在获取整数double值的小数部分，而不会丢失precision [英] Getting the fractional part of a double value in integer without losing precision

问题描述

我想用precision高达4位为整数的double值的小数部分转换。但是当我这样做，我失去了precision。有什么办法让我可以得到precise价值？

i want to convert the fractional part of a double value with precision upto 4 digits into integer. but when i do it, i lose precision. Is there any way so that i can get the precise value?

#include<stdio.h>
int main()
{
    double number;
    double fractional_part;
    int output;
    number = 1.1234;
    fractional_part = number-(int)number;
    fractional_part = fractional_part*10000.0;
    printf("%lf\n",fractional_part);
    output = (int)fractional_part;
    printf("%d\n",output);
    return 0;
}

我期待输出为1234，但它给了1233.请建议，这样我就可以得到所需要的输出方式。我想在C语言中的解决方案。

i am expecting output to be 1234 but it gives 1233. please suggest a way so that i can get desired output. i want the solution in C language.

推荐答案

假设你希望得到一个积极的分数甚至负值，我会去用

Assuming you want to get back a positive fraction even for negative values, I'd go with

(int)round(fabs(value - trunc(value)) * 1e4)

这应该给您预期的结果 1234 。

如果你不圆，只是截断值

If you do not round and just truncate the value

(int)(fabs(value - trunc(value)) * 1e4)

（基本上是一样的你原来的code），你会用意想不到的结果 1233 为 1.1234结束 - 1.0 = 0.12339999999999995 双precision。

(which is essentially the same as your original code), you'll end up with the unexpected result 1233 as 1.1234 - 1.0 = 0.12339999999999995 in double precision.

如果不使用 ROUND（），你还可以得到预期的结果，如果你改变操作的顺序，以

Without using round(), you'll also get the expected result if you change the order of operations to

(int)(fabs(value * 1e4 - trunc(value) * 1e4))

如果值的组成部分，是足够大的，浮点不准将再次踢当然的。

If the integral part of value is large enough, floating-point inaccuracies will of course kick in again.

您也可以使用 MODF（）而不是 TRUNC（）大卫指出，这可能是最好的方法尽可能浮点精度云：

You can also use modf() instead of trunc() as David suggests, which is probably the best approach as far as floating point accuracy goes:

double dummy;
(int)round(fabs(modf(value, &dummy)) * 1e4)

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