意外结果导致长/整数除法 [英] Unexpected result in long/int division

问题描述

我有这样的价值观:

long millis = 11400000;int 常数 = 86400000;双分辨率 = 毫/常数；

问题是:为什么 res 等于 0.0(而不是 ca. 0.131944)?它存储在 double 中，所以应该没有四舍五入吧?

解决方案

当你使用二元运算符时，两个参数的类型应该相同，结果也应该是它们的类型.当你想划分 (int)/(long) 它变成 (long)/(long) 结果是 (long).您应该将其设为 (double)/(long) 或 (int)/(double) 以获得双重结果.由于 double 大于 int 和 long，所以 int 和 long 在 (double)/(long) 和 (int)/(double)

I have values like this:

long millis = 11400000;
int consta = 86400000;
double res = millis/consta;

The question is: why res equals 0.0 (instead of ca. 0.131944)? It's stored in double so there should be no rounding right?

解决方案

When you are using a binary operator, both arguments should be of a same type and the result will be in their type too. When you want to divide (int)/(long) it turns into (long)/(long) and the result is (long). you shouldmake it (double)/(long) or (int)/(double) to get a double result. Since double is greater that int and long, int and long will be turned into double in (double)/(long) and (int)/(double)

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