在java中将double转换为integer

问题描述

在Java中，我想将double转换为整数，我知道如果你这样做：

  double x = 1.5; 
 int y =（int）x; 
  

你得到y = 1。如果你这样做：

  int y =（int）Math.round（x）; 
  

你可能会得到2.但是，我想知道：由于整数的双重表示有时看起来像1.9999999998或某事，是否有可能通过Math.round（）创建的双引号仍然会导致截断的数字，而不是我们正在寻找的舍入数（即：代表代码中的1而不是2）

（是的，我的意思是这样：x的任何值，其中y将显示一个结果截断而不是x的四舍五入表示？）

如果是的话：有没有更好的方法来使一个圆括号的double，而不会产生截断的风险？ p>

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图形的东西：Math.round（x）返回一个长而不是一个double。因此：Math.round（）返回一个看起来像3.9999998的数字是不可能的。因此，int（Math.round（））永远不需要截断任何东西，并且将始终工作。

解决方案

In Java, I want to convert a double to an integer, I know if you do this:

double x = 1.5;
int y = (int)x;

you get y=1. If you do this:

int y = (int)Math.round(x);

You'll likely get 2. However, I am wondering: since double representations of integers sometimes look like 1.9999999998 or something, is there a possibility that casting a double created via Math.round() will still result in a truncated down number, rather than the rounded number we are looking for (i.e.: 1 instead of 2 in the code as represented) ?

(and yes, I do mean it as such: Is there any value for x, where y will show a result that is a truncated rather than a rounded representation of x?)

If so: Is there a better way to make a double into a rounded int without running the risk of truncation?

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Figured something: Math.round(x) returns a long, not a double. Hence: it is impossible for Math.round() to return a number looking like 3.9999998. Therefore, int(Math.round()) will never need to truncate anything and will always work.

解决方案

No, round() will always round your double to the correct value, and then, it will be cast to an long which will truncate any decimal places. But after rounding, there will not be any fractional parts remaining.

Here are the docs from Math.round(double):

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