基于3个不同的整数创建唯一的整数

Question

I'd like to create an int called hash in Java, from the coordinates x , y and w .

The values of x and y are signed and often negative.

The size of an int is 32 bits.

I'd like to assign 4 bits to w (or 3 if the most significant bit isn't usable), 14 bits to x and 14 bits to y .

I have tried the following method and I do not understand why the values clash at all: w + x << 4 + y << 18 .

For example, x = 1 clashes with y = 1 when w == 0 .

The advantages of doing this are as follows:

• Quick to look up in a database
• Quicker to compare a single integer, rather than three of them
• The allocated number of bits for each integer are never exceeded anyway

Answer 1

The only problem here is operator precedence. + goes before <<, so you have to write it like this:

w + (x << 4) + (y << 18)

This doesn't confine w or x to their allotted fields, but that doesn't do bad things with the hash value. If you had used | to combine them, it would be a bad hash when w or x get negative, but it's fine with + .

Answer 2

Your problem is operator precedence : + takes precedence over << , so your expression

w + x << 4 + y << 18

is equivalent to

((w + x) << (4 + y)) << 18

Try this:

w + (x << 4) + (y << 18)

Answer 3

int hash =  w << 28 | (x & 0x3FFF) << 14 | y & 0x3FFF;

+ is not a bitwise operator. This will make it look like this:

11110000000000000011111111111111
w -^ X          -^ Y          -^