连接二进制数

Question

我正在尝试编写一个程序，该程序将采用以 10 为底的 float 并将其小数部分转换为以 2 为底的小数部分。在下面的代码中，我打算将我的转换函数调用到 printf 中，并格式化输出;我遇到的问题出在我的 fra_binary() 中，我无法找出返回整数的最佳方法，该整数分别由每个回合的转换结果组成(串联)。这是我现在所做的(代码未优化，因为我仍在努力):

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

int fra_binary(double fract) ;

int main() 
{
    long double n ;
    double fract, deci ;
    printf("base 10 :\n") ;
    scanf("%Lf", &n) ;
    fract = modf(n, &deci) ;
    int d = deci ;
    printf("base 2: %d.%d\n", d, fra_binary(fract)) ;

    return(0) ;
}

int fra_binary(double F) 
{
    double fl ;
    double decimal ;
    int array[30] ;

    for (int i = 0 ; i < 30 ; i++) {
        fl = F * 2 ;
        F = modf(fl, &decimal) ;
        array[i] = decimal ;
        if (F == 0) break ;
    }

    return array[0] ;
}

显然这会部分返回所需的输出，因为我需要将整个数组连接为一个 int 或 char 来显示我需要的 1 和 0 系列。所以每次，我都想使用我处理的数字的小数部分作为二进制数来连接(1 + 0 = 10 而不是 1)。我该怎么办？希望这是有道理的!

Answer 1

return array[0] ;只是 int array[30] 的第一个值设置在 fra_binary() .代码丢弃循环的第一个计算以外的所有 for (int i = 0 ; i < 30 ; i++) .

convert its fractional part in base 2

OP 的循环思想是一个很好的起点。然而int array[30]不足以对所有 double 的小数部分进行编码变成“二进制文件”。

can't figure out the best way to return an integer

返回 int将是不够的。而是考虑使用字符串 - 或者以类似的方式管理整数数组。

使用来自 <float.h> 的定义插入缓冲区需求。

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

char *fra_binary(char *dest, double x) {
  _Static_assert(FLT_RADIX == 2, "Unexpected FP base");
  double deci;
  double fract = modf(x, &deci);
  fract = fabs(fract);
  char *s = dest;
  do {
    double d;
    fract = modf(fract * 2.0, &d);
    *s++ = "01"[(int) d];
  } while (fract);
  *s = '\0';

  // For debug
  printf("%*.*g --> %.0f and .", DBL_DECIMAL_DIG + 8, DBL_DECIMAL_DIG, x,
      deci);

  return dest;
}

int main(void) {
  // Perhaps           53           - -1021       + 1
  char fraction_string[DBL_MANT_DIG - DBL_MIN_EXP + 1];
  puts(fra_binary(fraction_string, -0.0));
  puts(fra_binary(fraction_string, 1.0));
  puts(fra_binary(fraction_string, asin(-1)));  // machine pi
  puts(fra_binary(fraction_string, -0.1));
  puts(fra_binary(fraction_string, DBL_MAX));
  puts(fra_binary(fraction_string, DBL_MIN));
  puts(fra_binary(fraction_string, DBL_TRUE_MIN));
}

输出

                       -0 --> -0 and .0
                        1 --> 1 and .0
       3.1415926535897931 --> 3 and .001001000011111101101010100010001000010110100011
     -0.10000000000000001 --> -0 and .0001100110011001100110011001100110011001100110011001101
  1.7976931348623157e+308 --> 179769313486231570814527423731704356798070600000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 and .0
  2.2250738585072014e-308 --> 0 and .00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001
  4.9406564584124654e-324 --> 0 and .000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001

也不清楚为什么输入是 long double , 但处理是用 double .建议仅使用一种 FP 类型。