c - modf 功能无法正常工作

Question

我正在尝试存储十进制数的所有数字。我决定使用 modf 来实现此目的。我的代码段是；

struct high_precision scan_high(int *j)
{
    int i,a;
    struct high_precision mynum;
    double num1, fracpart, intpart;
    printf("Enter the values> ");
    scanf("%lf", &num1);
    if( num1 < 0 )
        mynum.sign = -1;
    else
        mynum.sign = 1;
    num1 = fabs(num1);
    fracpart = modf(num1, &intpart);
    if ( intpart > 0 && intpart < 10 )
        a = 1;
    while( intpart == 0 ) {
        fracpart *= 10;
        fracpart = modf(fracpart, &intpart);
        a -= 1;
    }
    for(i=0;fracpart > 0 && intpart != 0;i++){
        if( intpart > 0 ){
            mynum.digits[i] = intpart;
        }
        fracpart *= 10;
        fracpart = modf(fracpart, &intpart);
    }
    *j = i;
    mynum.decpt = a;

    return(mynum);
}

但不知怎的，它并没有按照我想要的方式工作。例如；

Enter the values> 0.009876
    0.876000 9.000000
    0.760000 8.000000
    0.600000 7.000000

它必须停在这一行。但是，它仍在继续计数；

1.000000 5.000000
1.000000 9.000000
1.000000 9.000000
1.000000 9.000000
1.000000 9.000000
1.000000 9.000000
0.999998 9.000000
0.999977 9.000000
0.999767 9.000000
0.997669 9.000000
0.976694 9.000000
0.766942 9.000000
0.669420 7.000000
0.694198 6.000000
0.941983 6.000000
0.419827 9.000000
0.198267 4.000000
0.982671 1.000000
0.826707 9.000000
0.267069 8.000000
0.000000 2.000000
0.000000 0.000000

Answer 1

这是因为 float 没有精确存储(例如，不能精确表示)。您可以使用以下示例来说明这一点:

#include <stdio.h>

int main() {
  double x = 0.009876;
  printf("%.20lf\n",x);
  return 0;
}

---------- Capture Output ----------
> "c:\windows\system32\cmd.exe" /c c:\temp\temp.exe
0.00987599999999999940

> Terminated with exit code 0.