Convert a float to a string

Question

How can I convert a floating point integer to a string in C/C++ without the library function sprintf?

I'm looking for a function, e.g. char *ftoa(float num) that converts num to a string and returns it.

ftoa(3.1415) should return "3.1415".

Latham

C has float to char conversion like sprintf(a, "%d", f) – Sunlight

One solution: edaboard.com/ftopic41714.html#160029 – Reynoso

Simply utilize the sprint function to format the floating point value – Reliquary

Answer 1

When you're dealing with fp numbers, it can get very compex but the algorithm is simplistic and similar to edgar holleis's answer; kudos! Its complex because when you're dealing with floating point numbers, the calculations will be a little off depending on the precision you've chosen. That's why its not good programming practice to compare a float to a zero.

But there is an answer and this is my attempt at implementing it. Here I've used a tolerance value so you don't end up calculating too many decimal places resulting in an infinite loop. I'm sure there might be better solutions out there but this should help give you a good understanding of how to do it.

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';

Epilepsy

There's a problem with this. If num is smaller than 1 you will calculate the string wrong. You need to change the 3rd line to be floor(log10(num)) – Chrysler

Another issue is if you try to convert a number that has a 0 in the ones place. For example 230.0. You will only get 23 in this case the way it is written now. You have to change the condition in the while loop to ((num > 0 + precision)||(m >= 0)) – Chrysler

Just to note that this implementation, and androider's improvement, is good. Hpwever, with a high precision (eg: 9 dp) the repeated pow() can make this rather slow. Not great if this is part of other actions providing user feedback I've found. – Toner

are precision and tolerance meant to be the same? precision is undefined, and tolerance is unused. – Cancroid

Answer 2

Based on Sophy Pal's answer, this is a slightly more complete solution that takes into account the number zero, NaN, infinite, negative numbers, and scientific notation. Albeit sprintf still provides a more accurate string representation.

/* 
   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]));
    }
}

Outputs:

1. printf: 0, dtoa: 0
2. printf: 42, dtoa: 42
3. printf: 1234567.8901234, dtoa: 1234567.89012344996444
4. printf: 1.8e-14, dtoa: 1.79999999999999e-14
5. printf: 555555.55555556, dtoa: 555555.55555555550381
6. printf: -8.8888888888889e+14, dtoa: -8.88888888888888e+14
7. printf: 1.1111111111111e+23, dtoa: 1.11111111111111e+23

Penney

I snagged this one for my (open source) library. Commenting everything and making it slightly more efficient. Once I'm done, would you rather I A) edit your answer with the tweaked code, or B) posted it separately and credited you with the original? I'm good either way. (If I don't hear, I'll assume B). – Pixie

Negative infinity prints inf right? Shouldn't it check the sign bit and return -inf if negative? – Olimpiaolin

This function turns encodes the number 1e-9 as :e-10, which is incorrect. Found this when porting this solution to another language. I'm not sure what the best fix is. – Veronicaveronika

Answer 3

You can use C++20 std::format or the {fmt} library, std::format is based on, to convert a floating-point number into a string, for example:

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

The advantage of this method compared to sprintf is that std::format gives you the shortest decimal representation with a round-trip guarantee.

Malpractice

Answer 4

1. Use the log-function to find out the magnitude m of your number. If the magnitude is negative print "0." and an appropriate amount of zeros.
2. Consecutively divide by 10^m and cast the result to int to get the decimal digits. m-- for the next digit.
3. If you came accross m==0, don't forget to print the decimal point ".".
4. Break off after a couple of digits. If m>0 when you break of, don't forget to print "E" and itoa(m).

Instead of the log-function you can also directly extract the exponent by bitshifting and correcting for the exponent's offset (see IEEE 754). Java has a double-to-bits function to get at the binary representation.

Bellflower

Answer 5

 /*
  * 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;
}

Cashew

Answer 6

You have two major problems:

1. Converting the bit representation into a string of characters
2. Allocating enough memory to store the characters.

The simplest way to solve the second part is to allocate a big enough chunk for every possible answer. Start with that. Later you'll want to be more clever, but don't bother until you've solved the numeric part of the problem.

You have two sets of tools available for dealing with the numeric part of the problem: direct bit manipulation (masking, shifting, etc) and arithmetic operation (*,+,/, plus possibly math functions link log()).

In principle you could tackle the bitwise representation directly, but that would not be portable in the event that floating point representation formats change in the future. The method suggested by edgar.holleis should be portable.

Sunsunbaked

Answer 7

Just found realy good implementation at 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);
}

Mv

Hmm, if you're already using string.h functions, might as well do strcpy(str,"nan"); instead of individually changing characters. Also, if you don't mind using math.h (and compile with -lm), you can do isnan(value) instead of the hacky test. – Shepley

Answer 8

Here's what I came up with; it's very efficient and very simple. It assumes your system has 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;
}

Try it online!

Shepley

Doesn't work for numbers with zeroes after the decimal point (e.g. 1.05->"1.5")... – Monolith

Answer 9

// This working code does:
// 1. Does not use sprintf as requested.
// 2. Gets some wide text from an editbox4 in VS2017
// 3. Converting that text to a double floating point number
// 4. Converts number to a wide string using ISO format, (StringCbPrintf replaced sprintf)
// 5. Transfers that text number back to another editbox5 for confirmation display as text
//
int const arraysize = 30;
wchar_t szItemName[arraysize]; // receives name of item
   if (!GetDlgItemText(hwnd, IDC_EDIT4, szItemName, arraysize )) *szItemName = 0;

double value = _wtof(szItemName);
wchar_t szname2[arraysize];

size_t cbDest = arraysize * sizeof(WCHAR);
LPCTSTR pszFormat = TEXT("%f");
HRESULT hr = StringCbPrintf(szname2, cbDest, pszFormat,value); //ISO format has buffer checking

SetDlgItemTextW(hwnd, IDC_EDIT5, szname2);

Crop

Answer 10

If you want a very fast implementation based on Grisu2 Algorithm I would recommend that you take a look at this file on Professor Lemire's github

Since link-only answers are not accepted, I will copy paste the code, credit: Florian Grisu for the algo and Professor Daniel Lemire for the port to C++ with the very interesting and enlightening comments:

#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

Be aware that it might not be the best algorithm as some internet benchmark suggest that both YY and Ryū algorithms outperform Grisu3. But nevertheless I hope you will find it interesting thanks to the clean code and explanations. Due to limits imposed by SO, I had to remove some comments, you can find explanations on cached powers in the raw file or here

Nevlin

Answer 11

My implementation

//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;
  }
}

Shiah

Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. – Subjugate

Answer 12

There is no ftoa in gcc for STM32, so there is mine:

#include <math.h>
void ftoa(float a, char* txt, char precision)
{
  char string[6];
  int roundoff = a;   // getting rounded integer
  float remainder = 0; // declaring the numbers to the right of the decimal point
  int whole_remainder = 0; // declaring remainder variable
  remainder = a - roundoff; // calculating remainder

  itoa(roundoff,txt,10); 
  strcat(txt,".");
  whole_remainder = abs(ceilf(remainder*(pow(10,precision)))); // making the decimals into whole integers by multiplying
  itoa(whole_remainder,string,10);
  strcat(txt,string);
}

Zanze

Answer 13

This gist might help : https://gist.github.com/psych0der/6319244 Basic idea is split the whole part and decimal part and then concatenate both of them with decimal in between.

Unguent

Link-only answers are not generally accepted on this site. – Shepley