c++ - floor() 行为异常

Question

我需要一个函数来将正 double 四舍五入到最接近的整数。潜伏 aorund 我发现这种非常优雅的方式

int x = floor(y + 0.5);

我写了一个简单的测试程序:

double a = 10.0;

for (int i = 0; i < 10; i++) {
  cout << a << "\t" << a + 0.5 << "\t" << floor(a + 0.5) << endl;
  a += 0.1;
}

但是我收到了一些奇怪的输出

10      10.5    10
10.1    10.6    10
10.2    10.7    10
10.3    10.8    10
10.4    10.9    10
10.5    11      10 <--- should be 11!
10.6    11.1    11
10.7    11.2    11
10.8    11.3    11
10.9    11.4    11

那是什么？

谢谢问候卢卡

Answer 1

通过添加 0.1，您确实添加了一个略低于 0.1 的值。

所以加0.1 5次和加0.5一次是不一样的；您没有完全达到该值。通过再次添加 .5，您不会超过 11，这会产生您观察到的行为。

---

一个C程序如

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

int main()
{
    double a = 10.0;
    int i;
    for (i = 0; i < 11; i++) {
        printf("%4.19f\t%4.19f\t%4.19f\n", a, a+.5, floor(a + 0.5));
        a += 0.1;
    }
    printf("\n");
    for (i = 0; i < 11; i++) {
        a = 10.0 + i/10.0;
        printf("%4.19f\t%4.19f\t%4.19f\n", a, a+.5, floor(a + 0.5));
    }
}

显示在它的输出上

10.0000000000000000000  10.5000000000000000000  10.0000000000000000000
10.0999999999999996447  10.5999999999999996447  10.0000000000000000000
10.1999999999999992895  10.6999999999999992895  10.0000000000000000000
10.2999999999999989342  10.7999999999999989342  10.0000000000000000000
10.3999999999999985789  10.8999999999999985789  10.0000000000000000000
10.4999999999999982236  10.9999999999999982236  10.0000000000000000000
10.5999999999999978684  11.0999999999999978684  11.0000000000000000000
10.6999999999999975131  11.1999999999999975131  11.0000000000000000000
10.7999999999999971578  11.2999999999999971578  11.0000000000000000000
10.8999999999999968026  11.3999999999999968026  11.0000000000000000000
10.9999999999999964473  11.4999999999999964473  11.0000000000000000000

10.0000000000000000000  10.5000000000000000000  10.0000000000000000000
10.0999999999999996447  10.5999999999999996447  10.0000000000000000000
10.1999999999999992895  10.6999999999999992895  10.0000000000000000000
10.3000000000000007105  10.8000000000000007105  10.0000000000000000000
10.4000000000000003553  10.9000000000000003553  10.0000000000000000000
10.5000000000000000000  11.0000000000000000000  11.0000000000000000000
10.5999999999999996447  11.0999999999999996447  11.0000000000000000000
10.6999999999999992895  11.1999999999999992895  11.0000000000000000000
10.8000000000000007105  11.3000000000000007105  11.0000000000000000000
10.9000000000000003553  11.4000000000000003553  11.0000000000000000000
11.0000000000000000000  11.5000000000000000000  11.0000000000000000000

区别:第一次运行是您的方法，累积误差和步长为 0.0999999999999996447，而第二次运行重新计算尽可能接近的 a，使其有可能准确地达到 10.5 和 11.0。