Ben Leggiero - 1 month ago 11

Swift Question

I want a function that takes in a bitmask

`Int`

`Int`

`func split(bitmask: Int) -> Set<Int> {`

// Do magic

}

such that

`split(bitmask: 0b01001110) == [0b1000000, 0b1000, 0b100, 0b10]`

Answer

One solution is to check each bit and add the corresponding mask if the bit is set.

```
func split(bitmask: Int) -> Set<Int> {
var results = Set<Int>()
// Change 31 to 63 or some other appropriate number based on how big your numbers can be
for shift in 0...31 {
let mask = 1 << shift
if bitmask & mask != 0 {
results.insert(mask)
}
}
return results
}
print(split(bitmask: 0b01001110))
```

For the binary number `0b01001110`

the results will be:

[64, 2, 4, 8]

which are the decimal equivalent of the results in your question.

For the hex number `0x01001110`

(which is `1000100010000`

in binary) the results will be:

[16, 256, 4096, 16777216]

Here's another solution that doesn't need to know the size of the value and it's slightly more efficient for smaller numbers:

```
func split(bitmask: Int) -> Set<Int> {
var results = Set<Int>()
var value = bitmask
var mask = 1
while value > 0 {
if value % 2 == 1 {
results.insert(mask)
}
value /= 2
mask *= 2
}
return results
}
```