当 double 转换为 int 时，最后一位数字丢失

Question

考虑以下程序：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(void)
{
  int i;
  long int rr;
  double   dd;
  double arr[10] = {55.550,55.551,55.552,55.553,55.554,55.555,55.556,55.557,55.558,55.559};
  printf("\n\nTest1");
  for(i=0;i<10;i++)
  {
    dd = 100 * arr[i];
    rr = (long int)dd;
    printf("\n 100 * %10.4lf == %10.4lf >>>>> %ld",arr[i], dd, rr);
  }
  printf("\n\nTest2");
  for(i=0;i<10;i++)
  {
    printf("\n 100 * %10.4lf == %10.4lf >>>>> %ld",
           arr[i], 100 * arr[i], (long int)(100 * arr[i]));
  }
  return 0;
}

在我执行这个东西后我得到：

Test1
 100 *    55.5500 ==  5555.0000 >>>>> 5555
 100 *    55.5510 ==  5555.1000 >>>>> 5555
 100 *    55.5520 ==  5555.2000 >>>>> 5555
 100 *    55.5530 ==  5555.3000 >>>>> 5555
 100 *    55.5540 ==  5555.4000 >>>>> 5555
 100 *    55.5550 ==  5555.5000 >>>>> 5555
 100 *    55.5560 ==  5555.6000 >>>>> 5555
 100 *    55.5570 ==  5555.7000 >>>>> 5555
 100 *    55.5580 ==  5555.8000 >>>>> 5555
 100 *    55.5590 ==  5555.9000 >>>>> 5555

Test2
 100 *    55.5500 ==  5555.0000 >>>>> 5554 <-- Look at here !
 100 *    55.5510 ==  5555.1000 >>>>> 5555
 100 *    55.5520 ==  5555.2000 >>>>> 5555
 100 *    55.5530 ==  5555.3000 >>>>> 5555
 100 *    55.5540 ==  5555.4000 >>>>> 5555
 100 *    55.5550 ==  5555.5000 >>>>> 5555
 100 *    55.5560 ==  5555.6000 >>>>> 5555
 100 *    55.5570 ==  5555.7000 >>>>> 5555
 100 *    55.5580 ==  5555.8000 >>>>> 5555
 100 *    55.5590 ==  5555.9000 >>>>> 5555
Process returned 0 (0x0)   execution time : -0.000 s
Press any key to continue.

好像当你像这样相乘然后投射 55.550 时，最后一位数字被遗漏了。这正常吗？

在那种情况下应该如何进行转换？

Answer 1

标准的float精度问题。 55.550 的值不能准确地存储在内存中，因此存储了一些不同但非常接近的值。因此，不同的操作顺序（甚至包括将中间结果存储到 double 中）都会影响结果。很可能在第一次测试中，dd 中的值变为 5555，而在第二种情况下 100 * arr[i] 变为 5554.9999999997.

当 double 转换为 int 时，最后一位数字丢失

The last digit is lost when the double is cast to int

c

double

integer

casting