Question

考虑以下计划：

#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;
}

执行完这件事之后我得到了：

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.

似乎当你像这样乘以然后投出55.550时，错过了最后一位数。这是正常的吗？

在这种情况下如何进行演员表？

Answer 1

这是浮动精度的标准问题。 55.550的值不能精确地存储在内存中，因此存储了一些不同但非常接近的值。因此，不同的操作序列（包括将中间结果存储到double中）可能会影响结果。很可能在第一次测试时，dd中的值变为5555，而在第二种情况下，100 * arr[i]变为5554.9999999997。

将double转换为int

1 个答案: