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package nom.tam.util;
/*
* #%L
* nom.tam FITS library
* %%
* Copyright (C) 2004 - 2024 nom-tam-fits
* %%
* This is free and unencumbered software released into the public domain.
*
* Anyone is free to copy, modify, publish, use, compile, sell, or
* distribute this software, either in source code form or as a compiled
* binary, for any purpose, commercial or non-commercial, and by any
* means.
*
* In jurisdictions that recognize copyright laws, the author or authors
* of this software dedicate any and all copyright interest in the
* software to the public domain. We make this dedication for the benefit
* of the public at large and to the detriment of our heirs and
* successors. We intend this dedication to be an overt act of
* relinquishment in perpetuity of all present and future rights to this
* software under copyright law.
*
* THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND,
* EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF
* MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT.
* IN NO EVENT SHALL THE AUTHORS BE LIABLE FOR ANY CLAIM, DAMAGES OR
* OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE,
* ARISING FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR
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* #L%
*/
/**
* This class provides mechanisms for efficiently formatting numbers and Strings. Data is appended to existing byte
* arrays. Note that the formatting of real or double values may differ slightly (in the last bit) from the standard
* Java packages since this routines are optimized for speed rather than accuracy.
*
* The methods in this class create no objects.
*
* If a number cannot fit into the requested space the truncateOnOverlow flag controls whether the formatter will
* attempt to append it using the available length in the output (a la C or Perl style formats). If this flag is set, or
* if the number cannot fit into space left in the buffer it is 'truncated' and the requested space is filled with a
* truncation fill character. A TruncationException may be thrown if the truncationThrow flag is set.
*
* This class does not explicitly support separate methods for formatting reals in exponential notation. Real numbers
* near one are by default formatted in decimal notation while numbers with large (or very negative) exponents are
* formatted in exponential notation. By setting the limits at which these transitions take place the user can force
* either exponential or decimal notation.
*
* @deprecated This class should not be exposed in the public API and is intended for internal use only in ASCII tables.
* Also, it may have overlapping functionality with other classes, which should probably be eliminated
* for simplicity's sake (and thus less chance of nasty bugs).
*
* @see ByteParser
*/
@Deprecated
public final class ByteFormatter {
/** String representation of NaN values */
public static final String NOT_A_NUMBER = "NaN";
/** String representation of (positive) infinity values */
public static final String INFINITY = "Infinity";
/** String representation of (negative) infinity values */
public static final String NEGATIVE_INFINITY = "-Infinity";
/**
* Maximum magnitude to print in non-scientific notation.
*/
private static final double DEFAULT_SIMPLE_MAX = 1.e6;
/**
* Minimum magnitude to print in non-scientific notation.
*/
private static final double DEFAULT_SIMPLE_MIN = 1.e-3;
/**
* Digits. We could handle other bases by extending or truncating this list and changing the division by 10 (and
* it's factors) at various locations.
*/
private static final byte[] DIGITS = {(byte) '0', (byte) '1', (byte) '2', (byte) '3', (byte) '4', (byte) '5',
(byte) '6', (byte) '7', (byte) '8', (byte) '9'};
private static final long DOUBLE_EXPONENT_BIT_MASK = 0x7FF0000000000000L;
private static final int DOUBLE_EXPONENT_EXCESS = 52;
/**
* The hidden bit in normalized double numbers.
*/
private static final long DOUBLE_EXPONENT_NORMALIZE_BIT = 0x0010000000000000L;
/**
* Used to check if double is normalized
*/
private static final int DOUBLE_MIN_EXPONENT = -1023;
private static final int DOUBLE_SHIFT_BASE = 17;
private static final int DOUBLE_SHIFT_LIMIT = 200;
private static final long DOUBLE_VALUE_BIT_MASK = 0x000FFFFFFFFFFFFFL;
private static final int FLOAT_EXPONENT_BIT_MASK = 0x7F800000;
private static final int FLOAT_EXPONENT_EXCESS = 23;
/**
* The hidden bit in normalized floating point numbers.
*/
private static final int FLOAT_EXPONENT_NORMALIZE_BIT = 0x00800000;
/**
* Used to check if float is normalized
*/
private static final int FLOAT_MIN_EXPONENT = -127;
private static final int FLOAT_SHIFT_BASE = 8;
private static final int FLOAT_SHIFT_LIMIT = 30;
private static final int FLOAT_VALUE_BIT_MASK = 0x007FFFFF;
private static final double I_LOG_10 = 1. / Math.log(ByteFormatter.NUMBER_BASE);
private static final long LONG_TO_INT_MODULO = 1000000000L;
private static final int MAX_LONG_LENGTH = 19;
/**
* The maximum single digit integer. Special case handling when rounding up and when incrementing the exponent.
*/
private static final int MAXIMUM_SINGLE_DIGIT_INTEGER = 9;
/**
* The maximum two digit integer. When we increment an exponent of this value, the exponent gets longer. Don't need
* to worry about 999 since exponents can't be that big.
*/
private static final int MAXIMUM_TWO_DIGIT_INTEGER = 99;
private static final int TEMP_BUFFER_SIZE = 32;
/**
* The underlying number base used in this class.
*/
private static final int NUMBER_BASE = 10;
/**
* Powers of 10. We over extend on both sides. These should perhaps be tabulated rather than computed though it may
* be faster to calculate them than to read in the extra bytes in the class file.
*/
private static final double[] NUMBER_BASE_POWERS;
/**
* What do we use to fill when we cannot print the number?Default is often used in Fortran.
*/
private static final byte TRUNCATION_FILL = (byte) '*';
/**
* What index of tenpow is 10^0
*/
private static final int ZERO_POW;
static { // Static initializer
int min = (int) Math.floor((int) (Math.log(Double.MIN_VALUE) * ByteFormatter.I_LOG_10));
int max = (int) Math.floor((int) (Math.log(Double.MAX_VALUE) * ByteFormatter.I_LOG_10));
max++;
NUMBER_BASE_POWERS = new double[max - min + 1];
for (int i = 0; i < ByteFormatter.NUMBER_BASE_POWERS.length; i++) {
ByteFormatter.NUMBER_BASE_POWERS[i] = Math.pow(ByteFormatter.NUMBER_BASE, i + min);
}
ZERO_POW = -min;
}
/**
* Internal buffers used in formatting fields
*/
private final byte[] tbuf1 = new byte[ByteFormatter.TEMP_BUFFER_SIZE];
private final byte[] tbuf2 = new byte[ByteFormatter.TEMP_BUFFER_SIZE];
/**
* This method formats a double given a decimal mantissa and exponent information.
*
* @param val The original number
* @param buf Output buffer
* @param off Offset into buffer
* @param len Maximum number of characters to use in buffer.
* @param mant A decimal mantissa for the number.
* @param lmant The number of characters in the mantissa
* @param shift The exponent of the power of 10 that we shifted val to get the given mantissa.
*
* @return Offset of next available character in buffer.
*/
private int combineReal(double val, byte[] buf, int off, int len, byte[] mant, int lmant, int shift) {
// First get the minimum size for the number
double pos = Math.abs(val);
boolean simple = false;
int minSize;
int maxSize;
if (pos >= ByteFormatter.DEFAULT_SIMPLE_MIN && pos <= ByteFormatter.DEFAULT_SIMPLE_MAX) {
simple = true;
}
int exp = lmant - shift - 1;
int lexp = 0;
if (!simple) {
lexp = format(exp, tbuf2, 0, ByteFormatter.TEMP_BUFFER_SIZE);
minSize = lexp + 2; // e.g., 2e-12
maxSize = lexp + lmant + 2; // add in "." and e
} else if (exp >= 0) {
minSize = exp + 1; // e.g. 32
// Special case. E.g., 99.9 has
// minumum size of 3.
int i;
for (i = 0; i < lmant && i <= exp; i++) {
if (mant[i] != (byte) '9') {
break;
}
}
if (i > exp && i < lmant && mant[i] >= (byte) '5') {
minSize++;
}
maxSize = lmant + 1; // Add in "."
if (maxSize <= minSize) { // Very large numbers.
maxSize = minSize + 1;
}
} else {
minSize = 2;
maxSize = 1 + Math.abs(exp) + lmant;
}
if (val < 0) {
minSize++;
}
// Can the number fit?
if (minSize > len || minSize > buf.length - off) {
truncationFiller(buf, off, len);
return off + len;
}
// Now begin filling in the buffer.
if (val < 0) {
buf[off] = (byte) '-';
off++;
len--;
}
if (simple) {
return Math.abs(mantissa(mant, lmant, exp, simple, buf, off, len));
}
off = mantissa(mant, lmant, 0, simple, buf, off, len - lexp - 1);
if (off < 0) {
off = -off;
len -= off;
// Handle the expanded exponent by filling
if (exp == MAXIMUM_SINGLE_DIGIT_INTEGER || exp == MAXIMUM_TWO_DIGIT_INTEGER) {
// Cannot fit...
if (off + len == minSize) {
truncationFiller(buf, off, len);
return off + len;
}
// Steal a character from the mantissa.
off--;
}
exp++;
lexp = format(exp, tbuf2, 0, ByteFormatter.TEMP_BUFFER_SIZE);
}
buf[off] = (byte) 'E';
off++;
System.arraycopy(tbuf2, 0, buf, off, lexp);
return off + lexp;
}
/**
* Format a boolean into an existing array.
*
* @param val value to write
* @param array the array to fill
*
* @return Offset of next available character in buffer.
*/
public int format(boolean val, byte[] array) {
return format(val, array, 0, array.length);
}
/**
* Format a boolean into an existing array
*
* @param val The boolean to be formatted
* @param array The buffer in which to format the data.
* @param off The starting offset within the buffer.
* @param len The maximum number of characters to use use in formatting the number.
*
* @return Offset of next available character in buffer.
*/
public int format(boolean val, byte[] array, int off, int len) {
if (len > 0) {
if (val) {
array[off] = (byte) 'T';
} else {
array[off] = (byte) 'F';
}
off++;
}
return off;
}
/**
* Format a double into an array.
*
* @param val The double to be formatted.
* @param array The array in which to place the result.
*
* @return The number of characters used. @ * if the value was truncated
*/
public int format(double val, byte[] array) {
return format(val, array, 0, array.length);
}
/**
* Format a double into an existing character array.
*
* This is hard to do exactly right... The JDK code does stuff with rational arithmetic and so forth. We use a much
* simpler algorithm which may give an answer off in the lowest order bit. Since this is pure Java, it should still
* be consistent from machine to machine.
*
* Recall that the binary representation of the double is of the form d = 0.bbbbbbbb x 2n
* where there are up to 53 binary digits in the binary fraction (including the assumed leading 1 bit for normalized
* numbers). We find a value m such that 10m d is between 253 and
* 263. This product will be exactly convertible to a long with no loss of precision.
* Getting the decimal representation for that is trivial (see formatLong). This is a decimal mantissa and we have
* an exponent (-m). All we have to do is manipulate the decimal point to where we want to see it.
* Errors can arise due to roundoff in the scaling multiplication, but should be no more than a single bit.
*
* @param val Double to be formatted
* @param buf Buffer in which result is to be stored
* @param off Offset within buffer
* @param len Maximum length of integer
*
* @return offset of next unused character in input buffer.
*/
public int format(double val, byte[] buf, int off, int len) {
double pos = Math.abs(val);
// Special cases -- It is OK if these get truncated.
if (pos == 0.) {
return format("0.0", buf, off, len);
}
if (Double.isNaN(val)) {
return format(NOT_A_NUMBER, buf, off, len);
}
if (Double.isInfinite(val)) {
if (val > 0) {
return format(INFINITY, buf, off, len);
}
return format(NEGATIVE_INFINITY, buf, off, len);
}
int power = (int) (Math.log(pos) * ByteFormatter.I_LOG_10);
int shift = DOUBLE_SHIFT_BASE - power;
double scale;
double scale2 = 1;
// Scale the number so that we get a number ~ n x 10^17.
if (shift < DOUBLE_SHIFT_LIMIT) {
scale = ByteFormatter.NUMBER_BASE_POWERS[shift + ByteFormatter.ZERO_POW];
} else {
// Can get overflow if the original number is
// very small, so we break out the shift
// into two multipliers.
scale2 = ByteFormatter.NUMBER_BASE_POWERS[DOUBLE_SHIFT_LIMIT + ByteFormatter.ZERO_POW];
scale = ByteFormatter.NUMBER_BASE_POWERS[shift - DOUBLE_SHIFT_LIMIT + ByteFormatter.ZERO_POW];
}
pos = pos * scale * scale2;
// Parse the double bits.
long bits = Double.doubleToLongBits(pos);
// The exponent should be a little more than 52.
int exp = (int) (((bits & DOUBLE_EXPONENT_BIT_MASK) >> DOUBLE_EXPONENT_EXCESS) + DOUBLE_MIN_EXPONENT);
long numb = bits & DOUBLE_VALUE_BIT_MASK;
// this can never be false because the min exponent of a double is -1022
// but lets keep it if we need it for big decimals
if (exp > DOUBLE_MIN_EXPONENT) {
// Normalized....
numb |= DOUBLE_EXPONENT_NORMALIZE_BIT;
} else {
// Denormalized
exp++;
}
// Multiple this number by the excess of the exponent
// over 52. This completes the conversion of double to long.
numb = numb << exp - DOUBLE_EXPONENT_EXCESS;
// Get a decimal mantissa.
int ndig = format(numb, tbuf1, 0, ByteFormatter.TEMP_BUFFER_SIZE);
// Now format the double.
return combineReal(val, buf, off, len, tbuf1, ndig, shift);
}
/**
* Format a float into an array.
*
* @param val The float to be formatted.
* @param array The array in which to place the result.
*
* @return The number of characters used. @ * if the value was truncated
*/
public int format(float val, byte[] array) {
return format(val, array, 0, array.length);
}
/**
* Format a float into an existing byteacter array.
*
* This is hard to do exactly right... The JDK code does stuff with rational arithmetic and so forth. We use a much
* simpler algorithm which may give an answer off in the lowest order bit. Since this is pure Java, it should still
* be consistent from machine to machine.
*
* Recall that the binary representation of the float is of the form d = 0.bbbbbbbb x 2n
* where there are up to 24 binary digits in the binary fraction (including the assumed leading 1 bit for normalized
* numbers). We find a value m such that 10m d is between 224 and
* 232. This product will be exactly convertible to an int with no loss of precision.
* Getting the decimal representation for that is trivial (see formatInteger). This is a decimal mantissa and we
* have an exponent ( -m). All we have to do is manipulate the decimal point to where we want to see
* it. Errors can arise due to roundoff in the scaling multiplication, but should be very small.
*
* @param val Float to be formatted
* @param buf Buffer in which result is to be stored
* @param off Offset within buffer
* @param len Maximum length of field
*
* @return Offset of next character in buffer. @ * if the value was truncated
*/
public int format(float val, byte[] buf, int off, int len) {
float pos = Math.abs(val);
// Special cases
if (pos == 0.) {
return format("0.0", buf, off, len);
}
if (Float.isNaN(val)) {
return format(NOT_A_NUMBER, buf, off, len);
}
if (Float.isInfinite(val)) {
if (val > 0) {
return format("Infinity", buf, off, len);
}
return format("-Infinity", buf, off, len);
}
int power = (int) Math.floor(Math.log(pos) * ByteFormatter.I_LOG_10);
int shift = FLOAT_SHIFT_BASE - power;
float scale;
float scale2 = 1;
// Scale the number so that we get a number ~ n x 10^8.
if (shift < FLOAT_SHIFT_LIMIT) {
scale = (float) ByteFormatter.NUMBER_BASE_POWERS[shift + ByteFormatter.ZERO_POW];
} else {
// Can get overflow if the original number is
// very small, so we break out the shift
// into two multipliers.
scale2 = (float) ByteFormatter.NUMBER_BASE_POWERS[FLOAT_SHIFT_LIMIT + ByteFormatter.ZERO_POW];
scale = (float) ByteFormatter.NUMBER_BASE_POWERS[shift - FLOAT_SHIFT_LIMIT + ByteFormatter.ZERO_POW];
}
pos = pos * scale * scale2;
// Parse the float bits.
int bits = Float.floatToIntBits(pos);
// The exponent should be a little more than 23
int exp = ((bits & FLOAT_EXPONENT_BIT_MASK) >> FLOAT_EXPONENT_EXCESS) + FLOAT_MIN_EXPONENT;
int numb = bits & FLOAT_VALUE_BIT_MASK;
if (exp > FLOAT_MIN_EXPONENT) {
// Normalized....
numb |= FLOAT_EXPONENT_NORMALIZE_BIT;
} else {
// Denormalized
exp++;
}
// Multiple this number by the excess of the exponent
// over 24. This completes the conversion of float to int
// (<<= did not work on Alpha TruUnix)
numb = numb << exp - FLOAT_EXPONENT_EXCESS;
// Get a decimal mantissa.
int ndig = format(numb, tbuf1, 0, ByteFormatter.TEMP_BUFFER_SIZE);
// Now format the float.
return combineReal(val, buf, off, len, tbuf1, ndig, shift);
}
/**
* Format an int into an array.
*
* @param val The int to be formatted.
* @param array The array in which to place the result.
*
* @return The number of characters used. @ * if the value was truncated
*/
public int format(int val, byte[] array) {
return format(val, array, 0, array.length);
}
/**
* Format an int into an existing array.
*
* @param val Integer to be formatted
* @param buf Buffer in which result is to be stored
* @param off Offset within buffer
* @param len Maximum length of integer
*
* @return offset of next unused character in input buffer.
*/
public int format(int val, byte[] buf, int off, int len) {
// Special case
if (val == Integer.MIN_VALUE) {
if (len > ByteFormatter.NUMBER_BASE) {
return format("-2147483648", buf, off, len);
}
truncationFiller(buf, off, len);
return off + len;
}
int pos = Math.abs(val);
// First count the number of characters in the result.
// Otherwise we need to use an intermediary buffer.
int ndig = 1;
int dmax = ByteFormatter.NUMBER_BASE;
while (ndig < ByteFormatter.NUMBER_BASE && pos >= dmax) {
ndig++;
dmax *= ByteFormatter.NUMBER_BASE;
}
if (val < 0) {
ndig++;
}
// Truncate if necessary.
if (ndig > len || ndig > buf.length - off) {
truncationFiller(buf, off, len);
return off + len;
}
// Now insert the actual characters we want -- backwards
// We use a do{} while() to handle the caByteFormatterse of 0.
off += ndig;
int xoff = off - 1;
do {
buf[xoff] = ByteFormatter.DIGITS[pos % ByteFormatter.NUMBER_BASE];
xoff--;
pos /= ByteFormatter.NUMBER_BASE;
} while (pos > 0);
if (val < 0) {
buf[xoff] = (byte) '-';
}
return off;
}
/**
* Format a long into an array.
*
* @param val The long to be formatted.
* @param array The array in which to place the result.
*
* @return The number of characters used. @ * if the value was truncated
*/
public int format(long val, byte[] array) {
return format(val, array, 0, array.length);
}
/**
* Format a long into an existing array.
*
* @param val Long to be formatted
* @param buf Buffer in which result is to be stored
* @param off Offset within buffer
* @param len Maximum length of integer
*
* @return offset of next unused character in input buffer. @ * if the value was truncated
*/
public int format(long val, byte[] buf, int off, int len) {
// Special case
if (val == Long.MIN_VALUE) {
if (len > MAX_LONG_LENGTH) {
return format("-9223372036854775808", buf, off, len);
}
truncationFiller(buf, off, len);
return off + len;
}
long pos = Math.abs(val);
// First count the number of characters in the result.
// Otherwise we need to use an intermediary buffer.
int ndig = 1;
long dmax = ByteFormatter.NUMBER_BASE;
// Might be faster to try to do this partially in ints
while (ndig < MAX_LONG_LENGTH && pos >= dmax) {
ndig++;
dmax *= ByteFormatter.NUMBER_BASE;
}
if (val < 0) {
ndig++;
}
// Truncate if necessary.
if (ndig > len || ndig > buf.length - off) {
truncationFiller(buf, off, len);
return off + len;
}
// Now insert the actual characters we want -- backwards.
off += ndig;
int xoff = off - 1;
buf[xoff] = (byte) '0';
boolean last = pos == 0;
while (!last) {
// Work on ints rather than longs.
int giga = (int) (pos % LONG_TO_INT_MODULO);
pos /= LONG_TO_INT_MODULO;
last = pos == 0;
for (int i = 0; i < MAXIMUM_SINGLE_DIGIT_INTEGER; i++) {
buf[xoff] = ByteFormatter.DIGITS[giga % ByteFormatter.NUMBER_BASE];
xoff--;
giga /= ByteFormatter.NUMBER_BASE;
if (last && giga == 0) {
break;
}
}
}
if (val < 0) {
buf[xoff] = (byte) '-';
}
return off;
}
/**
* Insert a string at the beginning of an array. * @return Offset of next available character in buffer.
*
* @param val The string to be inserted. A null string will insert len spaces.
* @param array The buffer in which to insert the string.
*
* @return Offset of next available character in buffer.
*/
public int format(String val, byte[] array) {
return format(val, array, 0, array.length);
}
/**
* Insert a String into an existing character array. If the String is longer than len, then only the the initial len
* characters will be inserted.
*
* @param val The string to be inserted. A null string will insert len spaces.
* @param array The buffer in which to insert the string.
* @param off The starting offset to insert the string.
* @param len The maximum number of characters to insert.
*
* @return Offset of next available character in buffer.
*/
public int format(String val, byte[] array, int off, int len) {
if (val == null) {
for (int i = 0; i < len; i++) {
array[off + i] = (byte) ' ';
}
return off + len;
}
int slen = val.length();
if (slen > len || slen > array.length - off) {
val = val.substring(0, len);
slen = len;
}
System.arraycopy(AsciiFuncs.getBytes(val), 0, array, off, slen);
return off + slen;
}
/**
* Write the mantissa of the number. This method addresses the subtleties involved in rounding numbers.
*/
private int mantissa(byte[] mant, int lmant, int exp, boolean simple, byte[] buf, int off, int len) {
// Save in case we need to extend the number.
int off0 = off;
int pos = 0;
if (exp < 0) {
buf[off] = (byte) '0';
len--;
off++;
if (len > 0) {
buf[off] = (byte) '.';
off++;
len--;
}
// Leading 0s in small numbers.
int cexp = exp;
while (cexp < -1 && len > 0) {
buf[off] = (byte) '0';
cexp++;
off++;
len--;
}
} else {
// Print out all digits to the left of the decimal.
while (exp >= 0 && pos < lmant) {
buf[off] = mant[pos];
off++;
pos++;
len--;
exp--;
}
// Trust we have enough space for this.
for (int i = 0; i <= exp; i++) {
buf[off] = (byte) '0';
off++;
len--;
}
// Add in a decimal if we have space.
if (len > 0) {
buf[off] = (byte) '.';
len--;
off++;
}
}
// Now handle the digits to the right of the decimal.
while (len > 0 && pos < lmant) {
buf[off] = mant[pos];
off++;
exp--;
len--;
pos++;
}
// Now handle rounding.
if (pos < lmant && mant[pos] >= (byte) '5') {
int i;
// Increment to the left until we find a non-9
for (i = off - 1; i >= off0; i--) {
if (buf[i] == (byte) '.' || buf[i] == (byte) '-') {
continue;
}
if (buf[i] != (byte) '9') {
buf[i]++;
break;
}
buf[i] = (byte) '0';
}
// Now we handle 99.99 case. This can cause problems
// in two cases. If we are not using scientific notation
// then we may want to convert 99.9 to 100., i.e.,
// we need to move the decimal point. If there is no
// decimal point, then we must not be truncating on overflow
// but we should be allowed to write it to the
// next character (i.e., we are not at the end of buf).
//
// If we are printing in scientific notation, then we want
// to convert 9.99 to 1.00, i.e. we do not move the decimal.
// However we need to signal that the exponent should be
// incremented by one.
//
// We cannot have aligned the number, since that requires
// the full precision number to fit within the requested
// length, and we would have printed out the entire
// mantissa (i.e., pos >= lmant)
if (i < off0) {
buf[off0] = (byte) '1';
boolean foundDecimal = false;
for (i = off0 + 1; i < off; i++) {
if (buf[i] == (byte) '.') {
foundDecimal = true;
if (simple) {
buf[i] = (byte) '0';
i++;
if (i < off) {
buf[i] = (byte) '.';
}
}
break;
}
}
if (simple && !foundDecimal) {
buf[off + 1] = (byte) '0'; // 99 went to 100
off++;
}
off = -off; // Signal to change exponent if necessary.
}
}
return off;
}
/**
* Fill the buffer with truncation characters. After filling the buffer, a TruncationException will be thrown if the
* appropriate flag is set.
*/
private void truncationFiller(byte[] buffer, int offset, int length) {
for (int i = offset; i < offset + length; i++) {
buffer[i] = ByteFormatter.TRUNCATION_FILL;
}
}
}