This code passes 3 as an argument to the arc4random_uniform() function and I guess that it returns true or false. Next you assign an enumerator to a variable. I don't understand what the function does, though.
let randomState = arc4random_uniform(3) == 2 ? CellState.Alive :
CellState.Empty
let cell = Cell(grid: self,
pos: (i, j),
state: randomState)
I am not following that logic.
Is it the "arc4random_uniform(3) == 2 ? CellState.Alive : CellState.Empty" format that you don't understand? It's written as a conditional (ternary) operator:
condition ? expr1 : expr2
Basically, if func arc4random_uniform(3) == 2, let randomState == CellState.Alive, otherwise let randomState == CellState.Empty.
Related
i have two values in bytes in two different variables . i want to perform a certain action whenever values are nearly equal to each other.
I there any method in swift in which i can perform any action on variables values nearly equal to.
If recommend me some code , tutorial or article to achieve this.
I am new to swift so please avoid down voting.
let string1 = "Hello World"
let string2 = "Hello"
let byteArrayOfString1: [UInt8] = string1.utf8.map{UInt8($0)} //Converting HELLO WORLD into Byte Type Array
let byteArrayOfString2: [UInt8] = string2.utf8.map{UInt8($0)} //Converting HELLO into Byte Type Array
if byteArrayOfString1 == byteArrayOfString2 {
print("Match")
}else {
print("Not Match")
}
For more Help, Visit https://medium.com/#gorjanshukov/working-with-bytes-in-ios-swift-4-de316a389a0c
well exactly i don't think so there is such method that compare approx values but if you discuss what exactly you want to do we can find a better alternative solution.
Here is the Solution:
func nearlyEqual(a: Float, b: Float, epsilon: Float) -> Bool {
let absA = abs(a)
let absB = abs(b)
let diff = abs(a - b)
if a == b {
return true
} else if (a == 0 || b == 0 || absA + absB < Float.leastNonzeroMagnitude) {
// a or b is zero or both are extremely close to it
// relative error is less meaningful here
return diff < (epsilon * Float.leastNonzeroMagnitude)
} else {
return diff / (absA + absB) < epsilon
}
}
Then you can use it like :
print(nearlyEqual(a: 1.2, b: 1.4, epsilon: 0.2))
This will return true.
I find out a lot of example to solve it, but nothing in SWIFT. Please help
smthng like this
Input : n = 4
Output : Yes
2^2 = 4
Input : n = 7
Output : No
Input : n = 32
Output : Yes
2^5 = 32
I needed algorithm for checking if a number is a power of 2. like 4, 8, 16 , 32 , 64 .... is number power of two
Determining if an integer is a power of 2
from the Bit Twiddling Hacks
is almost verbatim translated to Swift:
func isPowerOfTwo(_ n: Int) -> Bool {
return (n > 0) && (n & (n - 1) == 0)
}
Example:
print(isPowerOfTwo(4)) // true
print(isPowerOfTwo(5)) // false
Or as a generic function, so that it can be used with all binary
integer types:
func isPowerOfTwo<T: BinaryInteger> (_ n: T) -> Bool {
return (n > 0) && (n & (n - 1) == 0)
}
Example:
print(isPowerOfTwo(Int16(4))) // true
print(isPowerOfTwo(UInt8(5))) // false
Or as a protocol extension:
extension BinaryInteger {
var isPowerOfTwo: Bool {
return (self > 0) && (self & (self - 1) == 0)
}
}
Example:
print(1048576.isPowerOfTwo) // true
print(Int(50).isPowerOfTwo) // false
Partial answer:
If it's a FixedWidthInteger and it's positive and its non zero bit count is 1, then it is a power of 2.
let x = 128
if x > 0 && x.nonzeroBitCount == 1
{
// power of 2
}
For a floating point number, I think you can just test the significand. If it is exactly 1, the number is a power of 2.
let x: Double = 4
if x > 0 && x.significand == 1
{
// Power of 2
}
I haven't checked that in a Playground yet, so it might be wrong.
let numberToBeChecked = 4
result = numberToBeChecked.squareRoot()
If result%1 == 0 {
print(“4 is a power of 2”) } else {
print(“4 is not a power of 2”)
}
//note: result%1== 0 checks if result is a whole number.
Hope this works.
Per Codefighters:
Note: Write a solution with O(n) time complexity and O(1) additional space complexity, since this is what you would be asked to do during a real interview.
Given an array a that contains only numbers in the range from 1 to a.length, find the first duplicate number for which the second occurrence has the minimal index. In other words, if there are more than 1 duplicated numbers, return the number for which the second occurrence has a smaller index than the second occurrence of the other number does. If there are no such elements, return -1.
Example
For a = [2, 3, 3, 1, 5, 2], the output should be firstDuplicate(a) = 3.
There are 2 duplicates: numbers 2 and 3. The second occurrence of 3 has a smaller index than than second occurrence of 2 does, so the answer is 3.
For a = [2, 4, 3, 5, 1], the output should be firstDuplicate(a) = -1.
So here is what I came up with. It works but fails on the final test because it ran over 4000ms. I am stuck to what else I can do. Any Ideas to improve speed?
func firstDuplicate(a : [Int]) -> Int {
var duplicateIndexArray = [Int]()
for firstIndex in 0..<a.count {
for secondIndex in 0..<a.count {
if a[firstIndex] == a[secondIndex] && firstIndex != secondIndex {
print(firstIndex, secondIndex)
if !(duplicateIndexArray.contains(firstIndex)){
duplicateIndexArray.append(secondIndex)
break
}
}
}
}
// Check for duplicacy
if duplicateIndexArray.count > 0 {
print(duplicateIndexArray)
return a[duplicateIndexArray.min()!]
}
return -1
}
The O(n) time part is easy, but the O(1) additional space is a bit tricky. Usually, a hash set (or bit array in your case) can be used to check if a number occurred more than once, but that requires O(n) additional space. For O(1) additional space, we can use the source array itself as a bit array by making some of the numbers in it negative.
For example if the first number in the array is 3, then we make the number at position 3-1 negative. If one of the other numbers in the array is also 3, we can check if the number at position 3-1 is negative.
I don't have any experience with Swift, so I'll try to write a function in pseudocode:
function firstDuplicate(a)
result = -1
for i = 0 to a.count - 1
if a[abs(a[i])-1] < 0 then
result = a[i]
exit for loop
else
a[abs(a[i])-1] = -a[abs(a[i])-1]
// optional restore the negative numbers back to positive
for i = 0 to a.count - 1
if a[i] < 0 then
a[i] = -a[i]
return result
Replace this line
for secondIndex in 0..<a.count
with
for secondIndex in firstIndex..<a.count
There is no requirement of double checking
So Your Final code is
func firstDuplicate(a : [Int]) -> Int {
var duplicateIndexArray = [Int]()
for firstIndex in 0..<a.count {
for secondIndex in firstIndex..<a.count {
if a[firstIndex] == a[secondIndex] && firstIndex != secondIndex {
print(firstIndex, secondIndex)
if !(duplicateIndexArray.contains(firstIndex))
{
duplicateIndexArray.append(secondIndex)
break
}
}
}
}
// Check for duplicacy
if duplicateIndexArray.count > 0
{
print(duplicateIndexArray)
return a[duplicateIndexArray.min()!]
}
return -1
}
func firstDuplicate(input: [Int]) -> Int{
var map : [String : Int] = [:]
var result = -1
for i in 0 ..< input.count {
if map["\(input[i])"] != nil {
result = i
break
}
else {
map["\(input[i])"] = i
}
}
return result
}
I'd like to append an s to the end of a string, if the value of a variable is > 1.
There are obviously several ways to accomplish this - for example:
if(points == 1) {
points_as_string = "1 Point"
} else {
points_as_string = "\(points) Points"
}
A shorter version could be:
points_as_string = (points == 1 ? "1 Point" : "\(points) Points")
An even shorter version would be:
points_as_string = "\(points) Point" + (points == 1 ? "" : "s")
Is it possible to write something even shorter than that, or is that as good as it gets?
Only very slightly shorter:
points_as_string = "\(points) Point\(points == 1 ? "" : "s")"
Here is a shorter version that uses a dictionary lookup and the nil coalescing operator ??:
points_as_string = "\(points) Point\([1:""][points] ?? "s")"
Does anyone know if there is a way to use some kind shorthand in swift? more specifically, leaving out the braces in things like IF statements... eg
if num == 0
// Do something
instead of
if num == 0
{
// Do something
}
Those braces become rather space consuming when you have a few nested IF's.
PS. I do know I can do the following:
if num == 0 {
// Do something }
But I'm still curious if that sort of thing is possible
You can do that :
let x = 10, y = 20;
let max = (x < y) ? y : x ; // So max = 20
And so much interesting things :
let max = (x < y) ? "y is greater than x" : "x is greater than y" // max = "y is greater than x"
let max = (x < y) ? true : false // max = true
let max = (x > y) ? func() : anotherFunc() // max = anotherFunc()
(x < y) ? func() : anotherFunc() // code is running func()
This following stack : http://codereview.stackexchange.com can be better for your question ;)
Edit : ternary operators and compilation
By doing nothing more than replacing the ternary operator with an if else statement, the build time was reduced by 92.9%.
https://medium.com/#RobertGummesson/regarding-swift-build-time-optimizations-fc92cdd91e31#.42uncapwc
In swift you have to add braces even if there is just one statement in if:
if num == 0 {
// Do something
}
You cannot leave the braces, that how swift if statement work.
You could use a shorthand if statement like you would in objective-c:
num1 < num2 ? DO SOMETHING IF TRUE : DO SOMETHING IF FALSE
Swift 2.0 update
Method 1:
a != nil ? a! : b
Method 2: Shorthand if
b = a ?? ""
Referance: Apple Docs: Ternary Conditional Operator
and it does work,
u.dob = (userInfo["dob"] as? String) != nil ? (userInfo["dob"] as! String):""
I am replacing a json string with blank string if it is nil.
Edit: Adding Gerardo Medina`s suggestion...we can always use shorthand If
u.dob = userInfo["dob"] as? String ?? ""
It is called shorthand if-else condition. If you are into iOS development in Swift, then you can also manipulate your UI objects' behaviour with this property.
For e.g. - I want my button to be enabled only when there is some text in the textfield. In other words, should stay disabled when character count in textfield is zero.
button.enabled = (textField.characters.count > 0) ? true : false
its very simple :
in Swift 4
playButton.currentTitle == "Play" ? startPlay() : stopPlay()
Original Code is
if playButton.currentTitle == "Play"{
StartPlay()
}else{
StopPlay()
}
You could always put the entire if on one line:
if num == 0 { temp = 0 }