英文:
Java math.random query, rounding (concept understanding)
问题
我正在尝试理解math.random的逻辑，或者更具体地说，发生了什么以及为什么发生。
我理解，由于Java是一种以0为基础的语言，所以在索引等方面我们从0开始，如下面的代码所示，例如：
下面的代码会打印出一个介于0和10之间的值。
11是排除在外的，就像你理解的那样。
int result = (int) (Math.random() * 11);
我的问题是，为什么，例如，如果我们得到0.99 * 10然后等于10.99，为什么这个数字不会四舍五入到11？
这是否仅仅是Java的设计，总是向下舍入到0？还是我漏掉了某个概念？
感谢您的所有帮助。
英文:
I'm trying to understand the logic of the math.random or more specifically what is occurring and or why.
I understand essentially with the code below for example due to java being a 0 based language we start at 0 for indexing, etc.
the below code prints out a value between 0 and 10 inclusive.
11 is exclusive as understood.
int result = (int) (Math.random () * 11);
My question is why for example if we were to get 0.99 * 10 then = 10.99
Why does the number not round up to 11?
Is this simply by design java will always round down to 0? or am I missing a concept?
Thank you for all your help.
答案1
得分: 2
在评论中我提到过：
将其转换为int将导致向下取整。这意味着您的数字将始终被舍入为0，因此在转换为int后，10.99将变为10。
您还可以参考这个链接。
英文:
As I mentioned in the comments:
Casting to an int will result in a floor. This means that your number will always get rounded to 0, so 10.99 will become 10 after casting to int.
You can look at this reference as well
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