将浮点数转换为字符串
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【中文标题】将浮点数转换为字符串【英文标题】：Convert a float to a string 【发布时间】：2010-02-20 17:19:21 【问题描述】：如何在没有库函数sprintf 的情况下将浮点整数转换为 C/C++ 中的字符串？
我正在寻找一个函数，例如char *ftoa(float num) 将num 转换为字符串并返回。
ftoa(3.1415) 应该返回 "3.1415"。
【问题讨论】：
C 有浮点到字符的转换，如sprintf(a, "%d", f)
一种解决方案：edaboard.com/ftopic41714.html#160029
简单利用sprint函数格式化浮点值
【参考方案1】：
根据 Sophy Pal 的回答，这是一个稍微更完整的解决方案，它考虑了数字零、NaN、无穷大、负数和科学记数法。尽管 sprintf 仍然提供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
【讨论】：
我为我的（开源）库抓住了这个。评论所有内容并使其效率更高。完成后，您愿意 A) 使用调整后的代码编辑您的答案，还是 B) 单独发布并将原始代码归功于您？无论哪种方式我都很好。 （如果我没有听到，我会假设 B）。 负无穷大打印inf 对吗？它不应该检查符号位并返回-inf 如果为负吗？
这个函数将数字1e-9编码为:e-10，这是不正确的。在将此解决方案移植到另一种语言时发现了这一点。我不确定最好的解决方法是什么。【参考方案2】：
当您处理 fp 数字时，它可能会变得非常复杂，但算法很简单，类似于 edgar holleis 的答案；赞！它很复杂，因为当您处理浮点数时，根据您选择的精度，计算会有些偏差。这就是为什么将浮点数与零进行比较不是好的编程习惯。
但是有一个答案，这是我实现它的尝试。在这里，我使用了一个容差值，因此您最终不会计算太多小数位，从而导致无限循环。我确信那里可能有更好的解决方案，但这应该有助于您更好地了解如何做到这一点。
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';
【讨论】：
这个有问题。如果 num 小于 1，您将计算错误的字符串。您需要将第 3 行更改为 floor(log10(num)) 另一个问题是，如果您尝试转换一个位为 0 的数字。例如 230.0。在这种情况下，按照现在的编写方式，您只会得到 23。您必须将 while 循环中的条件更改为 ((num > 0 + precision)||(m >= 0)) 只是要注意这个实现，以及androider的改进，是好的。 Hpwever，以高精度（例如：9 dp）重复 pow() 可以使这相当慢。如果这是我发现的提供用户反馈的其他操作的一部分，那就不好了。 精度和公差是否相同？精度未定义，公差未使用。【参考方案3】：-
使用
log-函数找出你的号码的大小m。如果幅度为负，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 以获得十进制数字。 m-- 下一位。
如果遇到m==0，别忘了打印小数点"."。
几位数后断开。如果m>0在你break的时候别忘了打印"E"和itoa(m)。
除了log-函数，您还可以通过位移和校正指数偏移量来直接提取指数（参见 IEEE 754）。 Java 有一个双位函数来获取二进制表示。
【讨论】：
【参考方案4】：可以使用C++20的std::format或the fmt library，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，此方法的优势在于std::format 为您提供最短的十进制表示，并保证往返。
【讨论】：
【参考方案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;
【讨论】：
【参考方案6】：你有两个主要问题：
-
将位表示转换为字符串
分配足够的内存来存储字符。
解决第二部分的最简单方法是为每个可能的答案分配足够大的块。从那开始。以后你会想变得更聪明，但在你解决了问题的数字部分之前不要打扰。
您有两组工具可用于处理问题的数字部分：直接位操作（屏蔽、移位等）和算术运算（*、+、/，可能还有数学函数链接 log()）。
原则上，您可以直接处理按位表示，但如果将来浮点表示格式发生变化，那将无法移植。 method suggested by edgar.holleis 应该是可移植的。
【讨论】：
【参考方案7】：刚刚在https://code.google.com/p/stringencoders/找到了非常好的实现
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);
【讨论】：
嗯，如果您已经在使用string.h 函数，不妨使用strcpy(str,"nan");，而不是单独更改字符。此外，如果您不介意使用math.h（并使用-lm 编译），您可以使用isnan(value) 而不是hacky 测试。【参考方案8】：
这是我想出的；它非常高效且非常简单。它假定您的系统有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;
Try it online!
【讨论】：
不适用于小数点后为零的数字（例如 1.05->"1.5"）...【参考方案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);
【讨论】：
【参考方案10】：如果您想要基于 Grisu2 Algorithm 的快速实现，我建议您查看 Lemire 教授的 github 上的 this file
由于不接受仅链接的答案，我将复制粘贴代码，感谢：算法的 Florian Grisu 和教授 Daniel Lemire 的移植到 C++ 的非常有趣和启发性的 cmets：
#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_t1 << (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_t1
<< (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_t1 << -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_tp1 << -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_tpow10 << -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，它可能不是最好的算法，因为一些互联网基准表明 YY 和 Ryū 算法都优于 Grisu3。但是我希望你会因为干净的代码和解释而发现它很有趣。由于 SO 施加的限制，我不得不删除一些 cmets，您可以在原始文件或 here
中找到有关缓存权力的解释【讨论】：
【参考方案11】：这个要点可能会有所帮助：https://gist.github.com/psych0der/6319244 基本思想是将整个部分和小数部分分开，然后将它们之间的小数连接起来。
【讨论】：
本网站一般不接受仅链接的答案。以上是关于将浮点数转换为字符串的主要内容，如果未能解决你的问题，请参考以下文章