So I want to generate a random number between 0 and numberTextBox.text.
Let's say that the textbox text is 5, then i want to generate a random number between 0 and 5.
I have now:
#IBAction func nextButton(sender: AnyObject) {
let maxInteger = MacNumber.toInt()!
let randomNumber = arc4random_uniform(UInt32(maxNumber));
let mxNumber = numberTextBox.text
var textString = String(randomNumber)
numberLabel.text = textString
}
But when I test the application, and type 5 in the textBox, the numberLabel.text is something like 192819371.. It's very high.
Here you go:
#IBAction func nextButton(sender: AnyObject) {
if numberTextbox.text == ""{
return
}
var number:Int = 1 + numberTextbox.text.toInt()!
var randomNumber = random() % number
if randomNumber == 0{
randomNumber = 1
}
var textString = String(randomNumber)
numberLabel.text = textString
}
I also added that if the textbox is empty, it just returns, and if the randomNumber is 0, it goes to 1.
If you want to generate a random number in Swift, use the arc4random_uniform() function. The following code snippet generates a number from 0 to 9:
var i = Int(arc4random_uniform(10))
Do not use random(), use arc4random() or arc4random_uniform(). random() needs to be seeded and will return the same sequence for the same seed after every seeding.
let maxInteger = MaxNumber.toInt()!
let randomNumber = arc4random_uniform(UInt32(maxNumber));
arc4random() does not need to be seeded and returns cryptographic quality random numbers.
Use arc4random_uniform() when the range needs to be constrained instead of the modulus operate % to eliminate modulo bias.
See: ARC4RANDOM(3)
May be you should try let randomNumber = random() % MaxNumber
instead of
let randomNumber = random(); MaxNumber ?
Related
I am trying to solve the task
Using a standard for-in loop add all odd numbers less than or equal to 100 to the oddNumbers array
I tried the following:
var oddNumbers = [Int]()
var numbt = 0
for newNumt in 0..<100 {
var newNumt = numbt + 1; numbt += 2; oddNumbers.append(newNumt)
}
print(oddNumbers)
This results in:
1,3,5,7,9,...199
My question is: Why does it print numbers above 100 although I specify the range between 0 and <100?
You're doing a mistake:
for newNumt in 0..<100 {
var newNumt = numbt + 1; numbt += 2; oddNumbers.append(newNumt)
}
The variable newNumt defined inside the loop does not affect the variable newNumt declared in the for statement. So the for loop prints out the first 100 odd numbers, not the odd numbers between 0 and 100.
If you need to use a for loop:
var odds = [Int]()
for number in 0...100 where number % 2 == 1 {
odds.append(number)
}
Alternatively:
let odds = (0...100).filter { $0 % 2 == 1 }
will filter the odd numbers from an array with items from 0 to 100. For an even better implementation use the stride operator:
let odds = Array(stride(from: 1, to: 100, by: 2))
If you want all the odd numbers between 0 and 100 you can write
let oddNums = (0...100).filter { $0 % 2 == 1 }
or
let oddNums = Array(stride(from: 1, to: 100, by: 2))
Why does it print numbers above 100 although I specify the range between 0 and <100?
Look again at your code:
for newNumt in 0..<100 {
var newNumt = numbt + 1; numbt += 2; oddNumbers.append(newNumt)
}
The newNumt used inside the loop is different from the loop variable; the var newNumt declares a new variable whose scope is the body of the loop, so it gets created and destroyed each time through the loop. Meanwhile, numbt is declared outside the loop, so it keeps being incremented by 2 each time through the loop.
I see that this is an old question, but none of the answers specifically address looping over odd numbers, so I'll add another. The stride() function that Luca Angeletti pointed to is the right way to go, but you can use it directly in a for loop like this:
for oddNumber in stride(from:1, to:100, by:2) {
// your code here
}
stride(from:,to:,by:) creates a list of any strideable type up to but not including the from: parameter, in increments of the by: parameter, so in this case oddNumber starts at 1 and includes 3, 5, 7, 9...99. If you want to include the upper limit, there's a stride(from:,through:,by:) form where the through: parameter is included.
If you want all the odd numbers between 0 and 100 you can write
for i in 1...100 {
if i % 2 == 1 {
continue
}
print(i - 1)
}
For Swift 4.2
extension Collection {
func everyOther(_ body: (Element) -> Void) {
let start = self.startIndex
let end = self.endIndex
var iter = start
while iter != end {
body(self[iter])
let next = index(after: iter)
if next == end { break }
iter = index(after: next)
}
}
}
And then you can use it like this:
class OddsEvent: UIViewController {
override func viewDidLoad() {
super.viewDidLoad()
(1...900000).everyOther{ print($0) } //Even
(0...100000).everyOther{ print($0) } //Odds
}
}
This is more efficient than:
let oddNums = (0...100).filter { $0 % 2 == 1 } or
let oddNums = Array(stride(from: 1, to: 100, by: 2))
because supports larger Collections
Source: https://developer.apple.com/videos/play/wwdc2018/229/
I need to round stocks, indices and futures prices to the nearest tick. The first step is to look if the price is a multiple of the tick. Apple docs says "Unlike the remainder operator in C and Objective-C, Swift’s remainder operator can also operate on floating-point numbers".
If I write the following code in a playground or in a console app and I run it, I expect 0 as result but I get a remainder value equals to 0.00999999999999775:
var stringPrice = "17.66"
var price = Double(stringPrice)
var tickSize: Double = 0.01
let remainder = price! % ticksize
This problem breaks my rounding function when using values such 17.66 as aPrice and 0.01 as aTickSize:
func roundPriceToNearestTick(Price aPrice: Double, TickSize a TickSize: Double)-> Double{
let remainder = aPrice % aTickSize
let shouldRoundUp = remainder >= aTickSize/2 ? true : false
let multiple = floor(aPrice/aTickSize)
let returnPrice = !shouldRoundUp ? aTickSize*multiple : aTickSize*multiple + aTickSize
return returnPrice
}
What is the best way to fix this?
Following the comments about the broken floating point math and the need to avoid floats and doubles for all the operations concerning money I changed my code to perform the remainder operation using NSDecimalNumbers. This seems to solve the precision problem.
var stringPrice = "17.66"
var tickSizeDouble : Double = 0.01
var tickSizeDecimalNumber: NSDecimalNumber = 0.01
func decimalNumberRemainder(Dividend aDividend: NSDecimalNumber, Divisor aDivisor: NSDecimalNumber)->NSDecimalNumber{
let behaviour = NSDecimalNumberHandler(roundingMode: NSRoundingMode.RoundDown,
scale: 0,
raiseOnExactness: false ,
raiseOnOverflow: false,
raiseOnUnderflow: false,
raiseOnDivideByZero: false )
let quotient = aDividend.decimalNumberByDividingBy(aDivisor, withBehavior: behaviour)
let subtractAmount = quotient.decimalNumberByMultiplyingBy(aDivisor)
let remainder = aDividend.decimalNumberBySubtracting(subtractAmount)
return remainder
}
let doubleRemainder = Double(stringPrice)! % tickSizeDouble
let decimalRemainder = decimalNumberRemainder(Dividend: NSDecimalNumber(string: stringPrice), Divisor:tickSizeDecimalNumber)
print("Using Double: \(doubleRemainder)")
print("Using NSDecimalNumber: \(decimalRemainder)")
Hy,
I have a very Basic Question which is :
How can i create a random number with 20 digits no floats no negatives (basically an Int) in Swift ?
Thanks for all answers XD
Step 1
First of all we need an extension of Int to generate a random number in a range.
extension Int {
init(_ range: Range<Int> ) {
let delta = range.startIndex < 0 ? abs(range.startIndex) : 0
let min = UInt32(range.startIndex + delta)
let max = UInt32(range.endIndex + delta)
self.init(Int(min + arc4random_uniform(max - min)) - delta)
}
}
This can be used this way:
Int(0...9) // 4 or 1 or 1...
Int(10...99) // 90 or 33 or 11
Int(100...999) // 200 or 333 or 893
Step 2
Now we need a function that receive the number of digits requested, calculates the range of the random number and finally does invoke the new initializer of Int.
func random(digits:Int) -> Int {
let min = Int(pow(Double(10), Double(digits-1))) - 1
let max = Int(pow(Double(10), Double(digits))) - 1
return Int(min...max)
}
Test
random(1) // 8
random(2) // 12
random(3) // 829
random(4) // 2374
Swift 5: Simple Solution
func random(digits:Int) -> String {
var number = String()
for _ in 1...digits {
number += "\(Int.random(in: 1...9))"
}
return number
}
print(random(digits: 1)) //3
print(random(digits: 2)) //59
print(random(digits: 3)) //926
Note It will return value in String, if you need Int value then you can do like this
let number = Int(random(digits: 1)) ?? 0
Here is some pseudocode that should do what you want.
generateRandomNumber(20)
func generateRandomNumber(int numDigits){
var place = 1
var finalNumber = 0;
for(int i = 0; i < numDigits; i++){
place *= 10
var randomNumber = arc4random_uniform(10)
finalNumber += randomNumber * place
}
return finalNumber
}
Its pretty simple. You generate 20 random numbers, and multiply them by the respective tens, hundredths, thousands... place that they should be on. This way you will guarantee a number of the correct size, but will randomly generate the number that will be used in each place.
Update
As said in the comments you will most likely get an overflow exception with a number this long, so you'll have to be creative in how you'd like to store the number (String, ect...) but I merely wanted to show you a simple way to generate a number with a guaranteed digit length. Also, given the current code there is a small chance your leading number could be 0 so you should protect against that as well.
you can create a string number then convert the number to your required number.
func generateRandomDigits(_ digitNumber: Int) -> String {
var number = ""
for i in 0..<digitNumber {
var randomNumber = arc4random_uniform(10)
while randomNumber == 0 && i == 0 {
randomNumber = arc4random_uniform(10)
}
number += "\(randomNumber)"
}
return number
}
print(Int(generateRandomDigits(3)))
for 20 digit you can use Double instead of Int
Here is 18 decimal digits in a UInt64:
(Swift 3)
let sz: UInt32 = 1000000000
let ms: UInt64 = UInt64(arc4random_uniform(sz))
let ls: UInt64 = UInt64(arc4random_uniform(sz))
let digits: UInt64 = ms * UInt64(sz) + ls
print(String(format:"18 digits: %018llu", digits)) // Print with leading 0s.
16 decimal digits with leading digit 1..9 in a UInt64:
let sz: UInt64 = 100000000
let ld: UInt64 = UInt64(arc4random_uniform(9)+1)
let ms: UInt64 = UInt64(arc4random_uniform(UInt32(sz/10)))
let ls: UInt64 = UInt64(arc4random_uniform(UInt32(sz)))
let digits: UInt64 = ld * (sz*sz/10) + (ms * sz) + ls
print(String(format:"16 digits: %llu", digits))
Swift 3
appzyourlifz's answer updated to Swift 3
Step 1:
extension Int {
init(_ range: Range<Int> ) {
let delta = range.lowerBound < 0 ? abs(range.lowerBound) : 0
let min = UInt32(range.lowerBound + delta)
let max = UInt32(range.upperBound + delta)
self.init(Int(min + arc4random_uniform(max - min)) - delta)
}
}
Step 2:
func randomNumberWith(digits:Int) -> Int {
let min = Int(pow(Double(10), Double(digits-1))) - 1
let max = Int(pow(Double(10), Double(digits))) - 1
return Int(Range(uncheckedBounds: (min, max)))
}
Usage:
randomNumberWith(digits:4) // 2271
randomNumberWith(digits:8) // 65273410
Swift 4 version of Unome's validate response plus :
Guard it against overflow and 0 digit number
Adding support for Linux's device because "arc4random*" functions don't exit
With linux device don't forgot to do
#if os(Linux)
srandom(UInt32(time(nil)))
#endif
only once before calling random.
/// This function generate a random number of type Int with the given digits number
///
/// - Parameter digit: the number of digit
/// - Returns: the ramdom generate number or nil if wrong parameter
func randomNumber(with digit: Int) -> Int? {
guard 0 < digit, digit < 20 else { // 0 digit number don't exist and 20 digit Int are to big
return nil
}
/// The final ramdom generate Int
var finalNumber : Int = 0;
for i in 1...digit {
/// The new generated number which will be add to the final number
var randomOperator : Int = 0
repeat {
#if os(Linux)
randomOperator = Int(random() % 9) * Int(powf(10, Float(i - 1)))
#else
randomOperator = Int(arc4random_uniform(9)) * Int(powf(10, Float(i - 1)))
#endif
} while Double(randomOperator + finalNumber) > Double(Int.max) // Verification to be sure to don't overflow Int max size
finalNumber += randomOperator
}
return finalNumber
}
I have a function that produces a random number output (see below). I'd like for it to choose that output from either 0-50 or 150-400, but not in between. I couldn't find much about it on my own... so if any of you could post either resources or an answer that would be awesome! Thanks so much in advance!
The three question marks are where I assume some operator would go.
func random() -> UInt32 {
var range = UInt32(0)...UInt32(50) ??? UInt32(150)...UInt32(400)
return range.startIndex + arc4random_uniform(range.endIndex - range.startIndex + 1)
}
I tried using the + operator and the | operator, but no luck.
You cannot pass a union of ranges to arc4random_uniform().
What you can do is to create a random number in a single interval and then "adjust" all numbers which are not in the first range to the second range:
func random() -> UInt32 {
var x = arc4random_uniform(302) // 0 .. 301
if (x > 50) {
// map 51 .. 301 to 150 .. 400:
x += 99
}
return x
}
This can be generalized to multiple ranges (now updated for Swift 3):
func randomNumber(fromRanges ranges: Range<UInt32>...) -> UInt32 {
let totalLength = ranges.reduce(0) { $0 + ($1.upperBound - $1.lowerBound)}
var rnd = arc4random_uniform(totalLength)
for range in ranges {
if rnd < range.upperBound - range.lowerBound {
return rnd + range.lowerBound
}
rnd -= range.upperBound - range.lowerBound
}
fatalError("this should never be reached")
}
let x = randomNumber(fromRanges: 0 ..< 51, 150 ..< 401)
An alternative to ranges is to just use a while loop to make sure you get a value in the range you want.
func randomFunc() -> UInt32 {
var rand: UInt32 = 401
while rand < 0 || rand > 400 || (rand > 50 && rand < 150){
rand = arc4random_uniform(400)
}
return rand;
}
I'm trying to generate random values between two integers. I've tried this, which starts from 0,
let randomNumber = arc4random_uniform(10)
println(randomNumber)
But I need a value between 10 and 50.
try this
let randomNumber = arc4random_uniform(40) + 10
println(randomNumber)
in general form
let lower : UInt32 = 10
let upper : UInt32 = 50
let randomNumber = arc4random_uniform(upper - lower) + lower
println(randomNumber)
This is an option for Swift 4.2 and above using the random() method, which makes it easy!
let randomInt = Int.random(in: 10...50)
The range can be a closed (a...b) or half open (a..<b) range.
If you want a reusable function with simple parameters:
func generateRandomNumber(min: Int, max: Int) -> Int {
let randomNum = Int(arc4random_uniform(UInt32(max) - UInt32(min)) + UInt32(min))
return randomNum
}
more simple way of random number generator
func random(min: Int, max: Int) -> Int {
return Int(arc4random_uniform(UInt32(max - min + 1))) + min
}