问题

这个输出为什么会变成 '1'？

long number = 499_999_999_000_000_001L;
double converted = (double) number;
System.out.println(number - (long) converted);

long number = 499_999_999_000_000_001L;    
double converted = (double) number;
System.out.println(number - (long) converted);

答案1

得分: 3

TLDR: 这是由于溢出位导致的

如果你查阅Java文档 Double.MAX_VALUE，你会注意到Java支持的最大双精度整数值是2^53 ≅ 10^16，但是在进行类型转换后，你的值变成了(4.99999999 * 10^17)，超出了双精度范围，因此发生了溢出，被舍入。为了更好地理解，请运行以下代码：

public class Main
{
    public static void main(String[] args) {
        long longNumber = 499_999_999_000_000_001L;
        double doubleNumber = (double) longNumber;
        long longConverted = (long)doubleNumber;
        System.out.println(longNumber+" "+doubleNumber+" "+longConverted);
    }
}

其输出将会是：

499999999000000001 4.99999999E17 499999999000000000

public class Main
{
    public static void main(String[] args) {
        long longNumber = 499_999_999_000_000_001L;
        double doubleNumber = (double) longNumber;
        long longConverted = (long)doubleNumber;
        System.out.println(longNumber+" "+doubleNumber+" "+longConverted);
    }
}

499999999000000001 4.99999999E17 499999999000000000

答案2

得分: 2

以下是代码的翻译部分：

你只需进行一点“调试”。  
尝试运行此代码...
long number = 499_999_999_000_000_001L;
System.out.println(number);
double converted = (double) number;
System.out.println(converted);
System.out.println((long) converted);
System.out.println(number - (long) converted);

这是它显示的内容...

499999999000000001
4.99999999E17
499999999000000000
1

你想知道为什么从double转换回long会丢失1吗？
如果是的话，我建议你参考 Java 规范。

编辑

要更精确，参考 Widening Primitive Conversion 部分：

从... long 到 double 的扩展原始类型转换可能导致精度损失 - 即结果可能会丢失值的一些最不显著的位。在这种情况下，得到的浮点值将是整数值的正确舍入版本，使用 IEEE 754 最近舍入模式（§4.2.4）。

<details>
<summary>英文:</summary>

All you need to do is a little &quot;debugging&quot;.  
Try running this code...
```java
long number = 499_999_999_000_000_001L;
System.out.println(number);
double converted = (double) number;
System.out.println(converted);
System.out.println((long) converted);
System.out.println(number - (long) converted);

This is what it displays...

499999999000000001
4.99999999E17
499999999000000000
1

Do you want to know why the conversion back to long from double drops the 1?
If you do then I refer you to the java specifications.

EDIT

To be precise, refer to section Widening Primitive Conversion

> A widening primitive conversion from ... long to double, may result in loss of precision - that is, the result may lose some of the least significant bits of the value. In this case, the resulting floating-point value will be a correctly rounded version of the integer value, using IEEE 754 round-to-nearest mode (§4.2.4).