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[JBTM-4014] restore longToHexString method #2433

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Merged
merged 2 commits into from
Nov 10, 2025
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[JBTM-4014] restore longToHexString method #2433

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[JBTM-4014] restore longToHexString
marcosgopen Nov 6, 2025
ad4dfcc
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marcosgopen Nov 7, 2025
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209 changes: 108 additions & 101 deletions ArjunaCore/arjuna/classes/com/arjuna/ats/arjuna/utils/Utility.java
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Original file line number Diff line number Diff line change
Expand Up @@ -78,114 +78,121 @@ public static int hexStringToInt (String s) throws NumberFormatException
return val;
}

private static final byte[] HEX_DIGITS;

static {
// each byte contains 2 hex digits
HEX_DIGITS = new byte[16];
HEX_DIGITS[0] = (byte) '0';
HEX_DIGITS[1] = (byte) '1';
HEX_DIGITS[2] = (byte) '2';
HEX_DIGITS[3] = (byte) '3';
HEX_DIGITS[4] = (byte) '4';
HEX_DIGITS[5] = (byte) '5';
HEX_DIGITS[6] = (byte) '6';
HEX_DIGITS[7] = (byte) '7';
HEX_DIGITS[8] = (byte) '8';
HEX_DIGITS[9] = (byte) '9';
HEX_DIGITS[10] = (byte) 'a';
HEX_DIGITS[11] = (byte) 'b';
HEX_DIGITS[12] = (byte) 'c';
HEX_DIGITS[13] = (byte) 'd';
HEX_DIGITS[14] = (byte) 'e';
HEX_DIGITS[15] = (byte) 'f';
}

/**
* Determine the number of hex chars needed to represent the given int
* value. This is at least 1.
*
* @param value the int value
* @return the number of hex chars needed to represent the value
*/
public static int hexCharsOf(int value) {
int signLen = 0;
if (value < 0) {
value = -value;
signLen = 1;
}
int nonZeroBits = Integer.SIZE - Integer.numberOfLeadingZeros(value);
// each hex char represents 4 bits: align the non-zero bits to the next multiple of 4 and divide by 4
return signLen + Math.max(((nonZeroBits + 3) >> 2), 1);
}

/**
* Convert a int to hex chars. The byte array must be big enough to hold
* the expected number of hex chars. See {@link #hexCharsOf(int)} to determine
* the number of hex chars needed.
*
* @param value the int value to convert
* @param ascii the byte array to hold the hex chars
* @param offset the offset into the byte array to start writing
* @param expectedHexChars the expected number of hex chars to write
*/
public static void toHexChars(int value, byte[] ascii, int offset, int expectedHexChars) {
int numHexChars = expectedHexChars;
if (value < 0) {
value = -value;
ascii[offset++] = (byte) '-';
numHexChars--;
}
// write the hex digits in reverse order
for (int i = numHexChars - 1; i >= 0; i--) {
int digit = value & 0xF;
ascii[offset + i] = HEX_DIGITS[digit];
value >>>= 4;
}
}

/**
* Determine the number of hex chars needed to represent the given long
* value. This is at least 1.
*
* @param value the long value
* @return the number of hex chars needed to represent the value
/**
* Convert a long to a hex String.
*/
public static int hexCharsOf(long value) {
public static String longToHexString(long number) throws NumberFormatException {
return Long.toString(number, 16);
}

private static final byte[] HEX_DIGITS;

static {
// each byte contains 2 hex digits
HEX_DIGITS = new byte[16];
HEX_DIGITS[0] = (byte) '0';
HEX_DIGITS[1] = (byte) '1';
HEX_DIGITS[2] = (byte) '2';
HEX_DIGITS[3] = (byte) '3';
HEX_DIGITS[4] = (byte) '4';
HEX_DIGITS[5] = (byte) '5';
HEX_DIGITS[6] = (byte) '6';
HEX_DIGITS[7] = (byte) '7';
HEX_DIGITS[8] = (byte) '8';
HEX_DIGITS[9] = (byte) '9';
HEX_DIGITS[10] = (byte) 'a';
HEX_DIGITS[11] = (byte) 'b';
HEX_DIGITS[12] = (byte) 'c';
HEX_DIGITS[13] = (byte) 'd';
HEX_DIGITS[14] = (byte) 'e';
HEX_DIGITS[15] = (byte) 'f';
}

/**
* Determine the number of hex chars needed to represent the given int
* value. This is at least 1.
*
* @param value the int value
* @return the number of hex chars needed to represent the value
*/
public static int hexCharsOf(int value) {
int signLen = 0;
if (value < 0) {
value = -value;
signLen = 1;
}
int nonZeroBits = Long.SIZE - Long.numberOfLeadingZeros(value);
// each hex char represents 4 bits: align the non-zero bits to the next multiple of 4 and divide by 4
return signLen + Math.max(((nonZeroBits + 3) >> 2), 1);
}

/**
* Convert a long to hex chars. The byte array must be big enough to hold
* the expected number of hex chars. See {@link #hexCharsOf(long)} to determine
* the number of hex chars needed.
*
* @param value the long value to convert
* @param ascii the byte array to hold the hex chars
* @param offset the offset into the byte array to start writing
* @param expectedHexChars the expected number of hex chars to write
*/
public static void toHexChars(long value, byte[] ascii, int offset, int expectedHexChars) {
int numHexChars = expectedHexChars;
if (value < 0) {
int nonZeroBits = Integer.SIZE - Integer.numberOfLeadingZeros(value);
// each hex char represents 4 bits: align the non-zero bits to the next multiple of 4 and divide by 4
return signLen + Math.max(((nonZeroBits + 3) >> 2), 1);
}

/**
* Convert a int to hex chars. The byte array must be big enough to hold
* the expected number of hex chars. See {@link #hexCharsOf(int)} to determine
* the number of hex chars needed.
*
* @param value the int value to convert
* @param ascii the byte array to hold the hex chars
* @param offset the offset into the byte array to start writing
* @param expectedHexChars the expected number of hex chars to write
*/
public static void toHexChars(int value, byte[] ascii, int offset, int expectedHexChars) {
int numHexChars = expectedHexChars;
if (value < 0) {
value = -value;
ascii[offset++] = (byte) '-';
numHexChars--;
}
// write the hex digits in reverse order
for (int i = numHexChars - 1; i >= 0; i--) {
int digit = value & 0xF;
ascii[offset + i] = HEX_DIGITS[digit];
value >>>= 4;
}
}

/**
* Determine the number of hex chars needed to represent the given long
* value. This is at least 1.
*
* @param value the long value
* @return the number of hex chars needed to represent the value
*/
public static int hexCharsOf(long value) {
int signLen = 0;
if (value < 0) {
value = -value;
ascii[offset++] = (byte) '-';
numHexChars--;
}
// write the hex digits in reverse order
for (int i = numHexChars - 1; i >= 0; i--) {
int digit = (int) (value & 0xF);
ascii[offset + i] = HEX_DIGITS[digit];
value >>>= 4;
}
}
signLen = 1;
}
int nonZeroBits = Long.SIZE - Long.numberOfLeadingZeros(value);
// each hex char represents 4 bits: align the non-zero bits to the next multiple of 4 and divide by 4
return signLen + Math.max(((nonZeroBits + 3) >> 2), 1);
}

/**
* Convert a long to hex chars. The byte array must be big enough to hold
* the expected number of hex chars. See {@link #hexCharsOf(long)} to determine
* the number of hex chars needed.
*
* @param value the long value to convert
* @param ascii the byte array to hold the hex chars
* @param offset the offset into the byte array to start writing
* @param expectedHexChars the expected number of hex chars to write
*/
public static void toHexChars(long value, byte[] ascii, int offset, int expectedHexChars) {
int numHexChars = expectedHexChars;
if (value < 0) {
value = -value;
ascii[offset++] = (byte) '-';
numHexChars--;
}
// write the hex digits in reverse order
for (int i = numHexChars - 1; i >= 0; i--) {
int digit = (int) (value & 0xF);
ascii[offset + i] = HEX_DIGITS[digit];
value >>>= 4;
}
}


/**
Expand Down