for example, for 1, 2, 128, 256 the output can be (16 digits):
0000000000000001
0000000000000010
0000000010000000
0000000100000000
I tried
String.format("%16s", Integer.toBinaryString(1));
it puts spaces for left-padding:
` 1'
How to put 0s for padding. I couldn't find it in Formatter. Is there another way to do it?
P.S. this post describes how to format integers with left 0-padding, but it is not for the binary representation.
I think this is a suboptimal solution, but you could do
String.format("%16s", Integer.toBinaryString(1)).replace(' ', '0')
There is no binary conversion built into the java.util.Formatter, I would advise you to either use String.replace to replace space character with zeros, as in:
String.format("%16s", Integer.toBinaryString(1)).replace(" ", "0")
Or implement your own logic to convert integers to binary representation with added left padding somewhere along the lines given in this so.
Or if you really need to pass numbers to format, you can convert your binary representation to BigInteger and then format that with leading zeros, but this is very costly at runtime, as in:
String.format("%016d", new BigInteger(Integer.toBinaryString(1)))
Here a new answer for an old post.
To pad a binary value with leading zeros to a specific length, try this:
Integer.toBinaryString( (1 << len) | val ).substring( 1 )
If len = 4 and val = 1,
Integer.toBinaryString( (1 << len) | val )
returns the string "10001", then
"10001".substring( 1 )
discards the very first character. So we obtain what we want:
"0001"
If val is likely to be negative, rather try:
Integer.toBinaryString( (1 << len) | (val & ((1 << len) - 1)) ).substring( 1 )
You can use Apache Commons StringUtils. It offers methods for padding strings:
StringUtils.leftPad(Integer.toBinaryString(1), 16, '0');
I was trying all sorts of method calls that I haven't really used before to make this work, they worked with moderate success, until I thought of something that is so simple it just might work, and it did!
I'm sure it's been thought of before, not sure if it's any good for long string of binary codes but it works fine for 16Bit strings. Hope it helps!! (Note second piece of code is improved)
String binString = Integer.toBinaryString(256);
while (binString.length() < 16) { //pad with 16 0's
binString = "0" + binString;
}
Thanks to Will on helping improve this answer to make it work with out a loop.
This maybe a little clumsy but it works, please improve and comment back if you can....
binString = Integer.toBinaryString(256);
int length = 16 - binString.length();
char[] padArray = new char[length];
Arrays.fill(padArray, '0');
String padString = new String(padArray);
binString = padString + binString;
A simpler version of user3608934's idea "This is an old trick, create a string with 16 0's then append the trimmed binary string you got ":
private String toBinaryString32(int i) {
String binaryWithOutLeading0 = Integer.toBinaryString(i);
return "00000000000000000000000000000000"
.substring(binaryWithOutLeading0.length())
+ binaryWithOutLeading0;
}
I do not know "right" solution but I can suggest you a fast patch.
String.format("%16s", Integer.toBinaryString(1)).replace(" ", "0");
I have just tried it and saw that it works fine.
Starting with Java 11, you can use the repeat(...) method:
"0".repeat(Integer.numberOfLeadingZeros(i) - 16) + Integer.toBinaryString(i)
Or, if you need 32-bit representation of any integer:
"0".repeat(Integer.numberOfLeadingZeros(i != 0 ? i : 1)) + Integer.toBinaryString(i)
try...
String.format("%016d\n", Integer.parseInt(Integer.toBinaryString(256)));
I dont think this is the "correct" way to doing this... but it works :)
I would write my own util class with the method like below
public class NumberFormatUtils {
public static String longToBinString(long val) {
char[] buffer = new char[64];
Arrays.fill(buffer, '0');
for (int i = 0; i < 64; ++i) {
long mask = 1L << i;
if ((val & mask) == mask) {
buffer[63 - i] = '1';
}
}
return new String(buffer);
}
public static void main(String... args) {
long value = 0b0000000000000000000000000000000000000000000000000000000000000101L;
System.out.println(value);
System.out.println(Long.toBinaryString(value));
System.out.println(NumberFormatUtils.longToBinString(value));
}
}
Output:
5
101
0000000000000000000000000000000000000000000000000000000000000101
The same approach could be applied to any integral types. Pay attention to the type of mask
long mask = 1L << i;
A naive solution that work would be
String temp = Integer.toBinaryString(5);
while (temp.length() < Integer.SIZE) temp = "0"+temp; //pad leading zeros
temp = temp.substring(Integer.SIZE - Short.SIZE); //remove excess
One other method would be
String temp = Integer.toBinaryString((m | 0x80000000));
temp = temp.substring(Integer.SIZE - Short.SIZE);
This will produce a 16 bit string of the integer 5
// Below will handle proper sizes
public static String binaryString(int i) {
return String.format("%" + Integer.SIZE + "s", Integer.toBinaryString(i)).replace(' ', '0');
}
public static String binaryString(long i) {
return String.format("%" + Long.SIZE + "s", Long.toBinaryString(i)).replace(' ', '0');
}
This is an old trick, create a string with 16 0's then append the trimmed binary string you got from String.format("%s", Integer.toBinaryString(1)) and use the right-most 16 characters, lopping off any leading 0's. Better yet, make a function that lets you specify how long of a binary string you want. Of course there are probably a bazillion other ways to accomplish this including libraries, but I'm adding this post to help out a friend :)
public class BinaryPrinter {
public static void main(String[] args) {
System.out.format("%d in binary is %s\n", 1, binaryString(1, 4));
System.out.format("%d in binary is %s\n", 128, binaryString(128, 8));
System.out.format("%d in binary is %s\n", 256, binaryString(256, 16));
}
public static String binaryString( final int number, final int binaryDigits ) {
final String pattern = String.format( "%%0%dd", binaryDigits );
final String padding = String.format( pattern, 0 );
final String response = String.format( "%s%s", padding, Integer.toBinaryString(number) );
System.out.format( "\npattern = '%s'\npadding = '%s'\nresponse = '%s'\n\n", pattern, padding, response );
return response.substring( response.length() - binaryDigits );
}
}
This method converts an int to a String, length=bits. Either padded with 0s or with the most significant bits truncated.
static String toBitString( int x, int bits ){
String bitString = Integer.toBinaryString(x);
int size = bitString.length();
StringBuilder sb = new StringBuilder( bits );
if( bits > size ){
for( int i=0; i<bits-size; i++ )
sb.append('0');
sb.append( bitString );
}else
sb = sb.append( bitString.substring(size-bits, size) );
return sb.toString();
}
You can use lib https://github.com/kssource/BitSequence. It accept a number and return bynary string, padded and/or grouped.
String s = new BitSequence(2, 16).toBynaryString(ALIGN.RIGHT, GROUP.CONTINOUSLY));
return
0000000000000010
another examples:
[10, -20, 30]->00001010 11101100 00011110
i=-10->00000000000000000000000000001010
bi=10->1010
sh=10->00 0000 0000 1010
l=10->00000001 010
by=-10->1010
i=-10->bc->11111111 11111111 11111111 11110110
for(int i=0;i<n;i++)
{
for(int j=str[i].length();j<4;j++)
str[i]="0".concat(str[i]);
}
str[i].length() is length of number say 2 in binary is 01 which is length 2
change 4 to desired max length of number. This can be optimized to O(n).
by using continue.
import java.util.Scanner;
public class Q3{
public static void main(String[] args) {
Scanner scn=new Scanner(System.in);
System.out.println("Enter a number:");
int num=scn.nextInt();
int numB=Integer.parseInt(Integer.toBinaryString(num));
String strB=String.format("%08d",numB);//makes a 8 character code
if(num>=1 && num<=255){
System.out.println(strB);
}else{
System.out.println("Number should be in range between 1 and 255");
}
}
}
I want to calculate the size of the bitcoin block
And I saw the subject: link here,
and in one response someone says "merkle root hash - 32 bytes"
Then I went to this block: link here
merkle root is: 8f5318a45316efedf56ebcadd55889ab377ba78d9a8ac14a81a83e52ad45e07f
i tried to calculate the root, but it appeared with me that it was 64 bytes.
my way of calculating bytes:
byte[] bytes1 = "Here merkle root".getBytes();
System.out.println(bytes1.length);
Result 64 rather than 32
The merkle root is 32 bytes, presented with hex encoding (using 0-9a-z). That means that two characters will be used to represent each byte and the result string will have twice as many characters as the original data has bytes.
You can not convert a string to bytes by doing getBytes() as that gets the bytes representing the string.
You can parse the string into either a number (a big number) with BigInteger or into a byte array manually or using external library (not included in the JDK).
public static void main(String[] args) {
String merkle = "8f5318a45316efedf56ebcadd55889ab377ba78d9a8ac14a81a83e52ad45e07f";
System.out.println("String length: " + merkle.length());
System.out.println("Actual decimal value: " + new BigInteger(merkle, 16));
byte[] bytes = toBytes(merkle);
System.out.println("Total bytes: " + bytes.length);
System.out.println("Actual bytes: " + Arrays.toString(bytes));
}
private static byte[] toBytes(String input) {
byte[] result = new byte[input.length() / 2];
for (int i = 0; i < result.length; i++) {
result[i] = (byte)((Character.digit(input.charAt(i * 2), 16) << 4)
+ Character.digit(input.charAt(i * 2 + 1), 16));
}
return result;
}
How do I shift a byte array n positions to the right? For instance shifting a 16 byte array right 29 positions? I read somewhere it can be done using a long? Would using a long work like this:
Long k1 = byte array from 0 to 7
Long k2 = byte array from 8 to 15
Then right rotating these two longs using Long.rotateRight(Long x, number of rotations).How would the two longs be joined back into a byte array?
I believe you can do this using java.math.BigInteger which supports shifts on arbitrarily large numbers. This has advantage of simplicity, but disadvantage of not padding into original byte array size, i.e. input could be 16 bytes but output might only be 10 etc, requiring additional logic.
BigInteger approach
byte [] array = new byte[]{0x7F,0x11,0x22,0x33,0x44,0x55,0x66,0x77};
// create from array
BigInteger bigInt = new BigInteger(array);
// shift
BigInteger shiftInt = bigInt.shiftRight(4);
// back to array
byte [] shifted = shiftInt.toByteArray();
// print it as hex
for (byte b : shifted) {
System.out.print(String.format("%x", b));
}
Output
7f1122334455667 <== shifted 4 to the right. Looks OK
Long manipulation
I don't know why you'd want to do this as rotateRight() as this makes life more difficult, you have to blank at the bits that appear at the left hand side in K1 etc. You'd be better with using shift IMO as describe below. I've used a shift of 20 as divisible by 4 so easier to see the nibbles move in the output.
1) Use ByteBuffer to form two longs from 16 byte array
byte[] array = { 0x00, 0x00, 0x11, 0x11, 0x22, 0x22, 0x33, 0x33, 0x44, 0x44, 0x55, 0x55, 0x66, 0x66, 0x77, 0x77 };
ByteBuffer buffer = ByteBuffer.wrap(array);
long k1 = buffer.getLong();
long k2 = buffer.getLong();
2) Shift each long n bits to the right
int n = 20;
long k1Shift = k1 >> n;
long k2Shift = k2 >> n;
System.out.println(String.format("%016x => %016x", k1, k1Shift));
System.out.println(String.format("%016x => %016x", k2, k2Shift));
0000111122223333 => 0000000001111222
4444555566667777 => 0000044445555666
Determine bits from k1 that "got pushed off the edge"
long k1CarryBits = (k1 << (64 - n));
System.out.println(String.format("%016x => %016x", k1, k1CarryBits));
0000111122223333 => 2333300000000000
Join the K1 carry bits onto K2 on right hand side
long k2WithCarray = k2Shift | k1CarryBits;
System.out.println(String.format("%016x => %016x", k2Shift, k2WithCarray));
0000044445555666 => 2333344445555666
Write the two longs back into a ByteBuffer and extract as a byte array
buffer.position(0);
buffer.putLong(k1Shift);
buffer.putLong(k2WithCarray);
for (byte each : buffer.array()) {
System.out.print(Long.toHexString(each));
}
000011112222333344445555666
Here is what I came up with to shift a byte array by some arbitrary number of bits left:
/**
* Shifts input byte array len bits left.This method will alter the input byte array.
*/
public static byte[] shiftLeft(byte[] data, int len) {
int word_size = (len / 8) + 1;
int shift = len % 8;
byte carry_mask = (byte) ((1 << shift) - 1);
int offset = word_size - 1;
for (int i = 0; i < data.length; i++) {
int src_index = i+offset;
if (src_index >= data.length) {
data[i] = 0;
} else {
byte src = data[src_index];
byte dst = (byte) (src << shift);
if (src_index+1 < data.length) {
dst |= data[src_index+1] >>> (8-shift) & carry_mask;
}
data[i] = dst;
}
}
return data;
}
1. Manually implemented
Here are left and right shift implementation without using BigInteger (ie. without creating a copy of the input array) and with unsigned right shift (BigInteger only supports arithmetic shifts of course)
Left Shift <<
/**
* Left shift of whole byte array by shiftBitCount bits.
* This method will alter the input byte array.
*/
static byte[] shiftLeft(byte[] byteArray, int shiftBitCount) {
final int shiftMod = shiftBitCount % 8;
final byte carryMask = (byte) ((1 << shiftMod) - 1);
final int offsetBytes = (shiftBitCount / 8);
int sourceIndex;
for (int i = 0; i < byteArray.length; i++) {
sourceIndex = i + offsetBytes;
if (sourceIndex >= byteArray.length) {
byteArray[i] = 0;
} else {
byte src = byteArray[sourceIndex];
byte dst = (byte) (src << shiftMod);
if (sourceIndex + 1 < byteArray.length) {
dst |= byteArray[sourceIndex + 1] >>> (8 - shiftMod) & carryMask;
}
byteArray[i] = dst;
}
}
return byteArray;
}
Unsigned Right Shift >>>
/**
* Unsigned/logical right shift of whole byte array by shiftBitCount bits.
* This method will alter the input byte array.
*/
static byte[] shiftRight(byte[] byteArray, int shiftBitCount) {
final int shiftMod = shiftBitCount % 8;
final byte carryMask = (byte) (0xFF << (8 - shiftMod));
final int offsetBytes = (shiftBitCount / 8);
int sourceIndex;
for (int i = byteArray.length - 1; i >= 0; i--) {
sourceIndex = i - offsetBytes;
if (sourceIndex < 0) {
byteArray[i] = 0;
} else {
byte src = byteArray[sourceIndex];
byte dst = (byte) ((0xff & src) >>> shiftMod);
if (sourceIndex - 1 >= 0) {
dst |= byteArray[sourceIndex - 1] << (8 - shiftMod) & carryMask;
}
byteArray[i] = dst;
}
}
return byteArray;
}
Used in this class by this Project.
2. Using BigInteger
Be aware that BigInteger internally converts the byte array into an int[] array so this may not be the most optimized solution:
Arithmetic Left Shift <<:
byte[] result = new BigInteger(byteArray).shiftLeft(3).toByteArray();
Arithmetic Right Shift >>:
byte[] result = new BigInteger(byteArray).shiftRight(2).toByteArray();
3. External Library
Using the Bytes java library*:
Add to pom.xml:
<dependency>
<groupId>at.favre.lib</groupId>
<artifactId>bytes</artifactId>
<version>{latest-version}</version>
</dependency>
Code example:
Bytes b = Bytes.wrap(someByteArray);
b.leftShift(3);
b.rightShift(3);
byte[] result = b.array();
*Full Disclaimer: I am the developer.
The is an old post, but I want to update Adam's answer.
The long solution works with a few tweak.
In order to rotate, use >>> instead of >>, because >> will pad with significant bit, changing the original value.
second, the printbyte function seems to miss leading 00 when it prints.
use this instead.
private String getHexString(byte[] b) {
StringBuilder result = new StringBuilder();
for (int i = 0; i < b.length; i++)
result.append(Integer.toString((b[i] & 0xff) + 0x100, 16)
.substring(1));
return result.toString();
}
I am looking a way to convert a string to BCD equivalent. I use Java, but it is not a question of the language indeed. I am trying to understand step by step how to convert a string to BCD.
For example, suppose I have the following string;
"0200" (This string has four ASCII characters, if we were in java this string had been contained in a byte[4] where byte[0] = 48, byte[1] = 50, byte[2] = 48 and byte[3] = 48)
In BCD (according this page: http://es.wikipedia.org/wiki/Decimal_codificado_en_binario):
0 = 0000
2 = 0010
0 = 0000
0 = 0000
Ok, I think the conversion is correct but I have to save this in a byte[2]. What Should I have to do? After, I have to read the BCD and convert it to the original string "0200" but first I have to resolve String to BCD.
Find a utility class to do this for you. Surely someone out there has written a BCD conversion utility for Java.
Here you go. I Googled "BCD Java" and got this as the first result. Copying code here for future reference.
public class BCD {
/*
* long number to bcd byte array e.g. 123 --> (0000) 0001 0010 0011
* e.g. 12 ---> 0001 0010
*/
public static byte[] DecToBCDArray(long num) {
int digits = 0;
long temp = num;
while (temp != 0) {
digits++;
temp /= 10;
}
int byteLen = digits % 2 == 0 ? digits / 2 : (digits + 1) / 2;
boolean isOdd = digits % 2 != 0;
byte bcd[] = new byte[byteLen];
for (int i = 0; i < digits; i++) {
byte tmp = (byte) (num % 10);
if (i == digits - 1 && isOdd)
bcd[i / 2] = tmp;
else if (i % 2 == 0)
bcd[i / 2] = tmp;
else {
byte foo = (byte) (tmp << 4);
bcd[i / 2] |= foo;
}
num /= 10;
}
for (int i = 0; i < byteLen / 2; i++) {
byte tmp = bcd[i];
bcd[i] = bcd[byteLen - i - 1];
bcd[byteLen - i - 1] = tmp;
}
return bcd;
}
public static String BCDtoString(byte bcd) {
StringBuffer sb = new StringBuffer();
byte high = (byte) (bcd & 0xf0);
high >>>= (byte) 4;
high = (byte) (high & 0x0f);
byte low = (byte) (bcd & 0x0f);
sb.append(high);
sb.append(low);
return sb.toString();
}
public static String BCDtoString(byte[] bcd) {
StringBuffer sb = new StringBuffer();
for (int i = 0; i < bcd.length; i++) {
sb.append(BCDtoString(bcd[i]));
}
return sb.toString();
}
}
There's also this question: Java code or lib to decode a binary-coded decimal (BCD) from a String.
The first step would be to parse the string into an int so that you have the numeric value of it. Then, get the individual digits using division and modulus, and pack each pair of digits into a byte using shift and add (or shift and or).
Alternatively, you could parse each character of the string into an int individually, and avoid using division and modulus to get the numbers, but I would prefer to parse the entire string up front so that you discover right away if the string is invalid. (If you get a NumberFormatException, or if the value is less than 0 or greater than 9999 then it is invalid.)
Finally, once you have assembled the two individual bytes, you can put them into the byte[2].
You can use following:
//Convert BCD String to byte array
public static byte[] String2Bcd(java.lang.String bcdString) {
byte[] binBcd = new byte[bcdString.length() / 2];
for (int i = 0; i < binBcd.length; i++) {
String sByte = bcdString.substring(i*2, i*2+2);
binBcd[i] = Byte.parseByte(sByte, 16);
}
return binBcd;
}
You can try the following code:
public static byte[] hex2Bytes(String str) {
byte[] b = new byte[str.length() / 2];
int j = 0;
for (int i = 0; i < b.length; i++) {
char c0 = str.charAt(j++);
char c1 = str.charAt(j++);
b[i] = ((byte) (parse(c0) << 4 | parse(c1)));
}
return b;
}
My problem is that I've to write a code where a MAC direction must be written via keyboard and then, the code must recognize it to mount a "Magic Packet".
Let me show some code so you can understand better my problem and maybe reach some solution.
static String realMAC[] = new String [6];
static String mac; //that's the written string
private DatagramPacket buildWOLPacket(InetAddress address,String MAC) {
String splittedMAC[] = new String[12];
final int SIZE = 6 + 6 * 16; // See above for this magic number
byte data[] = new byte[SIZE];
for (int i = 0; i < 6; i++) {
data[i] = (byte) 0xff;
}
//Method where each character written
//is stored into a char array (realMAC[])
for (int i = 0 ; i<12; i++) {
String replacedMAC = MAC.replace(":", ""); //Here I delete all the : from the MAC adress introduced via keyboard.
if (i == 12)
splittedMAC[i] = replacedMAC.substring(i,(i));
else splittedMAC[i] = replacedMAC.substring(i,(i+1));}
And now the piece of code that is giving me problems
//All the (byte) 0x00 and so on are examples of a MAC direction and how to convert it if it is predefined
data[6 * i + 0] = Byte.parseByte(realMAC[0]); //(byte) 0x00;
data[6 * i + 1] = Byte.parseByte(realMAC[1]); //(byte) 0x1A;
data[6 * i + 2] = Byte.parseByte(realMAC[2]); //(byte) 0x09;
data[6 * i + 3] = Byte.parseByte(realMAC[3]); //(byte) 0x07;
data[6 * i + 4] = Byte.parseByte(realMAC[4]); //(byte) 0x8c;
data[6 * i + 5] = Byte.parseByte(realMAC[5]); //(byte) 0xe9;
My problem comes when converting the realMAC[] into bytes because of the x and letters from (0x8c, for example) because parseByte only accepts ints and ascii code. How could I tranform that string in hex form into bits?
Thank you so much.
Byte also has a .parseByte() method taking a radix as an argument. However that won't account for the initial 0x so you have to use substring(). Do:
Byte.parseByte(realMac[0].substring(2), 16)
//etc etc
More generally, all "numeric integral" classes (Integer, Short, Long) have such a method.
Byte.decode accepts a String that may begin with 0x or 0X. (It may also parse a decimal or octal number depending on what the string looks like.)