java Java将两个整数存储在一个long中
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Java storing two ints in a long
提问by LanguagesNamedAfterCofee
I want to store two ints in a long (instead of having to create a new Pointobject every time).
我想将两个整数存储在一个 long 中（而不是Point每次都创建一个新对象）。
Currently, I tried this. It's not working, but I don't know what is wrong with it:
目前，我试过这个。它不起作用，但我不知道它有什么问题：
// x and y are ints
long l = x;
l = (l << 32) | y;
And I'm getting the int values like so:
我得到这样的 int 值：
x = (int) l >> 32;
y = (int) l & 0xffffffff;
回答by harold
yis getting sign-extended in the first snippet, which would overwrite xwith -1whenever y < 0.
y越来越符号扩展的第一个片段，它会覆盖x与-1时y < 0。
In the second snippet, the cast to intis done before the shift, so xactually gets the value of y.
在第二个代码段中，转换int为在移位之前完成，因此x实际上获取了 的值y。
long l = (((long)x) << 32) | (y & 0xffffffffL);
int x = (int)(l >> 32);
int y = (int)l;
回答by Mingwei Samuel
Here is another option which uses a bytebuffer instead of bitwise operators. Speed-wise, it is slower, about 1/5 the speed, but it is much easier to see what is happening:
这是另一个使用字节缓冲区而不是按位运算符的选项。速度方面，它更慢，大约是速度的 1/5，但更容易看到正在发生的事情：
long l = ByteBuffer.allocate(8).putInt(x).putInt(y).getLong(0);
//
ByteBuffer buffer = ByteBuffer.allocate(8).putLong(l);
x = buffer.getInt(0);
y = buffer.getInt(4);
