c_cpp Redis数字字符串转换优化

Posted 2021-05-14

/* Return the number of digits of 'v' when converted to string in radix 10.
 * See ll2string() for more information. */
uint32_t digits10(uint64_t v) {
    if (v < 10) return 1;
    if (v < 100) return 2;
    if (v < 1000) return 3;
    if (v < 1000000000000UL) {
        if (v < 100000000UL) {
            if (v < 1000000) {
                if (v < 10000) return 4;
                return 5 + (v >= 100000);
            }
            return 7 + (v >= 10000000UL);
        }
        if (v < 10000000000UL) {
            return 9 + (v >= 1000000000UL);
        }
        return 11 + (v >= 100000000000UL);
    }
    return 12 + digits10(v / 1000000000000UL);
}

/* Like digits10() but for signed values. */
uint32_t sdigits10(int64_t v) {
    if (v < 0) {
        /* Abs value of LLONG_MIN requires special handling. */
        uint64_t uv = (v != LLONG_MIN) ?
                      (uint64_t)-v : ((uint64_t) LLONG_MAX)+1;
        return digits10(uv)+1; /* +1 for the minus. */
    } else {
        return digits10(v);
    }
}

/* Convert a long long into a string. Returns the number of
 * characters needed to represent the number.
 * If the buffer is not big enough to store the string, 0 is returned.
 *
 * Based on the following article (that apparently does not provide a
 * novel approach but only publicizes an already used technique):
 *
 * https://www.facebook.com/notes/facebook-engineering/three-optimization-tips-for-c/10151361643253920
 *
 * Modified in order to handle signed integers since the original code was
 * designed for unsigned integers. */
int ll2string(char *dst, size_t dstlen, long long svalue) {
    static const char digits[201] =
        "0001020304050607080910111213141516171819"
        "2021222324252627282930313233343536373839"
        "4041424344454647484950515253545556575859"
        "6061626364656667686970717273747576777879"
        "8081828384858687888990919293949596979899";
    int negative;
    unsigned long long value;

    /* The main loop works with 64bit unsigned integers for simplicity, so
     * we convert the number here and remember if it is negative. */
    if (svalue < 0) {
        if (svalue != LLONG_MIN) {
            value = -svalue;
        } else {
            value = ((unsigned long long) LLONG_MAX)+1;
        }
        negative = 1;
    } else {
        value = svalue;
        negative = 0;
    }

    /* Check length. */
    uint32_t const length = digits10(value)+negative;
    if (length >= dstlen) return 0;

    /* Null term. */
    uint32_t next = length;
    dst[next] = '\0';
    next--;
    while (value >= 100) {
        int const i = (value % 100) * 2;
        value /= 100;
        dst[next] = digits[i + 1];
        dst[next - 1] = digits[i];
        next -= 2;
    }

    /* Handle last 1-2 digits. */
    if (value < 10) {
        dst[next] = '0' + (uint32_t) value;
    } else {
        int i = (uint32_t) value * 2;
        dst[next] = digits[i + 1];
        dst[next - 1] = digits[i];
    }

    /* Add sign. */
    if (negative) dst[0] = '-';
    return length;
}

/* Convert a string into a long long. Returns 1 if the string could be parsed
 * into a (non-overflowing) long long, 0 otherwise. The value will be set to
 * the parsed value when appropriate.
 *
 * Note that this function demands that the string strictly represents
 * a long long: no spaces or other characters before or after the string
 * representing the number are accepted, nor zeroes at the start if not
 * for the string "0" representing the zero number.
 *
 * Because of its strictness, it is safe to use this function to check if
 * you can convert a string into a long long, and obtain back the string
 * from the number without any loss in the string representation. */
int string2ll(const char *s, size_t slen, long long *value) {
    const char *p = s;
    size_t plen = 0;
    int negative = 0;
    unsigned long long v;

    if (plen == slen)
        return 0;

    /* Special case: first and only digit is 0. */
    if (slen == 1 && p[0] == '0') {
        if (value != NULL) *value = 0;
        return 1;
    }

    if (p[0] == '-') {
        negative = 1;
        p++; plen++;

        /* Abort on only a negative sign. */
        if (plen == slen)
            return 0;
    }

    /* First digit should be 1-9, otherwise the string should just be 0. */
    if (p[0] >= '1' && p[0] <= '9') {
        v = p[0]-'0';
        p++; plen++;
    } else if (p[0] == '0' && slen == 1) {
        *value = 0;
        return 1;
    } else {
        return 0;
    }

    while (plen < slen && p[0] >= '0' && p[0] <= '9') {
        if (v > (ULLONG_MAX / 10)) /* Overflow. */
            return 0;
        v *= 10;

        if (v > (ULLONG_MAX - (p[0]-'0'))) /* Overflow. */
            return 0;
        v += p[0]-'0';

        p++; plen++;
    }

    /* Return if not all bytes were used. */
    if (plen < slen)
        return 0;

    if (negative) {
        if (v > ((unsigned long long)(-(LLONG_MIN+1))+1)) /* Overflow. */
            return 0;
        if (value != NULL) *value = -v;
    } else {
        if (v > LLONG_MAX) /* Overflow. */
            return 0;
        if (value != NULL) *value = v;
    }
    return 1;
}

/* Convert a string into a long. Returns 1 if the string could be parsed into a
 * (non-overflowing) long, 0 otherwise. The value will be set to the parsed
 * value when appropriate. */
int string2l(const char *s, size_t slen, long *lval) {
    long long llval;

    if (!string2ll(s,slen,&llval))
        return 0;

    if (llval < LONG_MIN || llval > LONG_MAX)
        return 0;

    *lval = (long)llval;
    return 1;
}

/* Convert a string into a double. Returns 1 if the string could be parsed
 * into a (non-overflowing) double, 0 otherwise. The value will be set to
 * the parsed value when appropriate.
 *
 * Note that this function demands that the string strictly represents
 * a double: no spaces or other characters before or after the string
 * representing the number are accepted. */
int string2ld(const char *s, size_t slen, long double *dp) {
    char buf[256];
    long double value;
    char *eptr;

    if (slen >= sizeof(buf)) return 0;
    memcpy(buf,s,slen);
    buf[slen] = '\0';

    errno = 0;
    value = strtold(buf, &eptr);
    if (isspace(buf[0]) || eptr[0] != '\0' ||
        (errno == ERANGE &&
            (value == HUGE_VAL || value == -HUGE_VAL || value == 0)) ||
        errno == EINVAL ||
        isnan(value))
        return 0;

    if (dp) *dp = value;
    return 1;
}

/* Convert a double to a string representation. Returns the number of bytes
 * required. The representation should always be parsable by strtod(3).
 * This function does not support human-friendly formatting like ld2string
 * does. It is intented mainly to be used inside t_zset.c when writing scores
 * into a ziplist representing a sorted set. */
int d2string(char *buf, size_t len, double value) {
    if (isnan(value)) {
        len = snprintf(buf,len,"nan");
    } else if (isinf(value)) {
        if (value < 0)
            len = snprintf(buf,len,"-inf");
        else
            len = snprintf(buf,len,"inf");
    } else if (value == 0) {
        /* See: http://en.wikipedia.org/wiki/Signed_zero, "Comparisons". */
        if (1.0/value < 0)
            len = snprintf(buf,len,"-0");
        else
            len = snprintf(buf,len,"0");
    } else {
#if (DBL_MANT_DIG >= 52) && (LLONG_MAX == 0x7fffffffffffffffLL)
        /* Check if the float is in a safe range to be casted into a
         * long long. We are assuming that long long is 64 bit here.
         * Also we are assuming that there are no implementations around where
         * double has precision < 52 bit.
         *
         * Under this assumptions we test if a double is inside an interval
         * where casting to long long is safe. Then using two castings we
         * make sure the decimal part is zero. If all this is true we use
         * integer printing function that is much faster. */
        double min = -4503599627370495; /* (2^52)-1 */
        double max = 4503599627370496; /* -(2^52) */
        if (value > min && value < max && value == ((double)((long long)value)))
            len = ll2string(buf,len,(long long)value);
        else
#endif
            len = snprintf(buf,len,"%.17g",value);
    }

    return len;
}

/* Convert a long double into a string. If humanfriendly is non-zero
 * it does not use exponential format and trims trailing zeroes at the end,
 * however this results in loss of precision. Otherwise exp format is used
 * and the output of snprintf() is not modified.
 *
 * The function returns the length of the string or zero if there was not
 * enough buffer room to store it. */
int ld2string(char *buf, size_t len, long double value, int humanfriendly) {
    size_t l;

    if (isinf(value)) {
        /* Libc in odd systems (Hi Solaris!) will format infinite in a
         * different way, so better to handle it in an explicit way. */
        if (len < 5) return 0; /* No room. 5 is "-inf\0" */
        if (value > 0) {
            memcpy(buf,"inf",3);
            l = 3;
        } else {
            memcpy(buf,"-inf",4);
            l = 4;
        }
    } else if (humanfriendly) {
        /* We use 17 digits precision since with 128 bit floats that precision
         * after rounding is able to represent most small decimal numbers in a
         * way that is "non surprising" for the user (that is, most small
         * decimal numbers will be represented in a way that when converted
         * back into a string are exactly the same as what the user typed.) */
        l = snprintf(buf,len,"%.17Lf", value);
        if (l+1 > len) return 0; /* No room. */
        /* Now remove trailing zeroes after the '.' */
        if (strchr(buf,'.') != NULL) {
            char *p = buf+l-1;
            while(*p == '0') {
                p--;
                l--;
            }
            if (*p == '.') l--;
        }
    } else {
        l = snprintf(buf,len,"%.17Lg", value);
        if (l+1 > len) return 0; /* No room. */
    }
    buf[l] = '\0';
    return l;
}