Lines Matching refs:digits
38 char digits[];
45 int precision; // number of digits to the right of decimal
51 private FormattedFloatingDecimal( boolean negSign, int decExponent, char []digits, int n, boolean e, int precision, Form form )
56 this.digits = digits;
304 * The only reason that we develop the digits here, rather than
311 char digits[];
337 digits = (char[])(perThreadBuffer.get());
347 digits[digitno--] = (char)(c+'0');
352 digits[digitno] = (char)(c+'0');
357 digits = (char[])(perThreadBuffer.get());
367 digits[digitno--] = (char)(c+'0');
372 digits[digitno] = (char)(c+'0');
377 System.arraycopy( digits, digitno, result, 0, ndigits );
378 this.digits = result;
393 int q = digits[ i = (nDigits-1)];
396 digits[i] = '0';
397 q = digits[--i];
402 digits[0] = '1';
407 digits[i] = (char)(q+1);
410 // Given the desired number of digits predict the result's exponent.
416 if (digits[i] != '9')
417 // a '9' anywhere in digits will absorb the round
419 return decExponent + (digits[length] >= '5' ? 1 : 0);
422 // Unlike roundup(), this method does not modify digits. It also
430 System.arraycopy(digits, 0, result, 0, nDigits);
435 // excludes all significant digits
436 if (digits[0] >= '5') {
443 int q = digits[i];
445 q = digits[--i];
448 q = digits[--i];
459 result[i] = digits[i];
496 digits = infinity;
498 digits = notANumber;
501 nDigits = digits.length;
513 digits = zero;
563 digits = infinity;
565 digits = notANumber;
568 nDigits = digits.length;
580 digits = zero;
647 * The following causes excess digits to be printed
702 int Bbits; // binary digits needed to represent B, approx.
703 int tenSbits; // binary digits needed to represent 10*S, approx.
755 char digits[] = this.digits = new char[18];
801 digits[ndigit++] = (char)('0' + q);
829 digits[ndigit++] = (char)('0' + q);
854 digits[ndigit++] = (char)('0' + q);
882 digits[ndigit++] = (char)('0' + q);
918 digits[ndigit++] = (char)('0' + q);
935 digits[ndigit++] = (char)('0' + q);
944 this.digits = digits;
953 // choose based on which digits we like.
954 if ( (digits[nDigits-1]&1) != 0 ) roundup();
970 result.append( digits, 0, nDigits );
973 result.append( digits, 0, nDigits );
995 System.arraycopy(digits, 0, result, i, nDigits);
998 char digits [] = this.digits;
1005 digits = applyPrecision(decExponent + precision);
1009 digits = applyPrecision(precision + 1);
1013 digits = applyPrecision(precision);
1014 // adjust precision to be the number of digits to right of decimal
1033 // print digits.digits.
1035 System.arraycopy(digits, 0, result, i, charLength);
1056 System.arraycopy(digits, charLength, result, i, t);
1065 System.arraycopy(digits, charLength, result, i, t);
1074 // print 0.0* digits
1077 // write '0' s before the significant digits
1085 int t = Math.min(digits.length, precision + exp);
1089 // copy only when significant digits are within the precision
1090 System.arraycopy(digits, 0, result, i, t);
1094 result[i++] = digits[0];
1098 System.arraycopy(digits, 1, result, i, nDigits-1);
1109 System.arraycopy(digits, 1, result, i, t);
1124 // decExponent has 1, 2, or 3, digits
1166 if(digits == infinity || digits == notANumber) {
1167 if(digits == notANumber)
1180 int iValue = (int)digits[0]-(int)'0';
1183 iValue = iValue*10 + (int)digits[i]-(int)'0';
1187 lValue = lValue*10L + (long)((int)digits[i]-(int)'0');
1193 * the first kDigits digits of the number.
1200 * We know that the digits can be represented
1269 * The sum of digits plus exponent is greater than
1377 FDBigInt bigD0 = new FDBigInt( lValue, digits, kDigits, nDigits );
1524 if(digits == infinity || digits == notANumber) {
1525 if(digits == notANumber)
1534 iValue = (int)digits[0]-(int)'0';
1536 iValue = iValue*10 + (int)digits[i]-(int)'0';
1542 * the first kDigits digits of the number.
1549 * We know that the digits can be represented
1601 * First, finish accumulating digits.
1607 lValue = lValue*10L + (long)((int)digits[i]-(int)'0');
1618 * The sum of digits plus exponent is greater than