i = 0; i < len; i += 2) {
data[i / 2] = (byte) ((Character.digit(s.charAt(i), 16) << 4)
+ Character.digit(s.charAt(i+1), 16));
}
return data;
}
Reasons why it is an improvement:
Safe with leading zeros (unlike BigInteger) and with negative byte values (unlike Byte.parseByte)
Doesn't convert the String into a char[], or create StringBuilder and String objects for every single byte.
No library dependencies that may not be available
Feel free to add argument checking via assert or exceptions if the argument is not known to be safe.
One-liners:
import javax.xml.bind.DatatypeConverter;
public static String toHexString(byte[] array) {
return DatatypeConverter.printHexBinary(array);
}
public static byte[] toByteArray(String s) {
return DatatypeConverter.parseHexBinary(s);
}
Warnings:
in Java 9 Jigsaw this is no longer part of the (default) java.se root
set so it will result in a ClassNotFoundException unless you specify
--add-modules java.se.ee (thanks to #eckes)
Not available on Android (thanks to Fabian for noting that), but you can just take the source code if your system lacks javax.xml for some reason. Thanks to #Bert Regelink for extracting the source.
The Hex class in commons-codec should do that for you.
http://commons.apache.org/codec/
import org.apache.commons.codec.binary.Hex;
...
byte[] decoded = Hex.decodeHex("00A0BF");
// 0x00 0xA0 0xBF
You can now use BaseEncoding in guava to accomplish this.
BaseEncoding.base16().decode(string);
To reverse it use
BaseEncoding.base16().encode(bytes);
Actually, I think the BigInteger is solution is very nice:
new BigInteger("00A0BF", 16).toByteArray();
Edit: Not safe for leading zeros, as noted by the poster.
One-liners:
import javax.xml.bind.DatatypeConverter;
public static String toHexString(byte[] array) {
return DatatypeConverter.printHexBinary(array);
}
public static byte[] toByteArray(String s) {
return DatatypeConverter.parseHexBinary(s);
}
For those of you interested in the actual code behind the One-liners from FractalizeR (I needed that since javax.xml.bind is not available for Android (by default)), this comes from com.sun.xml.internal.bind.DatatypeConverterImpl.java :
public byte[] parseHexBinary(String s) {
final int len = s.length();
// "111" is not a valid hex encoding.
if( len%2 != 0 )
throw new IllegalArgumentException("hexBinary needs to be even-length: "+s);
byte[] out = new byte[len/2];
for( int i=0; i<len; i+=2 ) {
int h = hexToBin(s.charAt(i ));
int l = hexToBin(s.charAt(i+1));
if( h==-1 || l==-1 )
throw new IllegalArgumentException("contains illegal character for hexBinary: "+s);
out[i/2] = (byte)(h*16+l);
}
return out;
}
private static int hexToBin( char ch ) {
if( '0'<=ch && ch<='9' ) return ch-'0';
if( 'A'<=ch && ch<='F' ) return ch-'A'+10;
if( 'a'<=ch && ch<='f' ) return ch-'a'+10;
return -1;
}
private static final char[] hexCode = "0123456789ABCDEF".toCharArray();
public String printHexBinary(byte[] data) {
StringBuilder r = new StringBuilder(data.length*2);
for ( byte b : data) {
r.append(hexCode[(b >> 4) & 0xF]);
r.append(hexCode[(b & 0xF)]);
}
return r.toString();
}
The HexBinaryAdapter provides the ability to marshal and unmarshal between String and byte[].
import javax.xml.bind.annotation.adapters.HexBinaryAdapter;
public byte[] hexToBytes(String hexString) {
HexBinaryAdapter adapter = new HexBinaryAdapter();
byte[] bytes = adapter.unmarshal(hexString);
return bytes;
}
That's just an example I typed in...I actually just use it as is and don't need to make a separate method for using it.
Here is a method that actually works (based on several previous semi-correct answers):
private static byte[] fromHexString(final String encoded) {
if ((encoded.length() % 2) != 0)
throw new IllegalArgumentException("Input string must contain an even number of characters");
final byte result[] = new byte[encoded.length()/2];
final char enc[] = encoded.toCharArray();
for (int i = 0; i < enc.length; i += 2) {
StringBuilder curr = new StringBuilder(2);
curr.append(enc[i]).append(enc[i + 1]);
result[i/2] = (byte) Integer.parseInt(curr.toString(), 16);
}
return result;
}
The only possible issue that I can see is if the input string is extremely long; calling toCharArray() makes a copy of the string's internal array.
EDIT: Oh, and by the way, bytes are signed in Java, so your input string converts to [0, -96, -65] instead of [0, 160, 191]. But you probably knew that already.
In android ,if you are working with hex, you can try okio.
simple usage:
byte[] bytes = ByteString.decodeHex("c000060000").toByteArray();
and result will be
[-64, 0, 6, 0, 0]
The BigInteger() Method from java.math is very Slow and not recommandable.
Integer.parseInt(HEXString, 16)
can cause problems with some characters without
converting to Digit / Integer
a Well Working method:
Integer.decode("0xXX") .byteValue()
Function:
public static byte[] HexStringToByteArray(String s) {
byte data[] = new byte[s.length()/2];
for(int i=0;i < s.length();i+=2) {
data[i/2] = (Integer.decode("0x"+s.charAt(i)+s.charAt(i+1))).byteValue();
}
return data;
}
Have Fun, Good Luck
EDIT: as pointed out by #mmyers, this method doesn't work on input that contains substrings corresponding to bytes with the high bit set ("80" - "FF"). The explanation is at Bug ID: 6259307 Byte.parseByte not working as advertised in the SDK Documentation.
public static final byte[] fromHexString(final String s) {
byte[] arr = new byte[s.length()/2];
for ( int start = 0; start < s.length(); start += 2 )
{
String thisByte = s.substring(start, start+2);
arr[start/2] = Byte.parseByte(thisByte, 16);
}
return arr;
}
For what it's worth, here's another version which supports odd length strings, without resorting to string concatenation.
public static byte[] hexStringToByteArray(String input) {
int len = input.length();
if (len == 0) {
return new byte[] {};
}
byte[] data;
int startIdx;
if (len % 2 != 0) {
data = new byte[(len / 2) + 1];
data[0] = (byte) Character.digit(input.charAt(0), 16);
startIdx = 1;
} else {
data = new byte[len / 2];
startIdx = 0;
}
for (int i = startIdx; i < len; i += 2) {
data[(i + 1) / 2] = (byte) ((Character.digit(input.charAt(i), 16) << 4)
+ Character.digit(input.charAt(i+1), 16));
}
return data;
}
I like the Character.digit solution, but here is how I solved it
public byte[] hex2ByteArray( String hexString ) {
String hexVal = "0123456789ABCDEF";
byte[] out = new byte[hexString.length() / 2];
int n = hexString.length();
for( int i = 0; i < n; i += 2 ) {
//make a bit representation in an int of the hex value
int hn = hexVal.indexOf( hexString.charAt( i ) );
int ln = hexVal.indexOf( hexString.charAt( i + 1 ) );
//now just shift the high order nibble and add them together
out[i/2] = (byte)( ( hn << 4 ) | ln );
}
return out;
}
I've always used a method like
public static final byte[] fromHexString(final String s) {
String[] v = s.split(" ");
byte[] arr = new byte[v.length];
int i = 0;
for(String val: v) {
arr[i++] = Integer.decode("0x" + val).byteValue();
}
return arr;
}
this method splits on space delimited hex values but it wouldn't be hard to make it split the string on any other criteria such as into groupings of two characters.
The Code presented by Bert Regelink simply does not work.
Try the following:
import javax.xml.bind.DatatypeConverter;
import java.io.*;
public class Test
{
#Test
public void testObjectStreams( ) throws IOException, ClassNotFoundException
{
ByteArrayOutputStream baos = new ByteArrayOutputStream();
ObjectOutputStream oos = new ObjectOutputStream(baos);
String stringTest = "TEST";
oos.writeObject( stringTest );
oos.close();
baos.close();
byte[] bytes = baos.toByteArray();
String hexString = DatatypeConverter.printHexBinary( bytes);
byte[] reconvertedBytes = DatatypeConverter.parseHexBinary(hexString);
assertArrayEquals( bytes, reconvertedBytes );
ByteArrayInputStream bais = new ByteArrayInputStream(reconvertedBytes);
ObjectInputStream ois = new ObjectInputStream(bais);
String readString = (String) ois.readObject();
assertEquals( stringTest, readString);
}
}
I found Kernel Panic to have the solution most useful to me, but ran into problems if the hex string was an odd number. solved it this way:
boolean isOdd(int value)
{
return (value & 0x01) !=0;
}
private int hexToByte(byte[] out, int value)
{
String hexVal = "0123456789ABCDEF";
String hexValL = "0123456789abcdef";
String st = Integer.toHexString(value);
int len = st.length();
if (isOdd(len))
{
len+=1; // need length to be an even number.
st = ("0" + st); // make it an even number of chars
}
out[0]=(byte)(len/2);
for (int i =0;i<len;i+=2)
{
int hh = hexVal.indexOf(st.charAt(i));
if (hh == -1) hh = hexValL.indexOf(st.charAt(i));
int lh = hexVal.indexOf(st.charAt(i+1));
if (lh == -1) lh = hexValL.indexOf(st.charAt(i+1));
out[(i/2)+1] = (byte)((hh << 4)|lh);
}
return (len/2)+1;
}
I am adding a number of hex numbers to an array, so i pass the reference to the array I am using, and the int I need converted and returning the relative position of the next hex number. So the final byte array has [0] number of hex pairs, [1...] hex pairs, then the number of pairs...
Based on the op voted solution, the following should be a bit more efficient:
public static byte [] hexStringToByteArray (final String s) {
if (s == null || (s.length () % 2) == 1)
throw new IllegalArgumentException ();
final char [] chars = s.toCharArray ();
final int len = chars.length;
final byte [] data = new byte [len / 2];
for (int i = 0; i < len; i += 2) {
data[i / 2] = (byte) ((Character.digit (chars[i], 16) << 4) + Character.digit (chars[i + 1], 16));
}
return data;
}
Because: the initial conversion to a char array spares the length checks in charAt
If you have a preference for Java 8 streams as your coding style then this can be achieved using just JDK primitives.
String hex = "0001027f80fdfeff";
byte[] converted = IntStream.range(0, hex.length() / 2)
.map(i -> Character.digit(hex.charAt(i * 2), 16) << 4 | Character.digit(hex.charAt((i * 2) + 1), 16))
.collect(ByteArrayOutputStream::new,
ByteArrayOutputStream::write,
(s1, s2) -> s1.write(s2.toByteArray(), 0, s2.size()))
.toByteArray();
The , 0, s2.size() parameters in the collector concatenate function can be omitted if you don't mind catching IOException.
If your needs are more than just the occasional conversion then you can use HexUtils.
Example:
byte[] byteArray = Hex.hexStrToBytes("00A0BF");
This is the most simple case. Your input may contain delimiters (think MAC addresses, certificate thumbprints, etc), your input may be streaming, etc. In such cases it gets easier to justify to pull in an external library like HexUtils, however small.
With JDK 17 the HexFormat class will fulfill most needs and the need for something like HexUtils is greatly diminished. However, HexUtils can still be used for things like converting very large amounts to/from hex (streaming) or pretty printing hex (think wire dumps) which the JDK HexFormat class cannot do.
(full disclosure: I'm the author of HexUtils)
public static byte[] hex2ba(String sHex) throws Hex2baException {
if (1==sHex.length()%2) {
throw(new Hex2baException("Hex string need even number of chars"));
}
byte[] ba = new byte[sHex.length()/2];
for (int i=0;i<sHex.length()/2;i++) {
ba[i] = (Integer.decode(
"0x"+sHex.substring(i*2, (i+1)*2))).byteValue();
}
return ba;
}
My formal solution:
/**
* Decodes a hexadecimally encoded binary string.
* <p>
* Note that this function does <em>NOT</em> convert a hexadecimal number to a
* binary number.
*
* #param hex Hexadecimal representation of data.
* #return The byte[] representation of the given data.
* #throws NumberFormatException If the hexadecimal input string is of odd
* length or invalid hexadecimal string.
*/
public static byte[] hex2bin(String hex) throws NumberFormatException {
if (hex.length() % 2 > 0) {
throw new NumberFormatException("Hexadecimal input string must have an even length.");
}
byte[] r = new byte[hex.length() / 2];
for (int i = hex.length(); i > 0;) {
r[i / 2 - 1] = (byte) (digit(hex.charAt(--i)) | (digit(hex.charAt(--i)) << 4));
}
return r;
}
private static int digit(char ch) {
int r = Character.digit(ch, 16);
if (r < 0) {
throw new NumberFormatException("Invalid hexadecimal string: " + ch);
}
return r;
}
Is like the PHP hex2bin() Function but in Java style.
Example:
String data = new String(hex2bin("6578616d706c65206865782064617461"));
// data value: "example hex data"
Late to the party, but I have amalgamated the answer above by DaveL into a class with the reverse action - just in case it helps.
public final class HexString {
private static final char[] digits = "0123456789ABCDEF".toCharArray();
private HexString() {}
public static final String fromBytes(final byte[] bytes) {
final StringBuilder buf = new StringBuilder();
for (int i = 0; i < bytes.length; i++) {
buf.append(HexString.digits[(bytes[i] >> 4) & 0x0f]);
buf.append(HexString.digits[bytes[i] & 0x0f]);
}
return buf.toString();
}
public static final byte[] toByteArray(final String hexString) {
if ((hexString.length() % 2) != 0) {
throw new IllegalArgumentException("Input string must contain an even number of characters");
}
final int len = hexString.length();
final byte[] data = new byte[len / 2];
for (int i = 0; i < len; i += 2) {
data[i / 2] = (byte) ((Character.digit(hexString.charAt(i), 16) << 4)
+ Character.digit(hexString.charAt(i + 1), 16));
}
return data;
}
}
And JUnit test class:
public class TestHexString {
#Test
public void test() {
String[] tests = {"0FA1056D73", "", "00", "0123456789ABCDEF", "FFFFFFFF"};
for (int i = 0; i < tests.length; i++) {
String in = tests[i];
byte[] bytes = HexString.toByteArray(in);
String out = HexString.fromBytes(bytes);
System.out.println(in); //DEBUG
System.out.println(out); //DEBUG
Assert.assertEquals(in, out);
}
}
}
I know this is a very old thread, but still like to add my penny worth.
If I really need to code up a simple hex string to binary converter, I'd like to do it as follows.
public static byte[] hexToBinary(String s){
/*
* skipped any input validation code
*/
byte[] data = new byte[s.length()/2];
for( int i=0, j=0;
i<s.length() && j<data.length;
i+=2, j++)
{
data[j] = (byte)Integer.parseInt(s.substring(i, i+2), 16);
}
return data;
}
I think will do it for you. I cobbled it together from a similar function that returned the data as a string:
private static byte[] decode(String encoded) {
byte result[] = new byte[encoded/2];
char enc[] = encoded.toUpperCase().toCharArray();
StringBuffer curr;
for (int i = 0; i < enc.length; i += 2) {
curr = new StringBuffer("");
curr.append(String.valueOf(enc[i]));
curr.append(String.valueOf(enc[i + 1]));
result[i] = (byte) Integer.parseInt(curr.toString(), 16);
}
return result;
}
For Me this was the solution, HEX="FF01" then split to FF(255) and 01(01)
private static byte[] BytesEncode(String encoded) {
//System.out.println(encoded.length());
byte result[] = new byte[encoded.length() / 2];
char enc[] = encoded.toUpperCase().toCharArray();
String curr = "";
for (int i = 0; i < encoded.length(); i=i+2) {
curr = encoded.substring(i,i+2);
System.out.println(curr);
if(i==0){
result[i]=((byte) Integer.parseInt(curr, 16));
}else{
result[i/2]=((byte) Integer.parseInt(curr, 16));
}
}
return result;
}
I am trying to make an Android app that encrypts any given string in ATOM-128
example : input "hello" / output "MIH3+/CC+qCC"
I already tried it with c# (windows desktop app) and it's working well when I tried to do the same thing with java in android studio I have this results :
input "hello"
outut "2662144270646427486464"
THIS IS THE CODE
public class ATOM {
public static String Encrypt(String clearText)
{
String key = "/128GhIoPQROSTeUbADfgHijKLM+n0pFWXY456xyzB7=39VaqrstJklmNuZvwcdEC";
StringBuilder result = new StringBuilder();
int i = 0;
int[] indexes = new int[4];
int[] chars = new int[3];
do
{
chars[0] = i + 1 > clearText.length() ? 0 : (int)clearText.toCharArray()[i++];
chars[1] = i + 2 > clearText.length() ? 0 : (int)clearText.toCharArray()[i++];
chars[2] = i + 3 > clearText.length() ? 0 : (int)clearText.toCharArray()[i++];
indexes[0] = chars[0] >> 2;
indexes[1] = ((chars[0] & 3) << 4) | (chars[1] >> 4);
indexes[2] = ((chars[1] & 15) << 2) | (chars[2] >> 6);
indexes[3] = chars[2] & 63;
if ((char)chars[1] == 0)
{
indexes[2] = 64;
indexes[3] = 64;
}
else if ((char)chars[2] == 0)
{
indexes[3] = 64;
}
for (int index : indexes)
{
result.append(index);
}
}
while (i < clearText.length());
return result.toString();
}
public static String Decrypt(String clearText)
{
String key = "/128GhIoPQROSTeUbADfgHijKLM+n0pFWXY456xyzB7=39VaqrstJklmNuZvwcdEC";
StringBuilder result = new StringBuilder();
int[] indexes = new int[4];
int[] chars = new int[3];
int i = 0;
do
{
indexes[0] = key.indexOf(i++);
indexes[1] = key.indexOf(i++);
indexes[2] = key.indexOf(i++);
indexes[3] = key.indexOf(i++);
chars[0] = (indexes[0] << 2) | (indexes[1] >> 4);
chars[1] = (indexes[1] & 15) << 4 | (indexes[2] >> 2);
chars[2] = (indexes[2] & 3) << 6 | indexes[3];
result.append((char)chars[0]);
if (indexes[2] != 64)
result.append((char)chars[1]);
if (indexes[3] != 64)
result.append((char)chars[2]);
}
while (i < clearText.length());
return result.toString();
}
}
Looking at the StringBuilder doc :
public StringBuilder append (int i)
Appends the string representation of the specified int value.
The int value is converted to a string according to the rule defined by valueOf(int).
So everytime you call
result.append(index);
You append the string representation of the actual number. What you need to do first is to parse your int into, I guess, the char ASCII representation.
result.append(((char)index));
should do it.
Link to doc :
http://developer.android.com/reference/java/lang/StringBuilder.html#append(int)
I am looking for a way to convert a long string (from a dump), that represents hex values into a byte array.
I couldn't have phrased it better than the person that posted the same question here.
But to keep it original, I'll phrase it my own way: suppose I have a string "00A0BF" that I would like interpreted as the
byte[] {0x00,0xA0,0xBf}
what should I do?
I am a Java novice and ended up using BigInteger and watching out for leading hex zeros. But I think it is ugly and I am sure I am missing something simple.
Update (2021) - Java 17 now includes java.util.HexFormat (only took 25 years):
HexFormat.of().parseHex(s)
For older versions of Java:
Here's a solution that I think is better than any posted so far:
/* s must be an even-length string. */
public static byte[] hexStringToByteArray(String s) {
int len = s.length();
byte[] data = new byte[len / 2];
for (int i = 0; i < len; i += 2) {
data[i / 2] = (byte) ((Character.digit(s.charAt(i), 16) << 4)
+ Character.digit(s.charAt(i+1), 16));
}
return data;
}
Reasons why it is an improvement:
Safe with leading zeros (unlike BigInteger) and with negative byte values (unlike Byte.parseByte)
Doesn't convert the String into a char[], or create StringBuilder and String objects for every single byte.
No library dependencies that may not be available
Feel free to add argument checking via assert or exceptions if the argument is not known to be safe.
One-liners:
import javax.xml.bind.DatatypeConverter;
public static String toHexString(byte[] array) {
return DatatypeConverter.printHexBinary(array);
}
public static byte[] toByteArray(String s) {
return DatatypeConverter.parseHexBinary(s);
}
Warnings:
in Java 9 Jigsaw this is no longer part of the (default) java.se root
set so it will result in a ClassNotFoundException unless you specify
--add-modules java.se.ee (thanks to #eckes)
Not available on Android (thanks to Fabian for noting that), but you can just take the source code if your system lacks javax.xml for some reason. Thanks to #Bert Regelink for extracting the source.
The Hex class in commons-codec should do that for you.
http://commons.apache.org/codec/
import org.apache.commons.codec.binary.Hex;
...
byte[] decoded = Hex.decodeHex("00A0BF");
// 0x00 0xA0 0xBF
You can now use BaseEncoding in guava to accomplish this.
BaseEncoding.base16().decode(string);
To reverse it use
BaseEncoding.base16().encode(bytes);
Actually, I think the BigInteger is solution is very nice:
new BigInteger("00A0BF", 16).toByteArray();
Edit: Not safe for leading zeros, as noted by the poster.
One-liners:
import javax.xml.bind.DatatypeConverter;
public static String toHexString(byte[] array) {
return DatatypeConverter.printHexBinary(array);
}
public static byte[] toByteArray(String s) {
return DatatypeConverter.parseHexBinary(s);
}
For those of you interested in the actual code behind the One-liners from FractalizeR (I needed that since javax.xml.bind is not available for Android (by default)), this comes from com.sun.xml.internal.bind.DatatypeConverterImpl.java :
public byte[] parseHexBinary(String s) {
final int len = s.length();
// "111" is not a valid hex encoding.
if( len%2 != 0 )
throw new IllegalArgumentException("hexBinary needs to be even-length: "+s);
byte[] out = new byte[len/2];
for( int i=0; i<len; i+=2 ) {
int h = hexToBin(s.charAt(i ));
int l = hexToBin(s.charAt(i+1));
if( h==-1 || l==-1 )
throw new IllegalArgumentException("contains illegal character for hexBinary: "+s);
out[i/2] = (byte)(h*16+l);
}
return out;
}
private static int hexToBin( char ch ) {
if( '0'<=ch && ch<='9' ) return ch-'0';
if( 'A'<=ch && ch<='F' ) return ch-'A'+10;
if( 'a'<=ch && ch<='f' ) return ch-'a'+10;
return -1;
}
private static final char[] hexCode = "0123456789ABCDEF".toCharArray();
public String printHexBinary(byte[] data) {
StringBuilder r = new StringBuilder(data.length*2);
for ( byte b : data) {
r.append(hexCode[(b >> 4) & 0xF]);
r.append(hexCode[(b & 0xF)]);
}
return r.toString();
}
The HexBinaryAdapter provides the ability to marshal and unmarshal between String and byte[].
import javax.xml.bind.annotation.adapters.HexBinaryAdapter;
public byte[] hexToBytes(String hexString) {
HexBinaryAdapter adapter = new HexBinaryAdapter();
byte[] bytes = adapter.unmarshal(hexString);
return bytes;
}
That's just an example I typed in...I actually just use it as is and don't need to make a separate method for using it.
Here is a method that actually works (based on several previous semi-correct answers):
private static byte[] fromHexString(final String encoded) {
if ((encoded.length() % 2) != 0)
throw new IllegalArgumentException("Input string must contain an even number of characters");
final byte result[] = new byte[encoded.length()/2];
final char enc[] = encoded.toCharArray();
for (int i = 0; i < enc.length; i += 2) {
StringBuilder curr = new StringBuilder(2);
curr.append(enc[i]).append(enc[i + 1]);
result[i/2] = (byte) Integer.parseInt(curr.toString(), 16);
}
return result;
}
The only possible issue that I can see is if the input string is extremely long; calling toCharArray() makes a copy of the string's internal array.
EDIT: Oh, and by the way, bytes are signed in Java, so your input string converts to [0, -96, -65] instead of [0, 160, 191]. But you probably knew that already.
In android ,if you are working with hex, you can try okio.
simple usage:
byte[] bytes = ByteString.decodeHex("c000060000").toByteArray();
and result will be
[-64, 0, 6, 0, 0]
The BigInteger() Method from java.math is very Slow and not recommandable.
Integer.parseInt(HEXString, 16)
can cause problems with some characters without
converting to Digit / Integer
a Well Working method:
Integer.decode("0xXX") .byteValue()
Function:
public static byte[] HexStringToByteArray(String s) {
byte data[] = new byte[s.length()/2];
for(int i=0;i < s.length();i+=2) {
data[i/2] = (Integer.decode("0x"+s.charAt(i)+s.charAt(i+1))).byteValue();
}
return data;
}
Have Fun, Good Luck
EDIT: as pointed out by #mmyers, this method doesn't work on input that contains substrings corresponding to bytes with the high bit set ("80" - "FF"). The explanation is at Bug ID: 6259307 Byte.parseByte not working as advertised in the SDK Documentation.
public static final byte[] fromHexString(final String s) {
byte[] arr = new byte[s.length()/2];
for ( int start = 0; start < s.length(); start += 2 )
{
String thisByte = s.substring(start, start+2);
arr[start/2] = Byte.parseByte(thisByte, 16);
}
return arr;
}
For what it's worth, here's another version which supports odd length strings, without resorting to string concatenation.
public static byte[] hexStringToByteArray(String input) {
int len = input.length();
if (len == 0) {
return new byte[] {};
}
byte[] data;
int startIdx;
if (len % 2 != 0) {
data = new byte[(len / 2) + 1];
data[0] = (byte) Character.digit(input.charAt(0), 16);
startIdx = 1;
} else {
data = new byte[len / 2];
startIdx = 0;
}
for (int i = startIdx; i < len; i += 2) {
data[(i + 1) / 2] = (byte) ((Character.digit(input.charAt(i), 16) << 4)
+ Character.digit(input.charAt(i+1), 16));
}
return data;
}
I like the Character.digit solution, but here is how I solved it
public byte[] hex2ByteArray( String hexString ) {
String hexVal = "0123456789ABCDEF";
byte[] out = new byte[hexString.length() / 2];
int n = hexString.length();
for( int i = 0; i < n; i += 2 ) {
//make a bit representation in an int of the hex value
int hn = hexVal.indexOf( hexString.charAt( i ) );
int ln = hexVal.indexOf( hexString.charAt( i + 1 ) );
//now just shift the high order nibble and add them together
out[i/2] = (byte)( ( hn << 4 ) | ln );
}
return out;
}
I've always used a method like
public static final byte[] fromHexString(final String s) {
String[] v = s.split(" ");
byte[] arr = new byte[v.length];
int i = 0;
for(String val: v) {
arr[i++] = Integer.decode("0x" + val).byteValue();
}
return arr;
}
this method splits on space delimited hex values but it wouldn't be hard to make it split the string on any other criteria such as into groupings of two characters.
The Code presented by Bert Regelink simply does not work.
Try the following:
import javax.xml.bind.DatatypeConverter;
import java.io.*;
public class Test
{
#Test
public void testObjectStreams( ) throws IOException, ClassNotFoundException
{
ByteArrayOutputStream baos = new ByteArrayOutputStream();
ObjectOutputStream oos = new ObjectOutputStream(baos);
String stringTest = "TEST";
oos.writeObject( stringTest );
oos.close();
baos.close();
byte[] bytes = baos.toByteArray();
String hexString = DatatypeConverter.printHexBinary( bytes);
byte[] reconvertedBytes = DatatypeConverter.parseHexBinary(hexString);
assertArrayEquals( bytes, reconvertedBytes );
ByteArrayInputStream bais = new ByteArrayInputStream(reconvertedBytes);
ObjectInputStream ois = new ObjectInputStream(bais);
String readString = (String) ois.readObject();
assertEquals( stringTest, readString);
}
}
I found Kernel Panic to have the solution most useful to me, but ran into problems if the hex string was an odd number. solved it this way:
boolean isOdd(int value)
{
return (value & 0x01) !=0;
}
private int hexToByte(byte[] out, int value)
{
String hexVal = "0123456789ABCDEF";
String hexValL = "0123456789abcdef";
String st = Integer.toHexString(value);
int len = st.length();
if (isOdd(len))
{
len+=1; // need length to be an even number.
st = ("0" + st); // make it an even number of chars
}
out[0]=(byte)(len/2);
for (int i =0;i<len;i+=2)
{
int hh = hexVal.indexOf(st.charAt(i));
if (hh == -1) hh = hexValL.indexOf(st.charAt(i));
int lh = hexVal.indexOf(st.charAt(i+1));
if (lh == -1) lh = hexValL.indexOf(st.charAt(i+1));
out[(i/2)+1] = (byte)((hh << 4)|lh);
}
return (len/2)+1;
}
I am adding a number of hex numbers to an array, so i pass the reference to the array I am using, and the int I need converted and returning the relative position of the next hex number. So the final byte array has [0] number of hex pairs, [1...] hex pairs, then the number of pairs...
Based on the op voted solution, the following should be a bit more efficient:
public static byte [] hexStringToByteArray (final String s) {
if (s == null || (s.length () % 2) == 1)
throw new IllegalArgumentException ();
final char [] chars = s.toCharArray ();
final int len = chars.length;
final byte [] data = new byte [len / 2];
for (int i = 0; i < len; i += 2) {
data[i / 2] = (byte) ((Character.digit (chars[i], 16) << 4) + Character.digit (chars[i + 1], 16));
}
return data;
}
Because: the initial conversion to a char array spares the length checks in charAt
If you have a preference for Java 8 streams as your coding style then this can be achieved using just JDK primitives.
String hex = "0001027f80fdfeff";
byte[] converted = IntStream.range(0, hex.length() / 2)
.map(i -> Character.digit(hex.charAt(i * 2), 16) << 4 | Character.digit(hex.charAt((i * 2) + 1), 16))
.collect(ByteArrayOutputStream::new,
ByteArrayOutputStream::write,
(s1, s2) -> s1.write(s2.toByteArray(), 0, s2.size()))
.toByteArray();
The , 0, s2.size() parameters in the collector concatenate function can be omitted if you don't mind catching IOException.
If your needs are more than just the occasional conversion then you can use HexUtils.
Example:
byte[] byteArray = Hex.hexStrToBytes("00A0BF");
This is the most simple case. Your input may contain delimiters (think MAC addresses, certificate thumbprints, etc), your input may be streaming, etc. In such cases it gets easier to justify to pull in an external library like HexUtils, however small.
With JDK 17 the HexFormat class will fulfill most needs and the need for something like HexUtils is greatly diminished. However, HexUtils can still be used for things like converting very large amounts to/from hex (streaming) or pretty printing hex (think wire dumps) which the JDK HexFormat class cannot do.
(full disclosure: I'm the author of HexUtils)
public static byte[] hex2ba(String sHex) throws Hex2baException {
if (1==sHex.length()%2) {
throw(new Hex2baException("Hex string need even number of chars"));
}
byte[] ba = new byte[sHex.length()/2];
for (int i=0;i<sHex.length()/2;i++) {
ba[i] = (Integer.decode(
"0x"+sHex.substring(i*2, (i+1)*2))).byteValue();
}
return ba;
}
My formal solution:
/**
* Decodes a hexadecimally encoded binary string.
* <p>
* Note that this function does <em>NOT</em> convert a hexadecimal number to a
* binary number.
*
* #param hex Hexadecimal representation of data.
* #return The byte[] representation of the given data.
* #throws NumberFormatException If the hexadecimal input string is of odd
* length or invalid hexadecimal string.
*/
public static byte[] hex2bin(String hex) throws NumberFormatException {
if (hex.length() % 2 > 0) {
throw new NumberFormatException("Hexadecimal input string must have an even length.");
}
byte[] r = new byte[hex.length() / 2];
for (int i = hex.length(); i > 0;) {
r[i / 2 - 1] = (byte) (digit(hex.charAt(--i)) | (digit(hex.charAt(--i)) << 4));
}
return r;
}
private static int digit(char ch) {
int r = Character.digit(ch, 16);
if (r < 0) {
throw new NumberFormatException("Invalid hexadecimal string: " + ch);
}
return r;
}
Is like the PHP hex2bin() Function but in Java style.
Example:
String data = new String(hex2bin("6578616d706c65206865782064617461"));
// data value: "example hex data"
Late to the party, but I have amalgamated the answer above by DaveL into a class with the reverse action - just in case it helps.
public final class HexString {
private static final char[] digits = "0123456789ABCDEF".toCharArray();
private HexString() {}
public static final String fromBytes(final byte[] bytes) {
final StringBuilder buf = new StringBuilder();
for (int i = 0; i < bytes.length; i++) {
buf.append(HexString.digits[(bytes[i] >> 4) & 0x0f]);
buf.append(HexString.digits[bytes[i] & 0x0f]);
}
return buf.toString();
}
public static final byte[] toByteArray(final String hexString) {
if ((hexString.length() % 2) != 0) {
throw new IllegalArgumentException("Input string must contain an even number of characters");
}
final int len = hexString.length();
final byte[] data = new byte[len / 2];
for (int i = 0; i < len; i += 2) {
data[i / 2] = (byte) ((Character.digit(hexString.charAt(i), 16) << 4)
+ Character.digit(hexString.charAt(i + 1), 16));
}
return data;
}
}
And JUnit test class:
public class TestHexString {
#Test
public void test() {
String[] tests = {"0FA1056D73", "", "00", "0123456789ABCDEF", "FFFFFFFF"};
for (int i = 0; i < tests.length; i++) {
String in = tests[i];
byte[] bytes = HexString.toByteArray(in);
String out = HexString.fromBytes(bytes);
System.out.println(in); //DEBUG
System.out.println(out); //DEBUG
Assert.assertEquals(in, out);
}
}
}
I know this is a very old thread, but still like to add my penny worth.
If I really need to code up a simple hex string to binary converter, I'd like to do it as follows.
public static byte[] hexToBinary(String s){
/*
* skipped any input validation code
*/
byte[] data = new byte[s.length()/2];
for( int i=0, j=0;
i<s.length() && j<data.length;
i+=2, j++)
{
data[j] = (byte)Integer.parseInt(s.substring(i, i+2), 16);
}
return data;
}
I think will do it for you. I cobbled it together from a similar function that returned the data as a string:
private static byte[] decode(String encoded) {
byte result[] = new byte[encoded/2];
char enc[] = encoded.toUpperCase().toCharArray();
StringBuffer curr;
for (int i = 0; i < enc.length; i += 2) {
curr = new StringBuffer("");
curr.append(String.valueOf(enc[i]));
curr.append(String.valueOf(enc[i + 1]));
result[i] = (byte) Integer.parseInt(curr.toString(), 16);
}
return result;
}
For Me this was the solution, HEX="FF01" then split to FF(255) and 01(01)
private static byte[] BytesEncode(String encoded) {
//System.out.println(encoded.length());
byte result[] = new byte[encoded.length() / 2];
char enc[] = encoded.toUpperCase().toCharArray();
String curr = "";
for (int i = 0; i < encoded.length(); i=i+2) {
curr = encoded.substring(i,i+2);
System.out.println(curr);
if(i==0){
result[i]=((byte) Integer.parseInt(curr, 16));
}else{
result[i/2]=((byte) Integer.parseInt(curr, 16));
}
}
return result;
}
I have tried many searches to find a way to search a string for byte code. Here is an example:
String stringThatHasBytes = "hello world hello world[B#9304b1";
If stringThatHasBytes . Does have bytes {
return true or false
}
Is there a method that can search a String for bytes?
You can't do this, in short. Because the printing of a byte changes everytime you print it out. Printing out the bytes doesn't print the actual bytes and it's meaningless if you're looking for exact comparison of bytes.
However, if you only look for any byte printing in the string, just control for [B#s in the string and return true.
String stringThatHasBytes = "hello world hello world[B#9304b1";
if (stringThatHasBytes.indexOf("[B#") >= 0)
return true;
} else return false;
EDIT:
If you need ways of printing bytes in a meaningful way, you should convert the bytes to some meaningful text such as:
public static String convertByteArrayToHexString(byte[] b) {
if (b != null) {
StringBuilder s = new StringBuilder(2 * b.length);
for (int i = 0; i < b.length; ++i) {
final String t = Integer.toHexString(b[i]);
final int l = t.length();
if (l > 2) {
s.append(t.substring(l - 2));
} else {
if (l == 1) {
s.append("0");
}
s.append(t);
}
}
return s.toString();
} else {
return "";
}
}
Would contains be too simple of an answer for you?
boolean hasBytes = str.contains("[B#");
If so let me know I'll show you some good regex! but that should be sufficient.
see if this help
byte[] myByteArray = new byte[5];
myByteArray[0] = 'a';
myByteArray[1] = 'b';
myByteArray[2] = 'c';
myByteArray[3] = 'd';
myByteArray[4] = 'e';
for (byte x : myByteArray)
System.out.println(x);
String myString = "abcde";
System.out.println(myString.equals(new String(myByteArray)));