Develop Reference

How to remove the nth hexadecimal digit of a integer number without using Strings?

Consider a hexadecimal integer value such as n = 0x12345, how to get 0x1235 as result by doing remove(n, 3) (big endian)?
For the inputs above I think this can be achieved by performing some bitwising steps:
partA = extract the part from index 0 to targetIndex - 1 (should return 0x123);
partB = extract the part from targetIndex + 1 to length(value) - 1 (0x5);
result, then, can be expressed by ((partA << length(partB) | partB), giving the 0x1235 result.
However I'm still confused in how to implement it, once each hex digit occupies 4 spaces. Also, I don't know a good way to retrieve the length of the numbers.
This can be easily done with strings however I need to use this in a context of thousands of iterations and don't think Strings is a good idea to choose.
So, what is a good way to this removing without Strings?

Similar to the idea you describe, this can be done by creating a mask for both the upper and the lower part, shifting the upper part, and then reassembling.
int remove(int x, int i) {
// create a mask covering the highest 1-bit and all lower bits
int m = x;
m |= (m >>> 1);
m |= (m >>> 2);
m |= (m >>> 4);
m |= (m >>> 8);
m |= (m >>> 16);
// clamp to 4-bit boundary
int l = m & 0x11111110;
m = l - (l >>> 4);
// shift to select relevant position
m >>>= 4 * i;
// assemble result
return ((x & ~(m << 4)) >>> 4) | (x & m);
}
where ">>>" is an unsigned shift.
As a note, if 0 indicates the highest hex digit in a 32-bit word independent of the input, this is much simpler:
int remove(int x, int i) {
int m = 0xffffffff >>> (4*i);
return ((x & ~m) >>> 4) | (x & (m >>> 4));
}

Solution:
Replace operations using 10 with operations using 16.
Demo
Using Bitwise Operator:
public class Main {
public static void main(String[] args) {
int n = 0x12345;
int temp = n;
int length = 0;
// Find length
while (temp != 0) {
length++;
temp /= 16;
}
System.out.println("Length of the number: " + length);
// Remove digit at index 3
int m = n;
int index = 3;
for (int i = index + 1; i <= length; i++) {
m /= 16;
}
m *= 1 << ((length - index - 1) << 2);
m += n % (1 << ((length - index - 1) << 2));
System.out.println("The number after removing digit at index " + index + ": 0x" + Integer.toHexString(m));
}
}
Output:
Length of the number: 5
The number after removing digit at index 3: 0x1235
Using Math::pow:
public class Main {
public static void main(String[] args) {
int n = 0x12345;
int temp = n;
int length = 0;
// Find length
while (temp != 0) {
length++;
temp /= 16;
}
System.out.println("Length of the number: " + length);
// Remove digit at index 3
int m = n;
int index = 3;
for (int i = index + 1; i <= length; i++) {
m /= 16;
}
m *= ((int) (Math.pow(16, length - index - 1)));
m += n % ((int) (Math.pow(16, length - index - 1)));
System.out.println("The number after removing digit at index " + index + ": 0x" + Integer.toHexString(m));
}
}
Output:
Length of the number: 5
The number after removing digit at index 3: 0x1235
JavaScript version:
n = parseInt(12345, 16);
temp = n;
length = 0;
// Find length
while (temp != 0) {
length++;
temp = Math.floor(temp / 16);
}
console.log("Length of the number: " + length);
// Remove digit at index 3
m = n;
index = 3;
for (i = index + 1; i <= length; i++) {
m = Math.floor(m / 16);
}
m *= 1 << ((length - index - 1) << 2);
m += n % (1 << ((length - index - 1) << 2));
console.log("The number after removing digit at index " + index + ": 0x" + m.toString(16));

This works by writing a method to remove from the right but adjusting the parameter to remove from the left. The bonus is that a remove from the right is also available for use. This method uses longs to maximize the length of the hex value.
long n = 0x12DFABCA12L;
int r = 3;
System.out.println("Supplied value: " + Long.toHexString(n).toUpperCase());
n = removeNthFromTheRight(n, r);
System.out.printf("Counting %d from the right: %X%n", r, n);
n = 0x12DFABCA12L;
n = removeNthFromTheLeft(n, r);
System.out.printf("Counting %d from the left: %X%n", r, n);
Prints
Supplied value: 12DFABCA12
Counting 3 from the right: 12DFABA12
Counting 3 from the left: 12DABCA12
This works by recursively removing a digit from the end until just before the one you want to remove. Then remove that and return thru the call stack, rebuilding the number with the original values.
This method counts from the right.
public static long removeNthFromTheRight(long v, int n) {
if (v <= 0) {
throw new IllegalArgumentException("Not enough digits");
}
// save hex digit
long k = v % 16;
while (n > 0) {
// continue removing digit until one
// before the one you want to remove
return removeNthFromTheRight(v / 16, n - 1) * 16 + k;
}
if (n == 0) {
// and ignore that digit.
v /= 16;
}
return v;
}
This method counts from the left. It simply adjusts the value of n and then calls removeFromTheRight.
public static long removeNthFromTheLeft(long v, int n) {
ndigits = (67-Long.numberOfLeadingZeros(v))>>2;
// Now just call removeNthFromTheRight with modified paramaters.
return removeNthFromTheRight(v, ndigits - n - 1);
}
Here is my version using bit manipulation with explanation.
the highest set bit helps find the offset for the mask. For a long that bit is 64-the number of leading zeroes. To get the number of hex digits, one must divide by 4. To account for numbers evenly divisible by 4, it is necessary to add 3 before dividing. So that makes the number of digits:
digits = (67-Long.numberOfLeadingZeros(i))>>2;
which then requires it to be adjusted to mask the appropriate parts of the number.
offset = digits-i - 1
m is the mask to mask off the digit to be removed. So start with a -1L (all hex 'F') and right shift 4*(16-offset) bits. This will result in a mask that masks everything to the right of the digit to be removed.
Note: If offset is 0 the shift operator will be 64 and no bits will be shifted. To accommodate this, the shift operation is broken up into two operations.
Now simply mask off the low order bits
v & m
And the high order bits right shifted 4 bits to eliminate the desired digit.
(v>>>4)^ ~m
and then the two parts are simply OR'd together.
static long remove(long v, int i) {
int offset = ((67 - Long.numberOfLeadingZeros(v))>>2) - i - 1;
long m = (-1L >>> (4*(16 - offset) - 1)) >> 1;
return ((v >>> 4) & ~m) | (v & m);
}

Converting Byte[4] to float - Integer[4] array works but byte[4] does not

This is probably a basic question for out more experienced programmers out there. I'm a bit of a noob and can't work this one out. I'm trying to unpack a binary file and the doco is not too clear on how floats are stored. I have found a routine that does this, but it will only work if I pass an integer array of the bytes. The correct answer is -1865.0. I need to be able to pass the byte array and get the correct answer. How do I need to change the code to make float4byte return -1865.0. Thanks in advance.
import java.nio.ByteBuffer;
import java.nio.ByteOrder;
public class HelloWorld {
public static void main(String[] args) {
byte[] bytes = {(byte) 0xC3,(byte) 0X74,(byte) 0X90,(byte) 0X00 };
int[] ints = {(int) 0xC3,(int) 0X74,(int) 0X90,(int) 0X00 };
// This give the wrong answer
float f = ByteBuffer.wrap(bytes).order(ByteOrder.BIG_ENDIAN).getFloat();
System.out.println("VAL ByteBuffer BI: " + f);
// This give the wrong answer
f = ByteBuffer.wrap(bytes).order(ByteOrder.LITTLE_ENDIAN).getFloat();
System.out.println("VAL ByteBuffer LI: " + f);
//This gives the RIGHT answer
f = float4int (ints[0], ints[1], ints[2], ints[3]);
System.out.println("VAL Integer : " + f);
// This gives the wrong answer
f = float4byte (bytes[0], bytes[1], bytes[2], bytes[3]);
System.out.println("VAL Bytes : " + f);
}
private static float float4int(int a, int b, int c, int d)
{
int sgn, mant, exp;
System.out.println ("IN Int: "+String.format("%02X ", a)+
String.format("%02X ", b)+String.format("%02X ", c)+String.format("%02X ", d));
mant = b << 16 | c << 8 | d;
if (mant == 0) return 0.0f;
sgn = -(((a & 128) >> 6) - 1);
exp = (a & 127) - 64;
return (float) (sgn * Math.pow(16.0, exp - 6) * mant);
}
private static float float4byte(byte a, byte b, byte c, byte d)
{
int sgn, mant, exp;
System.out.println ("IN Byte : "+String.format("%02X ", a)+
String.format("%02X ", b)+String.format("%02X ", c)+String.format("%02X ", d));
mant = b << 16 | c << 8 | d;
if (mant == 0) return 0.0f;
sgn = -(((a & 128) >> 6) - 1);
exp = (a & 127) - 64;
return (float) (sgn * Math.pow(16.0, exp - 6) * mant);
}
}

The reason why your solution with ByteBuffer doesn't work: the bytes do not match the (Java) internal representation of the float value.
The Java representation is
System.out.println(Integer.toHexString(Float.floatToIntBits(-1865.0f)));
which gives c4e92000

bytes are signed in Java. When calculating the mantissa mant, the bytes are implicitly converted from bytes to ints - with the sign "extended", i.e. (byte)0x90 (decimal -112) gets converted 0xFFFFFF90 (32 bits int). However what you want is just the original bytes' 8 bits (0x00000090).
In order to compensate for the effect of sign extension, it suffices to change one line:
mant = (b & 0xFF) << 16 | (c & 0xFF) << 8 | (d & 0xFF)
Here, in (c & 0xFF), the 1-bits caused by sign extension are stripped after (implicit) conversion to int.
Edit:
The repacking of floats could be done via the IEEE 754 representation which can be obtained by Float.floatToIntBits (which avoids using slow logarithms). Some complexity in the code is caused by the change of base from 2 to 16:
private static byte[] byte4float(float f) {
assert !Float.isNaN(f);
// see also JavaDoc of Float.intBitsToFloat(int)
int bits = Float.floatToIntBits(f);
int s = (bits >> 31) == 0 ? 1 : -1;
int e = (bits >> 23) & 0xFF;
int m = (e == 0) ? (bits & 0x7FFFFF) << 1 : (bits& 0x7FFFFF) | 0x800000;
int exp = (e - 150) / 4 + 6;
int mant;
int mantissaShift = (e - 150) % 4; // compensate for base 16
if (mantissaShift >= 0) mant = m << mantissaShift;
else { mant = m << (mantissaShift + 4); exp--; }
if (mant > 0xFFFFFFF) { mant >>= 4; exp++; } // loose of precision
byte a = (byte) ((1 - s) << 6 | (exp + 64));
return new byte[]{ a, (byte) (mant >> 16), (byte) (mant >> 8), (byte) mant };
}
The code does not take into account any rules that may exist for the packaging, e.g. for representing zero or normalization of the mantissa. But it might serve as a starting point.

Thanks to #halfbit and a bit of testing and minor changes, this routine appears convert IEEE 754 float into IBM float.
public static byte[] byte4float(float f) {
assert !Float.isNaN(f);
// see also JavaDoc of Float.intBitsToFloat(int)
int bits = Float.floatToIntBits(f);
int s = (bits >> 31) == 0 ? 1 : -1;
int e = (bits >> 23) & 0xFF;
int m = (e == 0) ? (bits & 0x7FFFFF) << 1 : (bits& 0x7FFFFF) | 0x800000;
int exp = (e - 150) / 4 + 6;
int mant;
int mantissaShift = (e - 150) % 4; // compensate for base 16
if (mantissaShift >= 0) mant = m >> mantissaShift;
else mant = m >> (Math.abs(mantissaShift));
if (mant > 0xFFFFFFF) { mant >>= 4; exp++; } // loose of precision */
byte a = (byte) ((1 - s) << 6 | (exp + 64));
return new byte[]{ a, (byte) (mant >> 16), (byte) (mant >> 8), (byte) mant };
}
I think this is right and appears to be working.

Vigenère cipher implementation

I have to implement a variant of the Vigenère cipher. I got the encryption part without issues, but I have a bug in the decryption code and I don't understand what I'm doing wrong.
The requirements are:
the key can only contain A - Z (uppercase)
code values for the key characters are 0 for A, 1 for B, ..., and 25 for Z
do not encode a character if the code is < 32 (preserve control characters)
encrypted character code = original character code + key character code
the final encrypted character must be between 32 and 126, exclusively so if the final encrypted character > 126 it must be brought back into the 32 - 126 range by adding 32 to the value and then subtracting 126
The encryption code:
// it works ok
// I have tested it with some provided strings and the results are as expected
public String encrypt(String plainText)
{
StringBuilder sb = new StringBuilder();
for (int i = 0; i < plainText.length(); i++) {
char c = plainText.charAt(i);
if (c >= 32) {
int keyCharValue = theKey.charAt(i % theKey.length()) - 'A';
c += keyCharValue;
if (c > 126) {
c = (char) (c + 32 - 126);
}
}
sb.append(c);
}
return sb.toString();
}
The decryption code:
// there probably is an off-by-one error somewhere
// everything is decrypted ok, except '~' which gets decrypted to ' ' (space)
public String decrypt(String cipherText)
{
StringBuilder sb = new StringBuilder();
for (int i = 0; i < cipherText.length(); i++) {
char c = cipherText.charAt(i);
if (c >= 32) {
int keyCharValue = theKey.charAt(i % theKey.length()) - 'A';
c -= keyCharValue;
if (c < 32) {
c = (char) (c + 126 - 32);
}
}
sb.append(c);
}
return sb.toString();
}
Example (with key ABCDEFGHIJKLMNOPQRSTUVWXYZ):
original ~~~~~~~~~~~~~~~~~~~~~~~~~~
encrypted ~!"#$%&'()*+,-./0123456789
decrypted ~ ('~' followed by spaces)
EDIT:
Here is the code I use for testing (it tests every character from 0 to 126 repeated as a string):
public static void main(String[] args) {
int passed = 0;
int failed = 0;
String key = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
for (int c = 0; c <= 126; c++) {
StringBuilder sbString = new StringBuilder();
for (int i = 0; i <= 25; i++) {
sbString.append((char) c);
}
String original = sbString.toString();
Cipher cipher = Cipher(key);
String encrypted = cipher.encrypt(original);
String decrypted = cipher.decrypt(encrypted);
if (!original.equals(decrypted)) {
failed++;
System.out.println("--FAILED--");
System.out.println(original);
System.out.println(encrypted);
System.out.println(decrypted);
} else {
passed++;
}
}
int tests = passed + failed;
System.out.println(tests + " tests");
System.out.println("passed: " + passed);
System.out.println("failed: " + failed);
}

I believe the If(c < 32) in the decryption needs to be If (c <= 32).
Reasoning: if you take the case of Char(126) or '~' then add one to it in the encryption you get 127, which goes through the encryption transform and becomes 33.
When decrypting that you get 33 minus that same 1 which leaves you with 32, which won't trigger the special decryption case. By including 32 in that statement it will trigger the special decryption and change the 32 (" ") to 126 ("~")
You are correct it is an off by one error, but it is kinda subtle
EDIT: there is a collision error because char(32) and char(126) are hashing to the same value. In my previous example that value would be 33, the equation needs to be changed such that Char(126) will hash to 32.
Changing the c = (char) (c + 32 - 126); to c = (char) (c + 32 - 127); should free up the extra space to prevent the collision from happening. THe decrypt will also have to be changed from c = (char) (c + 126 - 32); to c = (char) (c + 127 - 32);
And someone posted that in my comments.

The error about string encryption scheduling : (char) (ch + key) % 26

Problem one: Here are two code spinner. Code A runs wrong. But I do not know what is wrong.
Problem two: code B is right.but I do not understand why it need to delete 'A’. then add 'A' after fmod. What is the effect about 'A'? Why it has the error after delete?
Code A (ch + key) % 26 )
Code B ('A' + ((ch -'A' + key) % 26))
public void run() {
setFont("Arial-PLAIN-24");
String line = readLine ("Enter line: ");
int key = readInt ("Enter key: ");
String siphertext = encryptCaesar(line , key);
println("The result is: " + siphertext);
String newplain = encryptCaesar(siphertext , -key);
println("newplain:" + newplain);
}
private String encryptCaesar(String str , int key){
if(key < 0){
key = 26 - ( -key % 26 );
}
String result = "";
for(int i = 0; i < str.length(); i++){
char ch = str.charAt(i);
result += encryptChar(ch,key);
}
return result;
}
private char encryptChar(char ch, int key){
if(Character.isUpperCase(ch)){
return ( (char) ('A' + ((ch -'A' + key) % 26)) );
}
return ch;
}

'A' is added to make sure the result of "encryptChar" method, is a valid character in ASCII range 64 to 90, which is A (CAPITAL) to Z (CAPITAL). Refer the ASCII table here.
In your code subtracting of 'A' can also be ignored. That is the below will also work,
('A' + ((ch + key) % 26))

15.7.3 Remainder Operator %
...
It follows from this rule that the result of the remainder operation
can be negative only if the dividend is negative, and can be positive
only if the dividend is positive.
An example is then provided:
int e = (-5)%3; // -2
int f = (-5)/3; // -1
System.out.println("(-5)%3 produces " + e +
" (note that (-5)/3 produces " + f + ")");
If the result of ((ch -'A' + key) % 26)) is negative, then wouldn't the result of (char) ('A' + ((ch -'A' + key) % 26)) be some non-alphabet character? Perhaps you need to add 26 to any negative values or find the absolute value, so that they're positive and result in actual alphabet characters.

Java little endian order

I need to store data as LITTLE_ENDIAN instead of default BIG_ENDIAN.
Here's my sample code:
for (int i = 0; i < logo.length; i++) {
logoArray[i] = ((Integer) logo[i]).byteValue();
logoArray[i] = (byte) (((logoArray[i] & 1) << 7) + ((logoArray[i] & 2) << 5) + ((logoArray[i] & 4) << 3)
+ ((logoArray[i] & 8) << 1) + ((logoArray[i] & 16) >> 1) + ((logoArray[i] & 32) >> 3)
+ ((logoArray[i] & 64) >> 5) + ((logoArray[i] & 128) >> 7));
}
How should it be rewritten with ByteBuffer, for LITTLE_ENDIAN, as the following code doesn't work for me:
ByteBuffer record = ByteBuffer.allocate(logo.length);
record.order(ByteOrder.LITTLE_ENDIAN);
...
record.put(((Integer) logo[i]).byteValue());
...
record.array(); // get

ByteBuffer will work for you, if you use putInt and not put.
record.putInt((Integer) logo[i]);
A byte array (as Integer.byteValue()) has no "endianness" so it is stored as it is.

As andcoz says, the endian isn't taken into account when you put one byte at a time. Here is an example to show you how to do it:
import java.nio.*;
public class Test {
public static void main(String[] args) {
int[] logo = { 0xAABBCCDD, 0x11223344 };
byte[] logoLE = new byte[logo.length * 4];
ByteBuffer rec = ByteBuffer.wrap(logoLE).order(ByteOrder.LITTLE_ENDIAN);
for (int i = 0; i < logo.length; i++)
rec.putInt(logo[i]);
// Debug printouts...
System.out.println("logo:");
for (int b : logo)
System.out.println(Integer.toHexString((b < 0 ? b + 256 : b)));
System.out.println("\nlogoLE:");
int tmp = 0;
for (byte b : logoLE) {
System.out.print(Integer.toHexString((b < 0 ? b + 256 : b)));
if (++tmp % 4 == 0)
System.out.println();
}
}
}
Output:
logo:
aabbccdd
11223344
logoLE:
ddccbbaa
44332211