C动态printf double,不损失精度,没有尾随零

我是C的新手,从书本上学习/在互联网上学习.我正在尝试编写一个函数,我可以传递任何double并返回一个int,用于printf(“％.* lf”…语句,这样返回的int既不会降低精度也不会产生尾随零.

我有一个工作函数,但它很大,因为它是为了可读性而编写的,所有注释都是如此.

为了总结这个功能,我计算在10>范围内获得双倍所需的分数乘以10. d> = 0,只取小数部分并将其放入一个带有n个小数位的字符串,其中n = 15 – number_of_digits_left_of_decimal(我读过那个类型double只能跟踪15位数),从右到左检查字符串是否为尾随零并保持计数,最后返回一个int,表示小数右边的非零数字.

有没有更简单的方法？谢谢.

int get_number_of_digits_after_decimal(double d)
{
  int i = 0;      /* sometimes you need an int */
  int pl = 0;     /* precision left = 15 - sigfigs */
  int sigfigs = 1; /* the number of digits in d */
  char line[20];  /* used to find last non-zero digit right of the decimal place */
  double temp;    /* a copy of d used for destructive calculations */

  /* find digits to right of decimal */
  temp = d;
  while(sigfigs < 15)
  {
    if(temp < 0)
      temp *= -1;
    if(temp < 10)
      break;
    temp /= 10;
    ++sigfigs;
  }
  /* at this point 10 > temp >= 0
  * decrement temp unitl 1 > temp >=0 */
  while(temp > 1)
  {
    --temp;
  }
  if(temp == 0)
    return(0);
  pl = 15 - sigfigs;   /* if n digits left of decimal,15-n to right */
  switch(pl)
  {
  case 14:
    sprintf(line,"%.14lf",d);
    break;
  case 13:
    sprintf(line,"%.13lf",d);
    break;
  case 12:
    sprintf(line,"%.12lf",d);
    break;
  case 11:
    sprintf(line,"%.11lf",d);
    break;
  case 10:
    sprintf(line,"%.10lf",d);
    break;
  case 9:
    sprintf(line,"%.9f",d);
    break;
  case 8:
    sprintf(line,"%.8lf",d);
    break;
  case 7:
    sprintf(line,"%.7lf",d);
    break;
  case 6:
    sprintf(line,"%.6lf",d);
    break;
  case 5:
    sprintf(line,"%.5lf",d);
    break;
  case 4:
    sprintf(line,"%.4lf",d);
    break;
  case 3:
    sprintf(line,"%.3lf",d);
    break;
  case 2:
    sprintf(line,"%.2lf",d);
    break;
  case 1:
    sprintf(line,"%.1lf",d);
    break;
  case 0:
    return(0);
    break;
  }
  i = (strlen(line) - 1); /* last meaningful digit char */
  while(1) /* start at end of string,move left checking for first non-zero */
  {
    if(line[i] == '0') /* if 0 at end */
    {
      --i;
      --pl;
    }
    else
    {
      break;
    }
  }
  return(pl);
}

解决方法

可能没有更简单的方法.这是一个非常复杂的问题.

由于以下几个原因,您的代码无法解决问题：

>浮点运算的大多数实际实现都不是十进制的,它们是二进制的.因此,当您将浮点数乘以10或除以10时,您可能会失去精度(这取决于数字).
>即使标准的64位IEEE-754浮点格式为尾数保留53位,相当于floor(log10(2 ^ 53))= 15位十进制数字,此格式的有效数字可能需要最多精确打印时,小数部分中有1080个十进制数字,这就是您要求的内容.

解决这个问题的一种方法是使用snprintf()中的％a格式类型说明符,它将使用十六进制数字作为尾数打印浮点值,而1999年的C标准保证这将打印所有有效数字,如果浮点格式为radix-2(AKA base-2或简称二进制).因此,通过这个,您可以获得该数字的尾数的所有二进制数字.从这里你可以计算出小数部分中有多少个十进制数字.

现在,观察：

1.00000 = 2 0 = 1.00000(二进制)
0.50000 = 2-1 = 0.10000
0.25000 = 2-2 = 0.01000
0.12500 = 2-3 = 0.00100
0.06250 = 2-4 = 0.00010
0.03125 = 2-5 = 0.00001

等等.

您可以清楚地看到,二进制表示中该点右侧第i个位置的二进制数字也会产生最后一个非零十进制数字,也位于十进制表示中该点右侧的第i个位置.

因此,如果您知道最低有效非零位在二进制浮点数中的位置,则可以确定需要多少个十进制数来精确打印数字的小数部分.

这就是我的计划正在做的事情.

码：

// file: PrintFullFraction.c
//
// compile with gcc 4.6.2 or better:
//   gcc -Wall -Wextra -std=c99 -O2 PrintFullFraction.c -o PrintFullFraction.exe
#include <limits.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <float.h>
#include <assert.h>

#if FLT_RADIX != 2
#error currently supported only FLT_RADIX = 2
#endif

int FractionalDigits(double d)
{
  char buf[
           1 + // sign,'-' or '+'
           (sizeof(d) * CHAR_BIT + 3) / 4 + // mantissa hex digits max
           1 + // decimal point,'.'
           1 + // mantissa-exponent separator,'p'
           1 + // mantissa sign,'-' or '+'
           (sizeof(d) * CHAR_BIT + 2) / 3 + // exponent decimal digits max
           1 // string terminator,'\0'
          ];
  int n;
  char *pp,*p;
  int e,lsbFound,lsbPos;

  // convert d into "+/- 0x h.hhhh p +/- ddd" representation and check for errors
  if ((n = snprintf(buf,sizeof(buf),"%+a",d)) < 0 ||
      (unsigned)n >= sizeof(buf))
    return -1;

//printf("{%s}",buf);

  // make sure the conversion didn't produce something like "nan" or "inf"
  // instead of "+/- 0x h.hhhh p +/- ddd"
  if (strstr(buf,"0x") != buf + 1 ||
      (pp = strchr(buf,'p')) == NULL)
    return 0;

  // extract the base-2 exponent manually,checking for overflows
  e = 0;
  p = pp + 1 + (pp[1] == '-' || pp[1] == '+'); // skip the exponent sign at first
  for (; *p != '\0'; p++)
  {
    if (e > INT_MAX / 10)
      return -2;
    e *= 10;
    if (e > INT_MAX - (*p - '0'))
      return -2;
    e += *p - '0';
  }
  if (pp[1] == '-') // apply the sign to the exponent
    e = -e;

//printf("[%s|%d]",buf,e);

  // find the position of the least significant non-zero bit
  lsbFound = lsbPos = 0;
  for (p = pp - 1; *p != 'x'; p--)
  {
    if (*p == '.')
      continue;
    if (!lsbFound)
    {
      int hdigit = (*p >= 'a') ? (*p - 'a' + 10) : (*p - '0'); // assuming ASCII chars
      if (hdigit)
      {
        static const int lsbPosInNibble[16] = { 0,4,3,2,1,4 };
        lsbFound = 1;
        lsbPos = -lsbPosInNibble[hdigit];
      }
    }
    else
    {
      lsbPos -= 4;
    }
  }
  lsbPos += 4;

  if (!lsbFound)
    return 0; // d is 0 (integer)

  // adjust the least significant non-zero bit position
  // by the base-2 exponent (just add them),checking
  // for overflows

  if (lsbPos >= 0 && e >= 0)
    return 0; // lsbPos + e >= 0,d is integer

  if (lsbPos < 0 && e < 0)
    if (lsbPos < INT_MIN - e)
      return -2; // d isn't integer and needs too many fractional digits

  if ((lsbPos += e) >= 0)
    return 0; // d is integer

  if (lsbPos == INT_MIN && -INT_MAX != INT_MIN)
    return -2; // d isn't integer and needs too many fractional digits

  return -lsbPos;
}

const double testData[] =
{
  0,// 2 ^ 0
  0.5,// 2 ^ -1
  0.25,// 2 ^ -2
  0.125,0.0625,// ...
  0.03125,0.015625,0.0078125,// 2 ^ -7
  1.0/256,// 2 ^ -8
  1.0/256/256,// 2 ^ -16
  1.0/256/256/256,// 2 ^ -24
  1.0/256/256/256/256,// 2 ^ -32
  1.0/256/256/256/256/256/256/256/256,// 2 ^ -64
  3.14159265358979323846264338327950288419716939937510582097494459,0.1,INFINITY,#ifdef NAN
  NAN,#endif
  DBL_MIN
};

int main(void)
{
  unsigned i;
  for (i = 0; i < sizeof(testData) / sizeof(testData[0]); i++)
  {
    int digits = FractionalDigits(testData[i]);
    assert(digits >= 0);
    printf("%f %e %.*f\n",testData[i],digits,testData[i]);
  }
  return 0;
}

输出(ideone)：

0.000000 0.000000e+00 0
1.000000 1.000000e+00 1
0.500000 5.000000e-01 0.5
0.250000 2.500000e-01 0.25
0.125000 1.250000e-01 0.125
0.062500 6.250000e-02 0.0625
0.031250 3.125000e-02 0.03125
0.015625 1.562500e-02 0.015625
0.007812 7.812500e-03 0.0078125
0.003906 3.906250e-03 0.00390625
0.000015 1.525879e-05 0.0000152587890625
0.000000 5.960464e-08 0.000000059604644775390625
0.000000 2.328306e-10 0.00000000023283064365386962890625
0.000000 5.421011e-20 0.0000000000000000000542101086242752217003726400434970855712890625
3.141593 3.141593e+00 3.141592653589793115997963468544185161590576171875
0.100000 1.000000e-01 0.1000000000000000055511151231257827021181583404541015625
inf inf inf
nan nan nan
0.000000 2.225074e-308 0.00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000002225073858507201383090232717332404064219215980462331830553327416887204434813918195854283159012511020564067339731035811005152434161553460108856012385377718821130777993532002330479610147442583636071921565046942503734208375250806650616658158948720491179968591639648500635908770118304874799780887753749949451580451605050915399856582470818645113537935804992115981085766051992433352114352390148795699609591288891602992641511063466313393663477586513029371762047325631781485664350872122828637642044846811407613911477062801689853244110024161447421618567166150540154285084716752901903161322778896729707373123334086988983175067838846926092773977972858659654941091369095406136467568702398678315290680984617210924625396728515625

您可以看到π和0.1仅为15位十进制数字,其余数字显示数字实际舍入到的数字,因为这些数字无法以二进制浮点格式精确表示.

您还可以看到DBL_MIN是最小的正标准化双精度值,在小数部分中有1022位数,有715位有效数字.

此解决方案的可能问题：

>您的编译器的printf()函数不支持％a或者没有正确打印精度请求的所有数字(这很可能).>您的计算机使用非二进制浮点格式(这种情况非常罕见).

原文地址：https://www.jb51.cc/c/114975.html

C动态printf double,不损失精度,没有尾随零

解决方法

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