Swift Little endian hex value to decimal - ios

Question

I'm having difficulties implementing a function converting hex value(little endian) to decimal value.
So I'm writing function:
func convertHexLittleEndianToDeciaml(input:String) -> (int)
and Input is always 4 bytes(so 8 characters in input string)
value for convertHexLittleEndianToDeciaml("606d0000") should return 28,000

Answer 1

You can write something like this:
(See UPDATEs)
func convertHexLittleEndianToDeciaml(input:String) -> Int32 {
if let beValue = Int32(input, radix: 16) where input.characters.count == 8 {
return beValue.byteSwapped
} else {
//or `fatalError()` or `return 0` or ...
return Int32.min
}
}
print(convertHexLittleEndianToDeciaml("606d0000")) //->28000
UPDATE
Sorry, but the code above have some overflow issue, happens with something like "FF010000":
func convertHexLittleEndianToDeciaml(input:String) -> Int32 {
if let beValue = UInt32(input, radix: 16) where input.characters.count == 8 {
return Int32(bitPattern: beValue.byteSwapped)
} else {
//or `fatalError()` or `return 0` or ...
return Int32.min
}
}
UPDATE2
So, I have found that returning a valid Int32 in error case might cause a bug which cannot easily be found. I recommend you to change the return type to Optional and return nil in error case.
func convertHexLittleEndianToDeciaml(input:String) -> Int32? {
guard let beValue = UInt32(input, radix: 16) where input.characters.count == 8 else {
return nil
}
return Int32(bitPattern: beValue.byteSwapped)
}
if let value = convertHexLittleEndianToDeciaml("606d0000") {
print(value) //->28000
} else {
print("Hex format invalid")
}

Answer 2

Here's a solution that works for longer numbers that I wrote:
func convert(number: String) -> Int {
var formatted = number
if number.count <= 16 {
let difference = 16 - number.count
for _ in 0..<difference {
formatted.append("0")
}
} else {
return 0
}
let value = UInt64(formatted, radix: 16) ?? 0
let z = value.byteSwapped
let final = UInt64(bitPattern: Int64(z))
return Int(final)
}