Hy everyone,
I need to transform a Int to its hexadecimal value.
Example : -40 => D8
I have a working method for positive (or unsigned) Int but it doesn't work as expected with negatives. Here's my code.
class func encodeHex(data:[Int]) -> String {
let hexadecimal = data.reduce("") { (string , element) in
var append = String(element, radix:16 , uppercase : false)
if append.characters.count == 1 {
append = "0" + append
}
return string + append
}
return hexadecimal
}
If I pass -40 I get -28.
Can anyone help ? Thanks :)
I assume from your existing code that all integers are in the range
-128 ... 127. Then this would work:
func encodeHex(data:[Int]) -> String {
return data.map { String(format: "%02hhX", $0) }.joined()
}
The "%02hhX" format prints the least significant byte of the
given integer in base 16 with 2 digits.
Example:
print(encodeHex(data: [40, -40, 127, -128]))
// 28D87F80
D8 is the last byte of binary representation of -40. The remaining three bytes are all FFs.
If you are looking for a string that represents only the last byte, you can obtain by first converting your number to unsigned 8-bit integer, and then converting it to hex, like this:
let x = UInt8(bitPattern:Int8(data))
let res = String(format:"%02X", x)
Related
I have an array of UInt8 and I want to calculate CheckSum8 Modulo 256.
If sum of bytes is less than 255, checkSum function returns correct value.
e.g
let bytes1 : [UInt8] = [1, 0xa1]
let validCheck = checkSum(data : bytes1) // 162 = 0xa2
let bytes : [UInt8] = [6, 0xB1, 27,0xc5,0xf5,0x9d]
let invalidCheck = checkSum(data : bytes) // 41
Below function returns 41 but expected checksum is 35.
func checkSum(data: [UInt8]) -> UInt8 {
var sum = 0
for i in 0..<data.count {
sum += Int(data[i])
}
let retVal = sum & 0xff
return UInt8(retVal)
}
Your checkSum method is largely right. If you want, you could simplify it to:
func checkSum(_ values: [UInt8]) -> UInt8 {
let result = values.reduce(0) { ($0 + UInt32($1)) & 0xff }
return UInt8(result)
}
You point out a web site that reports the checksum8 for 06B127c5f59d is 35.
The problem is that in your array has 27, not 0x27. If you have hexadecimal values, you always need the 0x prefix for each value in your array literal (or, technically, at least if the value is larger than 9).
So, consider:
let values: [UInt8] = [0x06, 0xB1, 0x27, 0xc5, 0xf5, 0x9d]
let result = checkSum(values)
That’s 53. If you want to see that in hexadecimal (like that site you referred to):
let hex = String(result, radix: 16)
That shows us that the checksum is 0x35 in hexadecimal.
Convert Bytes values into GB/MB/KB, i am using ByteCountFormatter. Example code below.
func converByteToGB(_ bytes:Int64) -> String {
let formatter:ByteCountFormatter = ByteCountFormatter()
formatter.countStyle = .binary
return formatter.string(fromByteCount: Int64(bytes))
}
Now, my requirement is it should show only one digit after decimal point.
Example for 1.24 GB => 1.2 GB, not like 1.24 GB. force it for single digit after apply floor or ceil function.
ByteCountFormatter is not capable to show only one digit after decimal point. It shows by default 0 fraction digits for bytes and KB; 1 fraction digits for MB; 2 for GB and above. If isAdaptive is set to false it tries to show at least three significant digits, introducing fraction digits as necessary.
ByteCountFormatter also trims trailing zeros. To disable set zeroPadsFractionDigits to true.
I have adapted How to convert byte size into human readable format in java? to do what you want:
func humanReadableByteCount(bytes: Int) -> String {
if (bytes < 1000) { return "\(bytes) B" }
let exp = Int(log2(Double(bytes)) / log2(1000.0))
let unit = ["KB", "MB", "GB", "TB", "PB", "EB"][exp - 1]
let number = Double(bytes) / pow(1000, Double(exp))
return String(format: "%.1f %#", number, unit)
}
Notice this will format KB and MB differently than ByteCountFormatter. Here is a modification that removes trailing zero and does not show fraction digits for KB and for numbers larger than 100.
func humanReadableByteCount(bytes: Int) -> String {
if (bytes < 1000) { return "\(bytes) B" }
let exp = Int(log2(Double(bytes)) / log2(1000.0))
let unit = ["KB", "MB", "GB", "TB", "PB", "EB"][exp - 1]
let number = Double(bytes) / pow(1000, Double(exp))
if exp <= 1 || number >= 100 {
return String(format: "%.0f %#", number, unit)
} else {
return String(format: "%.1f %#", number, unit)
.replacingOccurrences(of: ".0", with: "")
}
}
Be also aware that this implementation does not take Locale into account. For example some locales use comma (",") instead of dot (".") as decimal separator.
After searching for a while, I can't figure out how to get this simple result:
let byte : UInt8 = 0xF3 //Should become "F3"
I have tried this method that won't compile when passing-in either a byte or a byte array.
Two ways:
let s1 = String(byte, radix: 16, uppercase: true) // does not do 0-padding but works with
// all radices between 2 and 36
let s2 = String(format: "%02X", byte) // very similar to C
In C# one can write the following one line to make an array of all the hex strings representing the values from 0 to 255:
using System.Linq;
static string[] HexTbl = Enumerable.Range(0, 256).Select(v => v.ToString("X2")).ToArray();
Is there a similarly compact way to do this in Swift?
Yes there is. The approach is the same: map each number in the range
0 ... 255 to a string using a hex format:
let hexTable = (0 ..< 256).map { v in String(format: "%02X", v) }
or slightly shorter:
let hexTable = (0 ..< 256).map { String(format: "%02X", $0) }
Result:
["00", "01", "02", ..., "FD", "FE", "FF"]
So I am trying to split a number in swift, I have tried searching for this on the internet but have had no success. So first I will start with a number like:
var number = 34.55
And from this number, I want to create two separate number by splitting from the dot. So the output should be something like:
var firstHalf = 34
var secondHalf = 55
Or the output can also be in an array of thats easier. How can I achieve this?
The easiest way would be to first cast the double to a string:
var d = 34.55
var b = "\(d)" // or String(d)
then split the string with the global split function:
var c = split(b) { $0 == "." } // [34, 55]
You can also bake this functionality into a double:
extension Double {
func splitAtDecimal() -> [Int] {
return (split("\(self)") { $0 == "." }).map({
return String($0).toInt()!
})
}
}
This would allow you to do the following:
var number = 34.55
print(number.splitAtDecimal()) // [34, 55]
Well, what you have there is a float, not a string. You can't really "split" it, and remember that a float is not strictly limited to 2 digits after the separator.
One solution is :
var firstHalf = Int(number)
var secondHalf = Int((number - firstHalf) * 100)
It's nasty but it'll do the right thing for your example (it will, however, fail when dealing with numbers that have more than two decimals of precision)
Alternatively, you could convert it into a string and then split that.
var stringified = NSString(format: "%.2f", number)
var parts = stringifed.componentsSeparatedByString(".")
Note that I'm explicitly calling the formatter here, to avoid unwanted behavior of standard float to string conversions.
Add the following extension:
extension Double {
func splitIntoParts(decimalPlaces: Int, round: Bool) -> (leftPart: Int, rightPart: Int) {
var number = self
if round {
//round to specified number of decimal places:
let divisor = pow(10.0, Double(decimalPlaces))
number = Darwin.round(self * divisor) / divisor
}
//convert to string and split on decimal point:
let parts = String(number).components(separatedBy: ".")
//extract left and right parts:
let leftPart = Int(parts[0]) ?? 0
let rightPart = Int(parts[1]) ?? 0
return(leftPart, rightPart)
}
Usage - Unrounded:
let number:Double = 95.99999999
let parts = number.splitIntoParts(decimalPlaces: 3, round: false)
print("LeftPart: \(parts.leftPart) RightPart: \(parts.rightPart)")
Outputs:
LeftPart: 95 RightPart: 999
Usage Rounded:
let number:Double = 95.199999999
let parts = number.splitIntoParts(decimalPlaces: 1, round: true)
Outputs:
LeftPart: 95 RightPart: 2
Actually, it's not possible to split a double by dot into two INTEGER numbers. All earlier offered solutions will produce a bug in nearly 10% of cases.
Here's why:
The breaking case will be numbers with decimal parts starting with one or more zeroes, for example: 1.05, 11.00698, etc.
Any decimal part that starts with one or more zeroes will have those zeroes discarded when converted to integers. The result of those conversions:
1.05 will become (1, 5)
11.00698 will become (11, 698)
An ugly and hard to find bug is guaranteed...
The only way to meaningfully split a decimal number is to convert it to (Int, Double). Below is a simple extension to Double that does that:
extension Double {
func splitAtDecimal() -> (Int, Double) {
let whole = Int(self)
let decimal = self - Darwin.floor(self)
return (whole, decimal)
}
}