C中将浮点数转换为二进制表示程序

Question

#include "ieee754.h"
#include <stdio.h>
#include <math.h>

//This program convert a floating number to its binary representation (IEEE754) in computer memory
int main() {
    long double f, binaryTotal, binaryFrac = 0.0, frac, fracFractor = 0.1;
    long int integer, binaryInt = 0;
    long int p = 0, rem, temp;

    printf("\nEnter floating number: ");
    scanf("%Lf", &f);

    //separate the integer part from the input floating number
    integer = (int)f;

    //separate the fractional part from the input floating number
    frac = f - integer;

    //loop to convert integer part to binary
    while (integer != 0) {
        rem = integer % 2;
        binaryInt = binaryInt + rem *pow(10, p);
        integer = integer / 2;
        p++;
    }

    //loop to convert fractional part to binary
    while (frac != 0) {
        frac = frac * 2;
        temp = frac;
        binaryFrac = binaryFrac + fracFractor * temp;
        if (temp == 1)
            frac = frac - temp;

        fracFractor = fracFractor / 10;
    }

    binaryTotal = binaryInt + binaryFrac;
    printf("binary equivalent = %Lf\n", binaryTotal);
}

我正在尝试将浮点数转换为二进制表示（64 位）。 此代码有效但并不完美。 例如，当我转换.575它给了我0.100100但是当我做使用本网站的转化http://www.exploringbinary.com/floating-point-converter/ ，正确的输出应该是0.1001001100110011001100110011001100110011001100110011

我无法理解是什么让我的代码截断了数字。 谁能帮我解决这个问题？ 我感谢您的帮助。

Answer 1

很多问题：

1. 使用(int)提取long double的整数部分严重限制了范围。 使用modfl(long double value, long double *iptr);

    long double f; long int integer; //separate the integer part from the input floating number // Weak code integer = (int)f; long double ipart; long double fpart = modfl(f, &ipart);

2. long p; pow(10,p); --> pow()返回值的精度损失，一旦p超过某个值，（示例 25）。 将pow()与采用long double的函数一起使用也很奇怪。 我希望powl() 。

3. 其他各种不精确的 FP 问题： fracFractor/10 ， long精度有限。

代码很奇怪，因为它试图将 FP 数（可能是某种二进制格式）转换为二进制表示。 它不应该在代码中的任何地方要求10 。

建议一些简单的东西

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<float.h>

static void print_ipart(long double x) {
  int digit = (int) (modfl(x/2, &x)*2.0) + '0';
  if (x) {
    print_ipart(x);
  }
  putchar(digit);
}

void print_bin(long double x) {
  // Some TBD code
  // Handle NAN with isnan()
  // Handle infinity with isfinite()

  putchar(signbit(x) ? '-' : '+');

  long double ipart;
  long double fpart = modfl(fabsl(x), &ipart);

  print_ipart(ipart);
  putchar('.');
  while (fpart) {
    long double ipart;
    fpart = modfl(fpart * 2, &ipart);
    putchar((int)ipart + '0');
  }
  putchar('\n');
}

int main() {
  print_bin(-4.25);
  print_bin(.575);
  print_bin(DBL_MAX);
  print_bin(DBL_MIN);
  print_bin(DBL_TRUE_MIN);
}

输出

-100.01
+0.1001001100110011001100110011001100110011001100110011001100110011
+1111111111111111111111111111111111111111111111111111100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000.
+0.00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001
+0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001

Answer 2

这就是为什么这不太可能奏效的原因：

fracFractor = 0.1
...
fracFractor = fracFractor/10

0.1 无法以任何二进制浮点格式准确表示。 您不能将 0.1 表示为 2 的负幂的倍数。 将其除以 10 将使其在每一步都收集舍入误差。 可能是因为您最终将一个重复分数与另一个重复分数进行了比较，因此您实际上退出了这个循环。

这将严重限制您可以实现的目标：

binaryTotal = binaryInt + binaryFrac;

在浮点中执行此操作将有严重的限制 - 至少表示 0.1 不能表示如上所述。 这就是为什么您得到的答案显示为二进制和十进制数字的混合。

要解决这个问题，您可能应该查看数字的各个位。 为了保持您的解决方案的整体思路完整，最简单的方法是不断从您的分数中减去 2 的负幂（0.5、0.25 等），测试它是否仍然为正，并基于此构建一个字符串。 然后对整数部分使用类似的逻辑。