由于类型转换，二进制值到浮点值的转换不起作用

Question

在这里我从我的代码中添加函数，用于从给定的二进制数生成浮点数。

代码：

double binary_float(double f) /* Function to convert binary to float.*/
{
   long     integral = 0, floatInt = 0, i = 1, temp1 = 0, k = 1;
   double   floatFract = 0, fractional = 0, floatTotal = 0;

   //Separating the integral value from the floating point variable
   integral = (long)f;

   //Separating the fractional value from the variable
   fractional = f - (long)f;

   //Converting binary to decimal
   floatInt = binary_decimal(integral);

   //Loop for converting binary to Fractional value
   while( k < 10000000 && fractional != (double)0 )
   {
      k = k * 10;
      i = i * 2;
      temp1 = (long)(fractional * k);

printf("temp: %ld, r: %lf\n", temp1, (fractional * k));

      floatFract = floatFract + (double)temp1/(double)i;

printf("fact: %lf, r: %lf\n", floatFract, ((double)temp1/(double)i));

      fractional = fractional - (double)temp1/(double)k;

printf("frac: %lf, r: %lf\n", fractional, ((double)temp1/(double)k));

   }

   //Combining both the integral and fractional binary value.
   floatTotal = floatInt + floatFract;

   return floatTotal;
}

long binary_decimal(long n)
{
   long decimal=0, i=0, rem;
   while (n!=0)
   {
      rem = n%10;
      n/=10;
      decimal += rem*pow(2,i);
      ++i;
   }
   return decimal;
}

输出：

Enter a binary number: 1010.001100
temp: 0, r: 0.011000
fact: 0.000000, r: 0.000000
frac: 0.001100, r: 0.000000
temp: 0, r: 0.110000
fact: 0.000000, r: 0.000000
frac: 0.001100, r: 0.000000
temp: 1, r: 1.100000
fact: 0.125000, r: 0.125000
frac: 0.000100, r: 0.001000

temp: 0, r: 1.000000

fact: 0.125000, r: 0.000000
frac: 0.000100, r: 0.000000
temp: 9, r: 10.000000
fact: 0.406250, r: 0.281250
frac: 0.000010, r: 0.000090
temp: 9, r: 10.000000
fact: 0.546875, r: 0.140625
frac: 0.000001, r: 0.000009
temp: 9, r: 10.000000
fact: 0.617188, r: 0.070312
frac: 0.000000, r: 0.000001
1010.001100 in binary = 10.617188 in float

在输出中您可以看到 temp: 0, r: 1.000000 表示 temp1 = 0 是 1.000000 的 long 类型转换。

谁能解释一下为什么这种类型转换不起作用？

Answer 1

OP: "temp: 0, r: 1.000000 表示 temp1 = 0 是 1.000000 的 long 类型转换。"
答：double 到 long 的转换是有效的，因为 double 小于 1.0。

代码正在打印double 的四舍五入 值，而不是其精确值，略小于 1.0。 (long) some_number_slightly_less_than_1 是 0。尝试使用更精确的"%.20le" 而不是"%lf" 打印double。

给定典型的 IEEE binary64 浮点，当代码以 double 或 1010.001100 开头时，代码实际上是以 1010.00109999999995125108.... 开头。

至于为什么10.617188的答案错误，我怀疑是未贴代码binary_decimal();

---

代码有错误：

// floatFract = floatFract + (double) temp1 / (double) i;
floatFract = floatFract + (double) temp1 / (double) k;
...
// wrong result of 10.61718750000000000000
// correct result follows
10.00109989999999982047....
// or to 6 decimal places
10.001100

---

次要：请注意，以下代码仅适用于 double f 范围内的 LONG_MIN 到 LONG_MAX。

integral = (long)f;

Answer 2

temp1 使用%ld 标志打印，即long int，而r 使用%lf 标志打印double，因此尾随0s。

Answer 3

在这里，我想出的东西并没有完全发挥作用。长数据类型有限制，所以我们不能在这个程序中输入比这个范围更大的数字，否则这个程序可以正常工作。

代码：

double binary_float(double f) /* Function to convert binary to float.*/
{
   long     integral = 0, floatInt = 0;
   double   floatFract = 0, fractional = 0, floatTotal = 0;

   //Separating the integral value from the floating point variable
   integral = (long)f;
   //Separating the fractional value from the variable
   fractional = f - (long)f;

   //Converting binary to decimal
   floatInt = binary_decimal(integral);
   //Converting float value from binary to float value
   floatFract = binary_float_float(fractional);

   //Combining both the integral and fractional binary value.
   floatTotal = floatInt + floatFract;

   return floatTotal;
}

long binary_decimal(long n) /* Function to convert binary to decimal.*/
{
   long decimal=0, i=0, rem;
   while (n!=0)
   {
      rem = n%10;
      n/=10;
      decimal += rem*pow(2,i);
      ++i;
   }
   return decimal;
}

double binary_float_float(double fractional)
{
   long     temp1 = 0, k = 1, i = 1;
   double   floatFract = 0;

   //Loop for converting binary to Fractional value
   while( k < 1000000000 && fractional != (double)0 )
   {
      k = k * 10;
      i = i * 2;
      temp1 = (long)(fractional * k);
      //If value is larger then 1 it means .001100 is represented as .00109994....
      //So it's last digit and so we can break loop from that point
      if (temp1 > 1)
      {
         temp1 = 1;
         floatFract = floatFract + (double)temp1/(double)i;
         break;
      }

      floatFract = floatFract + (double)temp1/(double)i;
      fractional = fractional - (double)temp1/(double)k;
   }

   return floatFract;
}

任何有更好的想法和更好的方法从二进制数中找到浮点值的人，请分享。