Convert [UInt8] to [UInt32] in Swift 3

Question

How to Convert [UInt8] to [UInt32] in Swift 3?

let u8: [UInt8] = [0x02, 0x02, 0x02]

What have you tried? `UInt32` is initializable with `UInt8`. Are you looking for something more sophisticated than mapping this array onto the resulting one? — Losiowaty, Apr 19 '17 at 13:56
See https://stackoverflow.com/questions/25267089/convert-a-two-byte-uint8-array-to-a-uint16-in-swift — Ortomala Lokni, Apr 19 '17 at 13:59
`let u32 = u8.map{ UInt32($0) }` or do you want `[UInt8]` to `UInt32` (no array)? — vadian, Apr 19 '17 at 14:01

score 5 · Answer 1 · edited Apr 19 '17 at 14:14

5

Try this

let u8: [UInt8] = [0x02, 0x02, 0x02]
let u32: [UInt32] = u8.map { UInt32($0)  }
print("u32: \(u32)")

edited Apr 19 '17 at 14:14

Paulo Mattos

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answered Apr 19 '17 at 14:01

RajeshKumar R

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Why do you call an array of 32-bit integers `u16`?? – The type annotation `[UInt32]` is not needed. – Martin R Apr 19 '17 at 14:10
@MartinR That was a typo. Yes. Type annotation `[UInt32]` isn't needed. – RajeshKumar R Apr 19 '17 at 14:15

Okan TEL · Answer 2 · 2022-05-14T17:25:06.827

I used bit shifting.You can try below code lines.

let bytes: [UInt8] = [41, 22, 91, 13, 85, 71, 12, 34]
func getU32Array(byteArr: [UInt8]) -> [UInt32]? {

let numBytes = byteArr.count
var byteArrSlice = byteArr[0..<numBytes]

guard numBytes % 4 == 0 else { print ("There is not a multiple of 4") ;return nil }

var arr =  [UInt32](repeating: 0, count: numBytes/4)
for i in (0..<numBytes/4).reversed() {
    arr[i] = UInt32(byteArrSlice.removeLast())  + UInt32(byteArrSlice.removeLast()) << 8 + UInt32(byteArrSlice.removeLast()) << 16 + UInt32(byteArrSlice.removeLast()) << 24
}
return arr
} 
print(getU32Array(byteArr: bytes)!)

//output: [689330957, 1430719522]

Convert [UInt8] to [UInt32] in Swift 3

2 Answers2