为什么将两个 int 分配给 double 时不会产生正确的值?
为什么会出现在下面的代码片段中
How come that in the following snippet
int a = 7;
int b = 3;
double c = 0;
c = a / b;
c 最终的值为 2，而不是预期的 2.3333.如果 a 和 b 是双精度数，那么答案确实会变成 2.333.但肯定是因为 c 已经是一个 double 它应该与整数一起使用?
c ends up having the value 2, rather than 2.3333, as one would expect. If a and b are doubles, the answer does turn to 2.333. But surely because c already is a double it should have worked with integers?
那么为什么 int/int=double 不起作用?
So how come int/int=double doesn't work?
推荐答案
这是因为你使用的是整数除法版本的operator/，需要2个ints并返回一个 int.为了使用返回 double 的 double 版本，必须将至少一个 int 显式转换为 双.
This is because you are using the integer division version of operator/, which takes 2 ints and returns an int. In order to use the double version, which returns a double, at least one of the ints must be explicitly casted to a double.
c = a/(double)b;
相关文章