将浮点数转换为字符串

Question

将浮点数转换为字符串

c++cfloating-point

20

如何在C/C++中将浮点整数转换为字符串而不使用库函数 sprintf？

我正在寻找一个函数，例如char *ftoa(float num)，它将num转换为字符串并返回它。

ftoa(3.1415)应该返回"3.1415"。

- SIVA

1

C语言中有浮点数转换为字符的功能，例如sprintf(a, "%d", f)。 - Hans W

一个解决方案：http://www.edaboard.com/ftopic41714.html#160029 - tur1ng

只需使用 sprint 函数来格式化浮点值。 - John Lawlor

12个回答

11

处理浮点数时，算法非常简单，类似于Edgar Holleis的答案; 并致以嘉奖！但这很复杂，因为在处理浮点数时，根据所选择的精度，计算结果可能会有一些偏差。这就是为什么将float与零进行比较不是良好的编程实践。

但是，有一个解决方案，这是我尝试实现它的方式。在这里，我使用了一个公差值，以便您不会计算太多小数位从而导致无限循环。我相信可能会有更好的解决方案，但这应该有助于让您对如何执行此操作有一个良好的了解。

char fstr[80];
float num = 2.55f;
int m = log10(num);
int digit;
float tolerance = .0001f;

while (num > 0 + precision)
{
    float weight = pow(10.0f, m);
    digit = floor(num / weight);
    num -= (digit*weight);
    *(fstr++)= '0' + digit;
    if (m == 0)
        *(fstr++) = '.';
    m--;
}
*(fstr) = '\0';

- Sophy Pal

这里有一个问题。如果num小于1，你会计算错误的字符串。你需要将第三行更改为floor(log10(num))。 - Andi Jay

另一个问题是，如果您尝试转换一个个位数为0的数字。例如230.0。在当前写法下，您只会得到23。您必须更改while循环中的条件为((num > 0 + precision)||(m >= 0))。 - Andi Jay

只是要注意，这个实现以及androider的改进都很不错。然而，使用高精度（例如：9位小数）时，重复的pow()可能会使其变得相当缓慢。如果这是提供用户反馈的其他操作的一部分，那就不太好了。 - Toby

精度和公差是指的同一个概念吗？精度未定义，而公差未被使用。 - RufusVS

6

你可以使用C++20的std::format或{fmt}库（基于std::format）将浮点数转换为字符串，例如：

std::string s = std::format("{}", M_PI);

与 sprintf 相比，std::format 的优点在于它可以提供最短的十进制表示，并保证往返转换。

- vitaut

5

使用log函数查找数字的幅值m。如果幅值为负，则打印"0."和适当数量的零。
连续除以10^m，并将结果转换为int以获得小数位数。m--用于下一个数字。
如果遇到m==0，不要忘记打印小数点"."。
在几个数字后中断。如果在中断时m>0，不要忘记打印"E"和itoa(m)。

您也可以通过位移并校正指数偏移（请参见IEEE 754）直接提取指数。Java具有double-to-bits函数以获取二进制表示。

- edgar.holleis

3

 /*
  * Program to convert float number to string without using sprintf
  */

#include "iostream"    
#include "string"    
#include "math.h"

# define PRECISION 5

using namespace std;

char*  floatToString(float num)
{
   int whole_part = num;
   int digit = 0, reminder =0;
   int log_value = log10(num), index = log_value;
   long wt =0;

   // String containg result
   char* str = new char[20];

   //Initilise stirng to zero
   memset(str, 0 ,20);

   //Extract the whole part from float num
   for(int  i = 1 ; i < log_value + 2 ; i++)
   {
       wt  =  pow(10.0,i);
       reminder = whole_part  %  wt;
       digit = (reminder - digit) / (wt/10);

       //Store digit in string
       str[index--] = digit + 48;              // ASCII value of digit  = digit + 48
       if (index == -1)
          break;    
   }

    index = log_value + 1;
    str[index] = '.';

   float fraction_part  = num - whole_part;
   float tmp1 = fraction_part,  tmp =0;

   //Extract the fraction part from  num
   for( int i= 1; i < PRECISION; i++)
   {
      wt =10; 
      tmp  = tmp1 * wt;
      digit = tmp;

      //Store digit in string
      str[++index] = digit +48;           // ASCII value of digit  = digit + 48
      tmp1 = tmp - digit;
   }    

   return str;
}


//Main program
void main()
{
    int i;
    float f = 123456.789;
    char* str =  floatToString(f);
    cout  << endl <<  str;
    cin >> i;
    delete [] str;
}

- Rahul Naik

1

你有两个主要问题：

将位表示转换为字符字符串
分配足够的内存来存储字符。

解决第二部分最简单的方法是为每个可能的答案分配足够大的空间。从这里开始。稍后你会想变得更聪明，但在解决数字部分之前不必费心。

你有两组工具可用于处理问题的数字部分：直接位操作（掩码、移位等）和算术运算（*，+，/ ，加上可能的数学函数如 log()）。

原则上，你可以直接处理位表示，但如果浮点表示格式在未来发生变化，那就不具备可移植性。 edgar.holleis建议的方法应该是可移植的。

- dmckee --- ex-moderator kitten

1

我刚刚发现了一个非常好的实现，位于https://code.google.com/p/stringencoders/。

size_t modp_dtoa(double value, char* str, int prec)
{
    /* Hacky test for NaN
     * under -fast-math this won't work, but then you also won't
     * have correct nan values anyways.  The alternative is
     * to link with libmath (bad) or hack IEEE double bits (bad)
     */
    if (! (value == value)) {
        str[0] = 'n'; str[1] = 'a'; str[2] = 'n'; str[3] = '\0';
        return (size_t)3;
    }
    /* if input is larger than thres_max, revert to exponential */
    const double thres_max = (double)(0x7FFFFFFF);

    double diff = 0.0;
    char* wstr = str;

    if (prec < 0) {
        prec = 0;
    } else if (prec > 9) {
        /* precision of >= 10 can lead to overflow errors */
        prec = 9;
    }


    /* we'll work in positive values and deal with the
       negative sign issue later */
    int neg = 0;
    if (value < 0) {
        neg = 1;
        value = -value;
    }


    int whole = (int) value;
    double tmp = (value - whole) * powers_of_10[prec];
    uint32_t frac = (uint32_t)(tmp);
    diff = tmp - frac;

    if (diff > 0.5) {
        ++frac;
        /* handle rollover, e.g.  case 0.99 with prec 1 is 1.0  */
        if (frac >= powers_of_10[prec]) {
            frac = 0;
            ++whole;
        }
    } else if (diff == 0.5 && ((frac == 0) || (frac & 1))) {
        /* if halfway, round up if odd, OR
           if last digit is 0.  That last part is strange */
        ++frac;
    }

    /* for very large numbers switch back to native sprintf for exponentials.
       anyone want to write code to replace this? */
    /*
      normal printf behavior is to print EVERY whole number digit
      which can be 100s of characters overflowing your buffers == bad
    */
    if (value > thres_max) {
        sprintf(str, "%e", neg ? -value : value);
        return strlen(str);
    }

    if (prec == 0) {
        diff = value - whole;
        if (diff > 0.5) {
            /* greater than 0.5, round up, e.g. 1.6 -> 2 */
            ++whole;
        } else if (diff == 0.5 && (whole & 1)) {
            /* exactly 0.5 and ODD, then round up */
            /* 1.5 -> 2, but 2.5 -> 2 */
            ++whole;
        }
    } else {
        int count = prec;
        // now do fractional part, as an unsigned number
        do {
            --count;
            *wstr++ = (char)(48 + (frac % 10));
        } while (frac /= 10);
        // add extra 0s
        while (count-- > 0) *wstr++ = '0';
        // add decimal
        *wstr++ = '.';
    }

    // do whole part
    // Take care of sign
    // Conversion. Number is reversed.
    do *wstr++ = (char)(48 + (whole % 10)); while (whole /= 10);
    if (neg) {
        *wstr++ = '-';
    }
    *wstr='\0';
    strreverse(str, wstr-1);
    return (size_t)(wstr - str);
}

- shapkin

嗯，如果你已经在使用 string.h 函数，不妨使用 strcpy(str,"nan"); 代替逐个更改字符。此外，如果你不介意使用 math.h（并且编译时加上 -lm），你可以使用 isnan(value) 而不是 hacky 测试。 - MD XF

0

这是我想出来的方案；它非常高效且非常简单。它假设您的系统具有 itoa。

#include <math.h>
#include <string.h>

/* return decimal part of val */
int dec(float val)
{
    int mult = floor(val);

    while (floor(val) != ceil(val)) {
        mult *= 10;
        val *= 10;
    }

    return floor(val) - mult;
}

/* convert a double to a string */
char *ftoa(float val, char *str)
{
    if (isnan(n)) {
        strcpy(str, "NaN");
        return str;
    } else if (isinf(n)) {
        strcpy(str, "inf");
        return str;
    }

    char leading_integer[31]  = {0};  // 63 instead of 31 for 64-bit systems
    char trailing_decimal[31] = {0};  // 63 instead of 31 for 64-bit systems

    /* fill string with leading integer */
    itoa(floor(val), leading_integer, 10);

    /* fill string with the decimal part */
    itoa(dec(val), trailing_decimal, 10);

    /* set given string to full decimal */
    strcpy(str, leading_integer);
    strcat(str, ".");
    strcat(str, trailing_decimal);

    return str;
}

在线尝试！

- MD XF

1

不适用于小数点后有零的数字（例如1.05->“1.5”）... - gregsmi

0

我的实现

//convert a floating point number to a C++ literal string
void ftoa(float value, char *str, int numberOfDecimals) {
  //find integer and fractional part
  float f_auxint, f_auxfract;
  f_auxfract = modff(value, &f_auxint);
  long intpart = long(f_auxint);
  long fractpart = long(floor((fabs(f_auxfract)*pow(10.0, float(numberOfDecimals)))+0.5));
  //covert integer part and copy it to literal string
  ltoa(intpart, &str[0], 10);
  //add dot (decimal separator)
  int lnhg = strlen(str);
  str[lnhg] = '.';
  lnhg += 1;
  //fill by '0' the rest of the destination string
  int aa = lnhg;
  for (size_t i = 0; i < numberOfDecimals; i++) {
    str[aa] = '0';
    aa += 1;
  }
  str[aa] = '\0';
  //covert fractional part in literal string
  char tempstr[10];
  ltoa(fractpart, &tempstr[0], 10);
  //shift cursor position based on how many zero there were
  int bb = strlen(tempstr);
  while (bb < numberOfDecimals) {
    lnhg += 1;
    bb+=1;
  }
  //paste fractional part to rest of the destination string
  for (size_t i = 0; i < strlen(tempstr); i++) {
    str[lnhg] = tempstr[i];
    lnhg+=1;
  }
}

- Lorenzo Brun

你的回答可以通过提供更多支持信息来改进。请编辑以添加进一步的细节，例如引用或文档，以便他人可以确认你的答案是正确的。您可以在帮助中心中找到有关如何编写良好答案的更多信息。 - Community

0

如果你想要一个基于Grisu2算法的非常快速的实现，我建议你看一下Lemire教授在github上的this file。

由于仅提供链接的答案是不被接受的，因此我将复制粘贴代码，并感谢Florian Grisu提供的算法以及Daniel Lemire教授对C++端口的支持和非常有趣和启发性的评论：

#include <cstring>
#include <cstdint>
#include <array>
namespace simdjson {
namespace internal {
// Skipped comments
namespace dtoa_impl {

template <typename Target, typename Source>
Target reinterpret_bits(const Source source) {
  static_assert(sizeof(Target) == sizeof(Source), "size mismatch");

  Target target;
  std::memcpy(&target, &source, sizeof(Source));
  return target;
}

struct diyfp // f * 2^e
{
  static constexpr int kPrecision = 64; // = q

  std::uint64_t f = 0;
  int e = 0;

  constexpr diyfp(std::uint64_t f_, int e_) noexcept : f(f_), e(e_) {}

  /*!
  @brief returns x - y
  @pre x.e == y.e and x.f >= y.f
  */
  static diyfp sub(const diyfp &x, const diyfp &y) noexcept {

    return {x.f - y.f, x.e};
  }

  /*!
  @brief returns x * y
  @note The result is rounded. (Only the upper q bits are returned.)
  */
  static diyfp mul(const diyfp &x, const diyfp &y) noexcept {
    static_assert(kPrecision == 64, "internal error");
// Skipped
    const std::uint64_t u_lo = x.f & 0xFFFFFFFFu;
    const std::uint64_t u_hi = x.f >> 32u;
    const std::uint64_t v_lo = y.f & 0xFFFFFFFFu;
    const std::uint64_t v_hi = y.f >> 32u;

    const std::uint64_t p0 = u_lo * v_lo;
    const std::uint64_t p1 = u_lo * v_hi;
    const std::uint64_t p2 = u_hi * v_lo;
    const std::uint64_t p3 = u_hi * v_hi;

    const std::uint64_t p0_hi = p0 >> 32u;
    const std::uint64_t p1_lo = p1 & 0xFFFFFFFFu;
    const std::uint64_t p1_hi = p1 >> 32u;
    const std::uint64_t p2_lo = p2 & 0xFFFFFFFFu;
    const std::uint64_t p2_hi = p2 >> 32u;

    std::uint64_t Q = p0_hi + p1_lo + p2_lo;

    // The full product might now be computed as
    //
    // p_hi = p3 + p2_hi + p1_hi + (Q >> 32)
    // p_lo = p0_lo + (Q << 32)
    //
    // But in this particular case here, the full p_lo is not required.
    // Effectively we only need to add the highest bit in p_lo to p_hi (and
    // Q_hi + 1 does not overflow).

    Q += std::uint64_t{1} << (64u - 32u - 1u); // round, ties up

    const std::uint64_t h = p3 + p2_hi + p1_hi + (Q >> 32u);

    return {h, x.e + y.e + 64};
  }

  /*!
  @brief normalize x such that the significand is >= 2^(q-1)
  @pre x.f != 0
  */
  static diyfp normalize(diyfp x) noexcept {

    while ((x.f >> 63u) == 0) {
      x.f <<= 1u;
      x.e--;
    }

    return x;
  }

  /*!
  @brief normalize x such that the result has the exponent E
  @pre e >= x.e and the upper e - x.e bits of x.f must be zero.
  */
  static diyfp normalize_to(const diyfp &x,
                            const int target_exponent) noexcept {
    const int delta = x.e - target_exponent;

    return {x.f << delta, target_exponent};
  }
};

struct boundaries {
  diyfp w;
  diyfp minus;
  diyfp plus;
};

/*!
Compute the (normalized) diyfp representing the input number 'value' and its
boundaries.
@pre value must be finite and positive
*/
template <typename FloatType> boundaries compute_boundaries(FloatType value) {

  // Convert the IEEE representation into a diyfp.
  //
  // If v is denormal:
  //      value = 0.F * 2^(1 - bias) = (          F) * 2^(1 - bias - (p-1))
  // If v is normalized:
  //      value = 1.F * 2^(E - bias) = (2^(p-1) + F) * 2^(E - bias - (p-1))

  static_assert(std::numeric_limits<FloatType>::is_iec559,
                "internal error: dtoa_short requires an IEEE-754 "
                "floating-point implementation");

  constexpr int kPrecision =
      std::numeric_limits<FloatType>::digits; // = p (includes the hidden bit)
  constexpr int kBias =
      std::numeric_limits<FloatType>::max_exponent - 1 + (kPrecision - 1);
  constexpr int kMinExp = 1 - kBias;
  constexpr std::uint64_t kHiddenBit = std::uint64_t{1}
                                       << (kPrecision - 1); // = 2^(p-1)

  using bits_type = typename std::conditional<kPrecision == 24, std::uint32_t,
                                              std::uint64_t>::type;

  const std::uint64_t bits = reinterpret_bits<bits_type>(value);
  const std::uint64_t E = bits >> (kPrecision - 1);
  const std::uint64_t F = bits & (kHiddenBit - 1);

  const bool is_denormal = E == 0;
  const diyfp v = is_denormal
                      ? diyfp(F, kMinExp)
                      : diyfp(F + kHiddenBit, static_cast<int>(E) - kBias);

  // Compute the boundaries m- and m+ of the floating-point value
  // v = f * 2^e.
  //
  // Determine v- and v+, the floating-point predecessor and successor if v,
  // respectively.
  //
  //      v- = v - 2^e        if f != 2^(p-1) or e == e_min                (A)
  //         = v - 2^(e-1)    if f == 2^(p-1) and e > e_min                (B)
  //
  //      v+ = v + 2^e
  //
  // Let m- = (v- + v) / 2 and m+ = (v + v+) / 2. All real numbers _strictly_
  // between m- and m+ round to v, regardless of how the input rounding
  // algorithm breaks ties.
  //
  //      ---+-------------+-------------+-------------+-------------+---  (A)
  //         v-            m-            v             m+            v+
  //
  //      -----------------+------+------+-------------+-------------+---  (B)
  //                       v-     m-     v             m+            v+

  const bool lower_boundary_is_closer = F == 0 && E > 1;
  const diyfp m_plus = diyfp(2 * v.f + 1, v.e - 1);
  const diyfp m_minus = lower_boundary_is_closer
                            ? diyfp(4 * v.f - 1, v.e - 2)  // (B)
                            : diyfp(2 * v.f - 1, v.e - 1); // (A)

  // Determine the normalized w+ = m+.
  const diyfp w_plus = diyfp::normalize(m_plus);

  // Determine w- = m- such that e_(w-) = e_(w+).
  const diyfp w_minus = diyfp::normalize_to(m_minus, w_plus.e);

  return {diyfp::normalize(v), w_minus, w_plus};
}

// Had to skip this

constexpr int kAlpha = -60;
constexpr int kGamma = -32;

struct cached_power // c = f * 2^e ~= 10^k
{
  std::uint64_t f;
  int e;
  int k;
};

/*!
For a normalized diyfp w = f * 2^e, this function returns a (normalized) cached
power-of-ten c = f_c * 2^e_c, such that the exponent of the product w * c
satisfies (Definition 3.2 from [1])
     alpha <= e_c + e + q <= gamma.
*/
inline cached_power get_cached_power_for_binary_exponent(int e) {

  constexpr int kCachedPowersMinDecExp = -300;
  constexpr int kCachedPowersDecStep = 8;

  static constexpr std::array<cached_power, 79> kCachedPowers = {{
      {0xAB70FE17C79AC6CA, -1060, -300}, {0xFF77B1FCBEBCDC4F, -1034, -292},
      {0xBE5691EF416BD60C, -1007, -284}, {0x8DD01FAD907FFC3C, -980, -276},
      {0xD3515C2831559A83, -954, -268},  {0x9D71AC8FADA6C9B5, -927, -260},
      {0xEA9C227723EE8BCB, -901, -252},  {0xAECC49914078536D, -874, -244},
      {0x823C12795DB6CE57, -847, -236},  {0xC21094364DFB5637, -821, -228},
      {0x9096EA6F3848984F, -794, -220},  {0xD77485CB25823AC7, -768, -212},
      {0xA086CFCD97BF97F4, -741, -204},  {0xEF340A98172AACE5, -715, -196},
      {0xB23867FB2A35B28E, -688, -188},  {0x84C8D4DFD2C63F3B, -661, -180},
      {0xC5DD44271AD3CDBA, -635, -172},  {0x936B9FCEBB25C996, -608, -164},
      {0xDBAC6C247D62A584, -582, -156},  {0xA3AB66580D5FDAF6, -555, -148},
      {0xF3E2F893DEC3F126, -529, -140},  {0xB5B5ADA8AAFF80B8, -502, -132},
      {0x87625F056C7C4A8B, -475, -124},  {0xC9BCFF6034C13053, -449, -116},
      {0x964E858C91BA2655, -422, -108},  {0xDFF9772470297EBD, -396, -100},
      {0xA6DFBD9FB8E5B88F, -369, -92},   {0xF8A95FCF88747D94, -343, -84},
      {0xB94470938FA89BCF, -316, -76},   {0x8A08F0F8BF0F156B, -289, -68},
      {0xCDB02555653131B6, -263, -60},   {0x993FE2C6D07B7FAC, -236, -52},
      {0xE45C10C42A2B3B06, -210, -44},   {0xAA242499697392D3, -183, -36},
      {0xFD87B5F28300CA0E, -157, -28},   {0xBCE5086492111AEB, -130, -20},
      {0x8CBCCC096F5088CC, -103, -12},   {0xD1B71758E219652C, -77, -4},
      {0x9C40000000000000, -50, 4},      {0xE8D4A51000000000, -24, 12},
      {0xAD78EBC5AC620000, 3, 20},       {0x813F3978F8940984, 30, 28},
      {0xC097CE7BC90715B3, 56, 36},      {0x8F7E32CE7BEA5C70, 83, 44},
      {0xD5D238A4ABE98068, 109, 52},     {0x9F4F2726179A2245, 136, 60},
      {0xED63A231D4C4FB27, 162, 68},     {0xB0DE65388CC8ADA8, 189, 76},
      {0x83C7088E1AAB65DB, 216, 84},     {0xC45D1DF942711D9A, 242, 92},
      {0x924D692CA61BE758, 269, 100},    {0xDA01EE641A708DEA, 295, 108},
      {0xA26DA3999AEF774A, 322, 116},    {0xF209787BB47D6B85, 348, 124},
      {0xB454E4A179DD1877, 375, 132},    {0x865B86925B9BC5C2, 402, 140},
      {0xC83553C5C8965D3D, 428, 148},    {0x952AB45CFA97A0B3, 455, 156},
      {0xDE469FBD99A05FE3, 481, 164},    {0xA59BC234DB398C25, 508, 172},
      {0xF6C69A72A3989F5C, 534, 180},    {0xB7DCBF5354E9BECE, 561, 188},
      {0x88FCF317F22241E2, 588, 196},    {0xCC20CE9BD35C78A5, 614, 204},
      {0x98165AF37B2153DF, 641, 212},    {0xE2A0B5DC971F303A, 667, 220},
      {0xA8D9D1535CE3B396, 694, 228},    {0xFB9B7CD9A4A7443C, 720, 236},
      {0xBB764C4CA7A44410, 747, 244},    {0x8BAB8EEFB6409C1A, 774, 252},
      {0xD01FEF10A657842C, 800, 260},    {0x9B10A4E5E9913129, 827, 268},
      {0xE7109BFBA19C0C9D, 853, 276},    {0xAC2820D9623BF429, 880, 284},
      {0x80444B5E7AA7CF85, 907, 292},    {0xBF21E44003ACDD2D, 933, 300},
      {0x8E679C2F5E44FF8F, 960, 308},    {0xD433179D9C8CB841, 986, 316},
      {0x9E19DB92B4E31BA9, 1013, 324},
  }};

  // This computation gives exactly the same results for k as
  //      k = ceil((kAlpha - e - 1) * 0.30102999566398114)
  // for |e| <= 1500, but doesn't require floating-point operations.
  // NB: log_10(2) ~= 78913 / 2^18
  const int f = kAlpha - e - 1;
  const int k = (f * 78913) / (1 << 18) + static_cast<int>(f > 0);

  const int index = (-kCachedPowersMinDecExp + k + (kCachedPowersDecStep - 1)) /
                    kCachedPowersDecStep;

  const cached_power cached = kCachedPowers[static_cast<std::size_t>(index)];

  return cached;
}

/*!
For n != 0, returns k, such that pow10 := 10^(k-1) <= n < 10^k.
For n == 0, returns 1 and sets pow10 := 1.
*/
inline int find_largest_pow10(const std::uint32_t n, std::uint32_t &pow10) {
  // LCOV_EXCL_START
  if (n >= 1000000000) {
    pow10 = 1000000000;
    return 10;
  }
  // LCOV_EXCL_STOP
  else if (n >= 100000000) {
    pow10 = 100000000;
    return 9;
  } else if (n >= 10000000) {
    pow10 = 10000000;
    return 8;
  } else if (n >= 1000000) {
    pow10 = 1000000;
    return 7;
  } else if (n >= 100000) {
    pow10 = 100000;
    return 6;
  } else if (n >= 10000) {
    pow10 = 10000;
    return 5;
  } else if (n >= 1000) {
    pow10 = 1000;
    return 4;
  } else if (n >= 100) {
    pow10 = 100;
    return 3;
  } else if (n >= 10) {
    pow10 = 10;
    return 2;
  } else {
    pow10 = 1;
    return 1;
  }
}

inline void grisu2_round(char *buf, int len, std::uint64_t dist,
                         std::uint64_t delta, std::uint64_t rest,
                         std::uint64_t ten_k) {

  //               <--------------------------- delta ---->
  //                                  <---- dist --------->
  // --------------[------------------+-------------------]--------------
  //               M-                 w                   M+
  //
  //                                  ten_k
  //                                <------>
  //                                       <---- rest ---->
  // --------------[------------------+----+--------------]--------------
  //                                  w    V
  //                                       = buf * 10^k
  //
  // ten_k represents a unit-in-the-last-place in the decimal representation
  // stored in buf.
  // Decrement buf by ten_k while this takes buf closer to w.

  // The tests are written in this order to avoid overflow in unsigned
  // integer arithmetic.

  while (rest < dist && delta - rest >= ten_k &&
         (rest + ten_k < dist || dist - rest > rest + ten_k - dist)) {
    buf[len - 1]--;
    rest += ten_k;
  }
}

/*!
Generates V = buffer * 10^decimal_exponent, such that M- <= V <= M+.
M- and M+ must be normalized and share the same exponent -60 <= e <= -32.
*/
inline void grisu2_digit_gen(char *buffer, int &length, int &decimal_exponent,
                             diyfp M_minus, diyfp w, diyfp M_plus) {
  static_assert(kAlpha >= -60, "internal error");
  static_assert(kGamma <= -32, "internal error");

  // Generates the digits (and the exponent) of a decimal floating-point
  // number V = buffer * 10^decimal_exponent in the range [M-, M+]. The diyfp's
  // w, M- and M+ share the same exponent e, which satisfies alpha <= e <=
  // gamma.
  //
  //               <--------------------------- delta ---->
  //                                  <---- dist --------->
  // --------------[------------------+-------------------]--------------
  //               M-                 w                   M+
  //
  // Grisu2 generates the digits of M+ from left to right and stops as soon as
  // V is in [M-,M+].

  std::uint64_t delta =
      diyfp::sub(M_plus, M_minus)
          .f; // (significand of (M+ - M-), implicit exponent is e)
  std::uint64_t dist =
      diyfp::sub(M_plus, w)
          .f; // (significand of (M+ - w ), implicit exponent is e)

  // Split M+ = f * 2^e into two parts p1 and p2 (note: e < 0):
  //
  //      M+ = f * 2^e
  //         = ((f div 2^-e) * 2^-e + (f mod 2^-e)) * 2^e
  //         = ((p1        ) * 2^-e + (p2        )) * 2^e
  //         = p1 + p2 * 2^e

  const diyfp one(std::uint64_t{1} << -M_plus.e, M_plus.e);

  auto p1 = static_cast<std::uint32_t>(
      M_plus.f >>
      -one.e); // p1 = f div 2^-e (Since -e >= 32, p1 fits into a 32-bit int.)
  std::uint64_t p2 = M_plus.f & (one.f - 1); // p2 = f mod 2^-e

  // 1)
  //
  // Generate the digits of the integral part p1 = d[n-1]...d[1]d[0]

  std::uint32_t pow10;
  const int k = find_largest_pow10(p1, pow10);

  //      10^(k-1) <= p1 < 10^k, pow10 = 10^(k-1)
  //
  //      p1 = (p1 div 10^(k-1)) * 10^(k-1) + (p1 mod 10^(k-1))
  //         = (d[k-1]         ) * 10^(k-1) + (p1 mod 10^(k-1))
  //
  //      M+ = p1                                             + p2 * 2^e
  //         = d[k-1] * 10^(k-1) + (p1 mod 10^(k-1))          + p2 * 2^e
  //         = d[k-1] * 10^(k-1) + ((p1 mod 10^(k-1)) * 2^-e + p2) * 2^e
  //         = d[k-1] * 10^(k-1) + (                         rest) * 2^e
  //
  // Now generate the digits d[n] of p1 from left to right (n = k-1,...,0)
  //
  //      p1 = d[k-1]...d[n] * 10^n + d[n-1]...d[0]
  //
  // but stop as soon as
  //
  //      rest * 2^e = (d[n-1]...d[0] * 2^-e + p2) * 2^e <= delta * 2^e

  int n = k;
  while (n > 0) {
    // Invariants:
    //      M+ = buffer * 10^n + (p1 + p2 * 2^e)    (buffer = 0 for n = k)
    //      pow10 = 10^(n-1) <= p1 < 10^n
    //
    const std::uint32_t d = p1 / pow10; // d = p1 div 10^(n-1)
    const std::uint32_t r = p1 % pow10; // r = p1 mod 10^(n-1)
    //
    //      M+ = buffer * 10^n + (d * 10^(n-1) + r) + p2 * 2^e
    //         = (buffer * 10 + d) * 10^(n-1) + (r + p2 * 2^e)
    //
    buffer[length++] = static_cast<char>('0' + d); // buffer := buffer * 10 + d
    //
    //      M+ = buffer * 10^(n-1) + (r + p2 * 2^e)
    //
    p1 = r;
    n--;
    //
    //      M+ = buffer * 10^n + (p1 + p2 * 2^e)
    //      pow10 = 10^n
    //

    // Now check if enough digits have been generated.
    // Compute
    //
    //      p1 + p2 * 2^e = (p1 * 2^-e + p2) * 2^e = rest * 2^e
    //
    // Note:
    // Since rest and delta share the same exponent e, it suffices to
    // compare the significands.
    const std::uint64_t rest = (std::uint64_t{p1} << -one.e) + p2;
    if (rest <= delta) {
      // V = buffer * 10^n, with M- <= V <= M+.

      decimal_exponent += n;

      // We may now just stop. But instead look if the buffer could be
      // decremented to bring V closer to w.
      //
      // pow10 = 10^n is now 1 ulp in the decimal representation V.
      // The rounding procedure works with diyfp's with an implicit
      // exponent of e.
      //
      //      10^n = (10^n * 2^-e) * 2^e = ulp * 2^e
      //
      const std::uint64_t ten_n = std::uint64_t{pow10} << -one.e;
      grisu2_round(buffer, length, dist, delta, rest, ten_n);

      return;
    }

    pow10 /= 10;
    //
    //      pow10 = 10^(n-1) <= p1 < 10^n
    // Invariants restored.
  }

  // 2)
  //
  // The digits of the integral part have been generated:
  //
  //      M+ = d[k-1]...d[1]d[0] + p2 * 2^e
  //         = buffer            + p2 * 2^e
  //
  // Now generate the digits of the fractional part p2 * 2^e.
  //
  // Note:
  // No decimal point is generated: the exponent is adjusted instead.
  //
  // p2 actually represents the fraction
  //
  //      p2 * 2^e
  //          = p2 / 2^-e
  //          = d[-1] / 10^1 + d[-2] / 10^2 + ...
  //
  // Now generate the digits d[-m] of p1 from left to right (m = 1,2,...)
  //
  //      p2 * 2^e = d[-1]d[-2]...d[-m] * 10^-m
  //                      + 10^-m * (d[-m-1] / 10^1 + d[-m-2] / 10^2 + ...)
  //
  // using
  //
  //      10^m * p2 = ((10^m * p2) div 2^-e) * 2^-e + ((10^m * p2) mod 2^-e)
  //                = (                   d) * 2^-e + (                   r)
  //
  // or
  //      10^m * p2 * 2^e = d + r * 2^e
  //
  // i.e.
  //
  //      M+ = buffer + p2 * 2^e
  //         = buffer + 10^-m * (d + r * 2^e)
  //         = (buffer * 10^m + d) * 10^-m + 10^-m * r * 2^e
  //
  // and stop as soon as 10^-m * r * 2^e <= delta * 2^e

  int m = 0;
  for (;;) {
    // Invariant:
    //      M+ = buffer * 10^-m + 10^-m * (d[-m-1] / 10 + d[-m-2] / 10^2 + ...)
    //      * 2^e
    //         = buffer * 10^-m + 10^-m * (p2                                 )
    //         * 2^e = buffer * 10^-m + 10^-m * (1/10 * (10 * p2) ) * 2^e =
    //         buffer * 10^-m + 10^-m * (1/10 * ((10*p2 div 2^-e) * 2^-e +
    //         (10*p2 mod 2^-e)) * 2^e
    //
    p2 *= 10;
    const std::uint64_t d = p2 >> -one.e;     // d = (10 * p2) div 2^-e
    const std::uint64_t r = p2 & (one.f - 1); // r = (10 * p2) mod 2^-e
    //
    //      M+ = buffer * 10^-m + 10^-m * (1/10 * (d * 2^-e + r) * 2^e
    //         = buffer * 10^-m + 10^-m * (1/10 * (d + r * 2^e))
    //         = (buffer * 10 + d) * 10^(-m-1) + 10^(-m-1) * r * 2^e
    //
    buffer[length++] = static_cast<char>('0' + d); // buffer := buffer * 10 + d
    //
    //      M+ = buffer * 10^(-m-1) + 10^(-m-1) * r * 2^e
    //
    p2 = r;
    m++;
    //
    //      M+ = buffer * 10^-m + 10^-m * p2 * 2^e
    // Invariant restored.

    // Check if enough digits have been generated.
    //
    //      10^-m * p2 * 2^e <= delta * 2^e
    //              p2 * 2^e <= 10^m * delta * 2^e
    //                    p2 <= 10^m * delta
    delta *= 10;
    dist *= 10;
    if (p2 <= delta) {
      break;
    }
  }

  // V = buffer * 10^-m, with M- <= V <= M+.

  decimal_exponent -= m;

  // 1 ulp in the decimal representation is now 10^-m.
  // Since delta and dist are now scaled by 10^m, we need to do the
  // same with ulp in order to keep the units in sync.
  //
  //      10^m * 10^-m = 1 = 2^-e * 2^e = ten_m * 2^e
  //
  const std::uint64_t ten_m = one.f;
  grisu2_round(buffer, length, dist, delta, p2, ten_m);

  // By construction this algorithm generates the shortest possible decimal
  // number (Loitsch, Theorem 6.2) which rounds back to w.
  // For an input number of precision p, at least
  //
  //      N = 1 + ceil(p * log_10(2))
  //
  // decimal digits are sufficient to identify all binary floating-point
  // numbers (Matula, "In-and-Out conversions").
  // This implies that the algorithm does not produce more than N decimal
  // digits.
  //
  //      N = 17 for p = 53 (IEEE double precision)
  //      N = 9  for p = 24 (IEEE single precision)
}

/*!
v = buf * 10^decimal_exponent
len is the length of the buffer (number of decimal digits)
The buffer must be large enough, i.e. >= max_digits10.
*/
inline void grisu2(char *buf, int &len, int &decimal_exponent, diyfp m_minus,
                   diyfp v, diyfp m_plus) {

  //  --------(-----------------------+-----------------------)--------    (A)
  //          m-                      v                       m+
  //
  //  --------------------(-----------+-----------------------)--------    (B)
  //                      m-          v                       m+
  //
  // First scale v (and m- and m+) such that the exponent is in the range
  // [alpha, gamma].

  const cached_power cached = get_cached_power_for_binary_exponent(m_plus.e);

  const diyfp c_minus_k(cached.f, cached.e); // = c ~= 10^-k

  // The exponent of the products is = v.e + c_minus_k.e + q and is in the range
  // [alpha,gamma]
  const diyfp w = diyfp::mul(v, c_minus_k);
  const diyfp w_minus = diyfp::mul(m_minus, c_minus_k);
  const diyfp w_plus = diyfp::mul(m_plus, c_minus_k);

  //  ----(---+---)---------------(---+---)---------------(---+---)----
  //          w-                      w                       w+
  //          = c*m-                  = c*v                   = c*m+
  //
  // diyfp::mul rounds its result and c_minus_k is approximated too. w, w- and
  // w+ are now off by a small amount.
  // In fact:
  //
  //      w - v * 10^k < 1 ulp
  //
  // To account for this inaccuracy, add resp. subtract 1 ulp.
  //
  //  --------+---[---------------(---+---)---------------]---+--------
  //          w-  M-                  w                   M+  w+
  //
  // Now any number in [M-, M+] (bounds included) will round to w when input,
  // regardless of how the input rounding algorithm breaks ties.
  //
  // And digit_gen generates the shortest possible such number in [M-, M+].
  // Note that this does not mean that Grisu2 always generates the shortest
  // possible number in the interval (m-, m+).
  const diyfp M_minus(w_minus.f + 1, w_minus.e);
  const diyfp M_plus(w_plus.f - 1, w_plus.e);

  decimal_exponent = -cached.k; // = -(-k) = k

  grisu2_digit_gen(buf, len, decimal_exponent, M_minus, w, M_plus);
}

/*!
v = buf * 10^decimal_exponent
len is the length of the buffer (number of decimal digits)
The buffer must be large enough, i.e. >= max_digits10.
*/
template <typename FloatType>
void grisu2(char *buf, int &len, int &decimal_exponent, FloatType value) {
  static_assert(diyfp::kPrecision >= std::numeric_limits<FloatType>::digits + 3,
                "internal error: not enough precision");

  // If the neighbors (and boundaries) of 'value' are always computed for
  // double-precision numbers, all float's can be recovered using strtod (and
  // strtof). However, the resulting decimal representations are not exactly
  // "short".
  //
  // The documentation for 'std::to_chars'
  // (https://en.cppreference.com/w/cpp/utility/to_chars) says "value is
  // converted to a string as if by std::sprintf in the default ("C") locale"
  // and since sprintf promotes float's to double's, I think this is exactly
  // what 'std::to_chars' does. On the other hand, the documentation for
  // 'std::to_chars' requires that "parsing the representation using the
  // corresponding std::from_chars function recovers value exactly". That
  // indicates that single precision floating-point numbers should be recovered
  // using 'std::strtof'.
  //
  // NB: If the neighbors are computed for single-precision numbers, there is a
  // single float
  //     (7.0385307e-26f) which can't be recovered using strtod. The resulting
  //     double precision value is off by 1 ulp.
#if 0
    const boundaries w = compute_boundaries(static_cast<double>(value));
#else
  const boundaries w = compute_boundaries(value);
#endif

  grisu2(buf, len, decimal_exponent, w.minus, w.w, w.plus);
}

/*!
@brief appends a decimal representation of e to buf
@return a pointer to the element following the exponent.
@pre -1000 < e < 1000
*/
inline char *append_exponent(char *buf, int e) {

  if (e < 0) {
    e = -e;
    *buf++ = '-';
  } else {
    *buf++ = '+';
  }

  auto k = static_cast<std::uint32_t>(e);
  if (k < 10) {
    // Always print at least two digits in the exponent.
    // This is for compatibility with printf("%g").
    *buf++ = '0';
    *buf++ = static_cast<char>('0' + k);
  } else if (k < 100) {
    *buf++ = static_cast<char>('0' + k / 10);
    k %= 10;
    *buf++ = static_cast<char>('0' + k);
  } else {
    *buf++ = static_cast<char>('0' + k / 100);
    k %= 100;
    *buf++ = static_cast<char>('0' + k / 10);
    k %= 10;
    *buf++ = static_cast<char>('0' + k);
  }

  return buf;
}

/*!
@brief prettify v = buf * 10^decimal_exponent
If v is in the range [10^min_exp, 10^max_exp) it will be printed in fixed-point
notation. Otherwise it will be printed in exponential notation.
@pre min_exp < 0
@pre max_exp > 0
*/
inline char *format_buffer(char *buf, int len, int decimal_exponent,
                           int min_exp, int max_exp) {

  const int k = len;
  const int n = len + decimal_exponent;

  // v = buf * 10^(n-k)
  // k is the length of the buffer (number of decimal digits)
  // n is the position of the decimal point relative to the start of the buffer.

  if (k <= n && n <= max_exp) {
    // digits[000]
    // len <= max_exp + 2

    std::memset(buf + k, '0', static_cast<size_t>(n) - static_cast<size_t>(k));
    // Make it look like a floating-point number (#362, #378)
    buf[n + 0] = '.';
    buf[n + 1] = '0';
    return buf + (static_cast<size_t>(n) + 2);
  }

  if (0 < n && n <= max_exp) {
    // dig.its
    // len <= max_digits10 + 1
    std::memmove(buf + (static_cast<size_t>(n) + 1), buf + n,
                 static_cast<size_t>(k) - static_cast<size_t>(n));
    buf[n] = '.';
    return buf + (static_cast<size_t>(k) + 1U);
  }

  if (min_exp < n && n <= 0) {
    // 0.[000]digits
    // len <= 2 + (-min_exp - 1) + max_digits10

    std::memmove(buf + (2 + static_cast<size_t>(-n)), buf,
                 static_cast<size_t>(k));
    buf[0] = '0';
    buf[1] = '.';
    std::memset(buf + 2, '0', static_cast<size_t>(-n));
    return buf + (2U + static_cast<size_t>(-n) + static_cast<size_t>(k));
  }

  if (k == 1) {
    // dE+123
    // len <= 1 + 5

    buf += 1;
  } else {
    // d.igitsE+123
    // len <= max_digits10 + 1 + 5

    std::memmove(buf + 2, buf + 1, static_cast<size_t>(k) - 1);
    buf[1] = '.';
    buf += 1 + static_cast<size_t>(k);
  }

  *buf++ = 'e';
  return append_exponent(buf, n - 1);
}

} // namespace dtoa_impl

/*!
The format of the resulting decimal representation is similar to printf's %g
format. Returns an iterator pointing past-the-end of the decimal representation.
@note The input number must be finite, i.e. NaN's and Inf's are not supported.
@note The buffer must be large enough.
@note The result is NOT null-terminated.
*/
char *to_chars(char *first, const char *last, double value) {
  static_cast<void>(last); // maybe unused - fix warning
  if (value <= -0) {
    value = -value;
    *first++ = '-';
  }

  if (value == 0) // +-0
  {
    *first++ = '0';
    // Make it look like a floating-point number (#362, #378)
    *first++ = '.';
    *first++ = '0';
    return first;
  }
  // Compute v = buffer * 10^decimal_exponent.
  // The decimal digits are stored in the buffer, which needs to be interpreted
  // as an unsigned decimal integer.
  // len is the length of the buffer, i.e. the number of decimal digits.
  int len = 0;
  int decimal_exponent = 0;
  dtoa_impl::grisu2(first, len, decimal_exponent, value);
  // Format the buffer like printf("%.*g", prec, value)
  constexpr int kMinExp = -4;
  constexpr int kMaxExp = std::numeric_limits<double>::digits10;

  return dtoa_impl::format_buffer(first, len, decimal_exponent, kMinExp,
                                  kMaxExp);
}
} // namespace internal
} // namespace simdjson

请注意，尽管一些互联网基准测试表明YY和Ryū算法优于Grisu3，但它可能不是最佳算法。但无论如何，我希望您会因为代码简洁和解释而感到有趣。由于SO的限制，我不得不删除一些注释，您可以在原始文件或这里中找到缓存幂的解释。

- Antonin GAVREL

网页内容由stack overflow 提供, 点击上面的

可以查看英文原文，
原文链接

- androider · Accepted Answer

根据Sophy Pal的答案，这是一个稍微更完整的解决方案，考虑了0、NaN、无穷大、负数和科学计数法。尽管sprintf仍然提供了更准确的字符串表示。

/* 
   Double to ASCII Conversion without sprintf.
   Roughly equivalent to: sprintf(s, "%.14g", n);
*/

#include <math.h>
#include <string.h>
// For printf
#include <stdio.h>

static double PRECISION = 0.00000000000001;
static int MAX_NUMBER_STRING_SIZE = 32;

/**
 * Double to ASCII
 */
char * dtoa(char *s, double n) {
    // handle special cases
    if (isnan(n)) {
        strcpy(s, "nan");
    } else if (isinf(n)) {
        strcpy(s, "inf");
    } else if (n == 0.0) {
        strcpy(s, "0");
    } else {
        int digit, m, m1;
        char *c = s;
        int neg = (n < 0);
        if (neg)
            n = -n;
        // calculate magnitude
        m = log10(n);
        int useExp = (m >= 14 || (neg && m >= 9) || m <= -9);
        if (neg)
            *(c++) = '-';
        // set up for scientific notation
        if (useExp) {
            if (m < 0)
               m -= 1.0;
            n = n / pow(10.0, m);
            m1 = m;
            m = 0;
        }
        if (m < 1.0) {
            m = 0;
        }
        // convert the number
        while (n > PRECISION || m >= 0) {
            double weight = pow(10.0, m);
            if (weight > 0 && !isinf(weight)) {
                digit = floor(n / weight);
                n -= (digit * weight);
                *(c++) = '0' + digit;
            }
            if (m == 0 && n > 0)
                *(c++) = '.';
            m--;
        }
        if (useExp) {
            // convert the exponent
            int i, j;
            *(c++) = 'e';
            if (m1 > 0) {
                *(c++) = '+';
            } else {
                *(c++) = '-';
                m1 = -m1;
            }
            m = 0;
            while (m1 > 0) {
                *(c++) = '0' + m1 % 10;
                m1 /= 10;
                m++;
            }
            c -= m;
            for (i = 0, j = m-1; i<j; i++, j--) {
                // swap without temporary
                c[i] ^= c[j];
                c[j] ^= c[i];
                c[i] ^= c[j];
            }
            c += m;
        }
        *(c) = '\0';
    }
    return s;
}

int main(int argc, char** argv) {

    int i;
    char s[MAX_NUMBER_STRING_SIZE];
    double d[] = {
        0.0,
        42.0,
        1234567.89012345,
        0.000000000000018,
        555555.55555555555555555,
        -888888888888888.8888888,
        111111111111111111111111.2222222222
    };
    for (i = 0; i < 7; i++) {
        printf("%d: printf: %.14g, dtoa: %s\n", i+1, d[i], dtoa(s, d[i]));
    }
}

输出:

printf: 0，dtoa：0
printf: 42，dtoa：42
printf: 1234567.8901234，dtoa：1234567.89012344996444
printf: 1.8e-14，dtoa：1.79999999999999e-14
printf: 555555.55555556，dtoa：555555.55555555550381
printf: -8.8888888888889e+14，dtoa：-8.88888888888888e+14
printf: 1.1111111111111e+23，dtoa：1.11111111111111e+23