Lines Matching refs:digits
37 char digits[];
45 private FloatingDecimal( boolean negSign, int decExponent, char []digits, int n, boolean e )
50 this.digits = digits;
296 * The only reason that we develop the digits here, rather than
303 char digits[];
329 digits = (char[])(perThreadBuffer.get());
339 digits[digitno--] = (char)(c+'0');
344 digits[digitno] = (char)(c+'0');
349 digits = (char[])(perThreadBuffer.get());
359 digits[digitno--] = (char)(c+'0');
364 digits[digitno] = (char)(c+'0');
369 System.arraycopy( digits, digitno, result, 0, ndigits );
370 this.digits = result;
385 int q = digits[ i = (nDigits-1)];
388 digits[i] = '0';
389 q = digits[--i];
394 digits[0] = '1';
399 digits[i] = (char)(q+1);
426 digits = infinity;
428 digits = notANumber;
431 nDigits = digits.length;
443 digits = zero;
486 digits = infinity;
488 digits = notANumber;
491 nDigits = digits.length;
503 digits = zero;
570 * The following causes excess digits to be printed
625 int Bbits; // binary digits needed to represent B, approx.
626 int tenSbits; // binary digits needed to represent 10*S, approx.
678 char digits[] = this.digits = new char[18];
724 digits[ndigit++] = (char)('0' + q);
752 digits[ndigit++] = (char)('0' + q);
777 digits[ndigit++] = (char)('0' + q);
805 digits[ndigit++] = (char)('0' + q);
841 digits[ndigit++] = (char)('0' + q);
858 digits[ndigit++] = (char)('0' + q);
867 this.digits = digits;
876 // choose based on which digits we like.
877 if ( (digits[nDigits-1]&1) != 0 ) roundup();
893 result.append( digits, 0, nDigits );
896 result.append( digits, 0, nDigits );
914 System.arraycopy(digits, 0, result, i, nDigits);
918 // print digits.digits.
920 System.arraycopy(digits, 0, result, i, charLength);
932 System.arraycopy(digits, charLength, result, i, t);
945 System.arraycopy(digits, 0, result, i, nDigits);
948 result[i++] = digits[0];
951 System.arraycopy(digits, 1, result, i, nDigits-1);
964 // decExponent has 1, 2, or 3, digits
1067 char[] digits = new char[ l ];
1093 digits[nDigits++] = '0';
1096 digits[nDigits++] = c;
1115 * At this point, we've scanned all the digits and decimal
1121 * nLeadZero of them were zeros before any other digits.
1122 * nTrailZero of them were zeros after any other digits.
1129 * special hack: if we saw no non-zero digits, then the
1134 digits = zero;
1218 // after the [Ee][+-], without seeing any digits at all
1219 // this is certainly an error. If we saw some digits,
1239 return new FloatingDecimal( isNegative, decExp, digits, nDigits, false );
1260 if(digits == infinity || digits == notANumber) {
1261 if(digits == notANumber)
1274 int iValue = (int)digits[0]-(int)'0';
1277 iValue = iValue*10 + (int)digits[i]-(int)'0';
1281 lValue = lValue*10L + (long)((int)digits[i]-(int)'0');
1287 * the first kDigits digits of the number.
1294 * We know that the digits can be represented
1363 * The sum of digits plus exponent is greater than
1471 FDBigInt bigD0 = new FDBigInt( lValue, digits, kDigits, nDigits );
1618 if(digits == infinity || digits == notANumber) {
1619 if(digits == notANumber)
1628 iValue = (int)digits[0]-(int)'0';
1630 iValue = iValue*10 + (int)digits[i]-(int)'0';
1636 * the first kDigits digits of the number.
1643 * We know that the digits can be represented
1695 * First, finish accumulating digits.
1701 lValue = lValue*10L + (long)((int)digits[i]-(int)'0');
1712 * The sum of digits plus exponent is greater than
1933 * letters are used in indicate nonzero digits:
1955 int leftDigits = 0; // number of meaningful digits to
1958 int rightDigits = 0; // number of digits to right of
2031 // digits. Therefore, if an NumberFormatException
2119 // hex digits.
2130 // Now, we must copy any partial hex digits into the
2179 // current digit and test for any nonzero digits in
2279 * least significant bit, the digits to the right of
2453 int limit = nd-5; // slurp digits 5 at a time.
2795 // digits. If S is properly normalized and q < 10 then