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Separate out parsing of decimal number. Use this to only allocate
memory for the significand once up-front. Also ignore insignificant trailing zeroes; this saves unnecessary multiplications later. git-svn-id: https://llvm.org/svn/llvm-project/llvm/trunk@42964 91177308-0d34-0410-b5e6-96231b3b80d8
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Neil Booth
parent
1074878fbf
commit
1870f29c6c
@ -80,16 +80,11 @@ namespace {
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return ((bits) + integerPartWidth - 1) / integerPartWidth;
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}
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unsigned int
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digitValue(unsigned int c)
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/* Returns 0U-9U. Return values >= 10U are not digits. */
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inline unsigned int
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decDigitValue(unsigned int c)
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{
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unsigned int r;
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r = c - '0';
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if(r <= 9)
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return r;
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return -1U;
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return c - '0';
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}
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unsigned int
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@ -112,6 +107,47 @@ namespace {
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return -1U;
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}
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/* Return the value of a decimal exponent of the form
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[+-]ddddddd.
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If the exponent overflows, returns a large exponent with the
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appropriate sign. */
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static int
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readExponent(const char *p)
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{
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bool isNegative;
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unsigned int absExponent;
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const unsigned int overlargeExponent = 24000; /* FIXME. */
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isNegative = (*p == '-');
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if (*p == '-' || *p == '+')
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p++;
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absExponent = decDigitValue(*p++);
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assert (absExponent < 10U);
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for (;;) {
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unsigned int value;
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value = decDigitValue(*p);
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if (value >= 10U)
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break;
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p++;
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value += absExponent * 10;
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if (absExponent >= overlargeExponent) {
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absExponent = overlargeExponent;
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break;
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}
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absExponent = value;
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}
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if (isNegative)
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return -(int) absExponent;
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else
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return (int) absExponent;
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}
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/* This is ugly and needs cleaning up, but I don't immediately see
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how whilst remaining safe. */
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static int
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@ -132,8 +168,8 @@ namespace {
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for(;;) {
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unsigned int value;
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value = digitValue(*p);
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if(value == -1U)
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value = decDigitValue(*p);
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if(value >= 10U)
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break;
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p++;
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@ -176,6 +212,62 @@ namespace {
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return p;
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}
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/* Given a normal decimal floating point number of the form
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dddd.dddd[eE][+-]ddd
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where the decimal point and exponent are optional, fill out the
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structure D. If the value is zero, V->firstSigDigit
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points to a zero, and the return exponent is zero. */
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struct decimalInfo {
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const char *firstSigDigit;
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const char *lastSigDigit;
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int exponent;
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};
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void
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interpretDecimal(const char *p, decimalInfo *D)
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{
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const char *dot;
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p = skipLeadingZeroesAndAnyDot (p, &dot);
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D->firstSigDigit = p;
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D->exponent = 0;
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for (;;) {
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if (*p == '.') {
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assert(dot == 0);
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dot = p++;
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}
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if (decDigitValue(*p) >= 10U)
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break;
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p++;
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}
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/* If number is all zerooes accept any exponent. */
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if (p != D->firstSigDigit) {
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if (*p == 'e' || *p == 'E')
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D->exponent = readExponent(p + 1);
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/* Implied decimal point? */
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if (!dot)
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dot = p;
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/* Drop insignificant trailing zeroes. */
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do
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do
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p--;
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while (*p == '0');
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while (*p == '.');
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/* Adjust the specified exponent for any decimal point. */
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D->exponent += (dot - p) - (dot > p);
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}
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D->lastSigDigit = p;
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}
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/* Return the trailing fraction of a hexadecimal number.
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DIGITVALUE is the first hex digit of the fraction, P points to
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the next digit. */
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@ -1981,104 +2073,65 @@ APFloat::roundSignificandWithExponent(const integerPart *decSigParts,
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APFloat::opStatus
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APFloat::convertFromDecimalString(const char *p, roundingMode rounding_mode)
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{
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const char *dot, *firstSignificantDigit;
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integerPart val, maxVal, decValue;
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decimalInfo D;
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opStatus fs;
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/* Skip leading zeroes and any decimal point. */
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p = skipLeadingZeroesAndAnyDot(p, &dot);
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firstSignificantDigit = p;
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/* Scan the text. */
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interpretDecimal(p, &D);
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/* The maximum number that can be multiplied by ten with any digit
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added without overflowing an integerPart. */
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maxVal = (~ (integerPart) 0 - 9) / 10;
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val = 0;
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while (val <= maxVal) {
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if (*p == '.') {
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assert(dot == 0);
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dot = p++;
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}
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decValue = digitValue(*p);
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if (decValue == -1U)
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break;
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p++;
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val = val * 10 + decValue;
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}
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integerPart *decSignificand;
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unsigned int partCount, maxPartCount;
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partCount = 0;
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maxPartCount = 4;
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decSignificand = new integerPart[maxPartCount];
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decSignificand[partCount++] = val;
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/* Now continue to do single-part arithmetic for as long as we can.
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Then do a part multiplication, and repeat. */
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while (decValue != -1U) {
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integerPart multiplier;
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val = 0;
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multiplier = 1;
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while (multiplier <= maxVal) {
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if (*p == '.') {
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assert(dot == 0);
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dot = p++;
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}
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decValue = digitValue(*p);
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if (decValue == -1U)
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break;
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p++;
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multiplier *= 10;
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val = val * 10 + decValue;
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}
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if (partCount == maxPartCount) {
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integerPart *newDecSignificand;
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newDecSignificand = new integerPart[maxPartCount = partCount * 2];
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APInt::tcAssign(newDecSignificand, decSignificand, partCount);
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delete [] decSignificand;
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decSignificand = newDecSignificand;
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}
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APInt::tcMultiplyPart(decSignificand, decSignificand, multiplier, val,
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partCount, partCount + 1, false);
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/* If we used another part (likely), increase the count. */
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if (decSignificand[partCount] != 0)
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partCount++;
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}
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/* Now decSignificand contains the supplied significand ignoring the
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decimal point. Figure out our effective exponent, which is the
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specified exponent adjusted for any decimal point. */
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if (p == firstSignificantDigit) {
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/* Ignore the exponent if we are zero - we cannot overflow. */
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if (*D.firstSigDigit == '0') {
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category = fcZero;
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fs = opOK;
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} else {
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int decimalExponent;
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integerPart *decSignificand;
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unsigned int partCount;
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if (dot)
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decimalExponent = dot + 1 - p;
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else
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decimalExponent = 0;
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/* A tight upper bound on number of bits required to hold an
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N-digit decimal integer is N * 256 / 77. Allocate enough space
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to hold the full significand, and an extra part required by
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tcMultiplyPart. */
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partCount = (D.lastSigDigit - D.firstSigDigit) + 1;
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partCount = partCountForBits(1 + 256 * partCount / 77);
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decSignificand = new integerPart[partCount + 1];
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partCount = 0;
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/* Add the given exponent. */
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if (*p == 'e' || *p == 'E')
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decimalExponent = totalExponent(p, decimalExponent);
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/* Convert to binary efficiently - we do almost all multiplication
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in an integerPart. When this would overflow do we do a single
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bignum multiplication, and then revert again to multiplication
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in an integerPart. */
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do {
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integerPart decValue, val, multiplier;
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val = 0;
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multiplier = 1;
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do {
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if (*p == '.')
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p++;
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decValue = decDigitValue(*p++);
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multiplier *= 10;
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val = val * 10 + decValue;
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/* The maximum number that can be multiplied by ten with any
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digit added without overflowing an integerPart. */
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} while (p <= D.lastSigDigit && multiplier <= (~ (integerPart) 0 - 9) / 10);
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/* Multiply out the current part. */
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APInt::tcMultiplyPart(decSignificand, decSignificand, multiplier, val,
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partCount, partCount + 1, false);
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/* If we used another part (likely but not guaranteed), increase
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the count. */
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if (decSignificand[partCount])
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partCount++;
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} while (p <= D.lastSigDigit);
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category = fcNormal;
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fs = roundSignificandWithExponent(decSignificand, partCount,
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decimalExponent, rounding_mode);
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}
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D.exponent, rounding_mode);
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delete [] decSignificand;
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delete [] decSignificand;
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}
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return fs;
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}
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