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?
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?
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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