问题:

I want to store two ints in a long (instead of having to create a new Point object every time).

Currently, I tried this. It's not working, but I don't know what is wrong with it:

// x and y are ints
long l = x;
l = (l << 32) | y;

And I'm getting the int values like so:

x = (int) l >> 32;
y = (int) l & 0xffffffff;

回答1:

y is getting sign-extended in the first snippet, which would overwrite x with -1 whenever y < 0.

In the second snippet, the cast to int is done before the shift, so x actually gets the value of y.

long l = (((long)x) << 32) | (y & 0xffffffffL);
int x = (int)(l >> 32);
int y = (int)l;

回答2:

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:

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(1);