Jacob Parker - 7 months ago 28

Swift Question

Say I have a

`struct`

`struct Cube {`

var x: Int

var y: Int

var z: Int

var width: Int

// ...

}

How do I then create a

`Set`

`let points: Set<Cube> = Set()`

// Type ‘Cube’ does not conform to protocol ‘Hashable’

But it’s not immediately obvious how to implement hashable. From what I read, I need to make a hash function, but that doesn’t look easily possible with the amount of properties I have in the struct.

Answer

First of all, `Hashable`

extends `Equatable`

, so you must implement
a `==`

operator which compares two values, using all properties
which uniquely identify a cube:

```
func ==(lhs: Cube, rhs: Cube) -> Bool {
return lhs.x == rhs.x && lhs.y == rhs.y && lhs.z == rhs.z && lhs.width == rhs.width
}
```

The `Hashable`

protocol *only* requires that

`x == y`

implies`x.hashValue == y.hashValue`

so

```
var hashValue: Int {
return 0
}
```

would be a valid (and working) implementation. However, this would put all objects in the same hash bucket of a set (or dictionary), which is not effective. A better implementation is for example

```
struct Cube: Hashable {
var x: Int
var y: Int
var z: Int
var width: Int
var hashValue: Int {
return x.hashValue ^ y.hashValue ^ z.hashValue ^ width.hashValue
}
}
```

Here the "XOR" operator `^`

is chosen because it cannot overflow.
You could also use the "overflow operator" `&+`

.

More sophisticated hash functions would be able to distinguish different values better, so that the set operations become faster. On the other hand, the computation of the hash function itself would be slower. Therefore I would look for a "better" hash function only if the set operations turn out to be a performance bottleneck in your program.