Convert array of bytes to Int64 in Swift 5

Question

I am processing various arrays of UInt8 (little endian) and need to convert them to Int64. In Swift 4 I used

let array: [UInt8] = [13,164,167,80,4,0]
let raw = Int64(littleEndian: Data(array).withUnsafeBytes { $0.pointee })
print(raw) //18533032973

which worked fine. However in Swift 5 this way is deprecated so I switched to

let array: [UInt8] = [13,164,167,80,4,0]
let raw = array.withUnsafeBytes { $0.load(as: Int64.self) }
print(raw)

which gives an error message:

Fatal error: UnsafeRawBufferPointer.load out of bounds

Is there a way in Swift 5 to convert this without filling the array with additional 0s until the conversion works?

Thanks!

How about traditional adding the pure byte values, and shifting in a loop? — jazz64, Jun 03 '19 at 14:41
`Int64` is made of 8 bytes, you need to give the missing 2 bytes explicitly. — OOPer, Jun 03 '19 at 14:46

score 4 · Accepted Answer · answered Jun 03 '19 at 14:58

4

Alternatively you can compute the number by repeated shifting and adding, as suggested in the comments:

let array: [UInt8] = [13, 164, 167, 80, 4, 0]
let raw = array.reversed().reduce(0) { $0 << 8 + UInt64($1) }
print(raw) // 18533032973

answered Jun 03 '19 at 14:58

Martin R

score 3 · Answer 2 · answered Jun 03 '19 at 14:46

withUnsafeBytes { $0.load works only if the array contains exactly 64 bit (8 bytes) for example

let array: [UInt8] = [13,164,167,80,4,0,0,0]
let raw = array.withUnsafeBytes { $0.load(as: Int64.self) }
print(raw)

With your 6 bytes array you can use

var raw : Int64 = 0
withUnsafeMutableBytes(of: &raw, { array.copyBytes(to: $0)} )
print(raw)

2 Answers2