When to calculate key's hashcode, spread() method is called:

```
static final int spread(int h) {
return (h ^ (h >>> 16)) & HASH_BITS;
}
```

where `HASH_BITS`

equals `0x7fffffff`

, so, what is the purpose of `HASH_BITS`

? Some one says it make the sign bit to 0, I am not sure about that.

asked on Stack Overflow May 30, 2018 by lambdie

The `index`

of KV Node in hash buckets is calculated by following formula:

```
index = (n - 1) & hash
```

`hash`

is the result of`spread()`

`n`

is the length of hash buckets which maximum is 2^30`private static final int MAXIMUM_CAPACITY = 1 << 30;`

So the maximum of `n - 1`

is 2^30 - 1 which means the top bit of `hash`

will never be used in index calculation.

But i still don't understand is it necessary to clear the top bit of `hash`

to 0.It seems that there are more reasons to do so.

```
/**
* Spreads (XORs) higher bits of hash to lower and also forces top
* bit to 0. Because the table uses power-of-two masking, sets of
* hashes that vary only in bits above the current mask will
* always collide. (Among known examples are sets of Float keys
* holding consecutive whole numbers in small tables.) So we
* apply a transform that spreads the impact of higher bits
* downward. There is a tradeoff between speed, utility, and
* quality of bit-spreading. Because many common sets of hashes
* are already reasonably distributed (so don't benefit from
* spreading), and because we use trees to handle large sets of
* collisions in bins, we just XOR some shifted bits in the
* cheapest possible way to reduce systematic lossage, as well as
* to incorporate impact of the highest bits that would otherwise
* never be used in index calculations because of table bounds.
*/
static final int spread(int h) {
return (h ^ (h >>> 16)) & HASH_BITS;
}
```

answered on Stack Overflow May 31, 2018 by Wangsla

I think it is to avoid collision with the preserved hashcodes: MOVED(-1), TREEBIN(-2) and RESERVED(-3) of which symbol bits are always 1.

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