Question

我有以下测试用例

#include <stdio.h>

int main() { 

    double x = 3.987;

    printf("x = %lf\n", x);

    printf("(double) (long) (x) = %lf\n", (double) (long) (x));

    printf("(x*100)/100 = %lf\n", (x*100)/100);

    printf("(double) (long) (x*100)/100 = %lf\n", (double) (long) (x*100)/100);

    printf("(double) (long) (x*10)/10 = %lf\n", (double) (long) (x*10)/10);

    return 0;
}

输出结果为：

x = 3.987000
(double) (long) (x) = 3.000000
(x*100)/100 = 3.987000
(double) (long) (x*100)/100 = 3.980000
(double) (long) (x*10)/10 = 3.900000

在我看来，乘以100并除以100会相互抵消？但它实际上降低了精度。这是如何工作的？

Answer 1

在某些地方，你正在施展为“长”，而是整数类型。因此，例如在最后一种情况下，您将3.987乘以10，得到39.87。然后，你将它投射到“长”所以它变为39.将它除以10后得到3.9。

您希望使用(long)

来实现什么目标？

Answer 2

类型转换为long的优先级高于除以100.

(double) (long) (x*100)/100

实际上相当于

((double) (long) (x*100)) / 100

不要

(double) (long) ((x*100)/100)

数字的精确度，乘以除以100

2 个答案: