浮点算术的最佳实践 [英] Best practices for floating-point arithmetics

问题描述

我在表演的加法和减法是precise到小数点后第二位小数的输入。我试图把它们转换成整数，以提高准确性通过乘以100，但取得了相反的效果。

I was performing addition and subtraction on an input of decimals that are precise to the second decimal place. I tried to improve accuracy by converting them to integers through multiplying 100, but achieved opposite effects.

考虑以下code和输出：

Consider the following code and output:

double d = 2.01;
int a = (int) (d * 100.0);
cout << a << endl;

输出是：

200

一劳永逸，什么是一些与浮点算术的最佳实践？是它在所有可行的第一个双击转换为 INT 使用code的一些变种以上，然后铸造回来？

Once and for all, what are some of the best practices regarding floating-point arithmetics? Is it at all feasible to first convert the double to an int using some variant of the code above, and then casting it back?

推荐答案

如果您打印出 D * 100.0 来（说）20位小数，问题的结果将很快变得明显：

If you print out the result of d*100.0 to (say) 20 decimal places, the problem will quickly become apparent:

200.99999999999997

由于这是（有一每分钟）不足201，它被截断到200，当你转换为 INT 。

最明显的解决方法是强制四舍五入。至少如果你输入的都是积极的，可以作为增加0.5到结果一样简单：

The obvious cure is to force rounding. At least if your inputs are all positive that can be as simple as adding 0.5 to the result:

int a = (int)(d*100.0 + 0.5);

如果你可以在你的编译器支持指望它，它更容易使用标准库的圆形：

If you can count on your compiler supporting it, it's even easier to use the standard library's round:

long a = lround(d*100.0);

这正常工作为正数和负数。

This works correctly for both positive and negative numbers.

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