The following works. It prints: At least one true
let arr = [false, false, false, true, false, false, true, false]
// Variable set by executing a closure...
let hasAtLeastOneTrue: Bool = {
var x = false
arr.forEach {
x = x || $0
}
return x
}()
if hasAtLeastOneTrue {
print("At least one true")
} else {
print("All false")
}
Is it possible to compress the code by eliminating the constant hasAtLeastOneTrue and replace the conditional with the closure? Something like:
if {CLOSURE CODE}() {
print("At least one true")
} else {
print("All false")
}
Yes. The key is to wrap the entire closure invocation with brackets:
if ({
var x = false
arr.forEach {
x = x || $0
}
return x
}()) {
print("At least one true")
} else {
print("All false")
}
Though I think this is a lot less readable than if hasAtLeastOneTrue { .... I suggest that you still name your closures, or just use inner functions.
Note that in this case, you can just do:
if arr.contains(true) {
print("At least one true")
} else {
print("All false")
}
How to make the closures shorter？ I want to know the simple programming of the closures.
let closures = { (fillBefore: Bool, fillAfter: Bool) -> String in
if fillBefore && fillAfter {
return kCAFillModeBoth
} else if !fillBefore && fillAfter {
return kCAFillModeBackwards
} else if fillBefore && !fillAfter {
return kCAFillModeForwards
} else {
return kCAFillModeRemoved
}
}
anim?.fillMode = closures((item?.fillBefore)!, (item?.fillAfter)!)
How to make the closures shorter？
Based on your case, I think that at some point you have to evaluate both of booleans, so I would assume that there is no "shorter" code for handling it.
However, you might be looking for a "neater" approach, so I would suggest to evaluate them as pair of booleans (tuple), with a switch statement:
let closure = { (fillBefore: Bool, fillAfter: Bool) -> String in
switch (fillBefore, fillAfter) {
case (true, true):
return kCAFillModeBoth
case (false, true):
return kCAFillModeBackwards
case (true, false):
return kCAFillModeForwards
default: // on your case, it would be the same as (false, false)
return kCAFillModeRemoved
}
}
let myClosure = closure(false,false)
myClosure // removed
This question already has an answer here:
Why is Equatable not defined for optional arrays
(1 answer)
Closed 6 years ago.
I'd like a function to compare [String]? with [String]? where either, both or neither are nil. If both are nil then the function returns true
Is there notation where I can avoid checking a value is equal to nil?
This:
func compare_colours(a1:[String]?, a2:[String]?) -> Bool {
return a1 == a2
}
isn't valid in Swift because I'd need to add ! first:
return a1! == a2!
if let a1Unwrapped = a1{
if let a2Unwrapped = a2{
return a1Unwrapped == a2Unwrapped
}
}
if (a1 == nil && a2 == nil){
return true
}
return false
I prefer using guard over if statements so I can avoid the pyramid of doom:
func compare_colours(a1:[String]?, a2:[String]?) -> Bool {
if a1 == nil && a2 == nil { return true }
guard let unwrapped1 = a1 else { return false }
guard let unwrapped2 = a2 else { return false }
return unwrapped1 == unwrapped2
}
You don't need to check for nil. Instead you can check to see if the Optional enum is .None:
switch (a1, a2) {
case (.None, .None): print("Both unset Optional")
}
Either way you have to do the same basic thing, check for nil/.None. For more details on how this works you can read this answer: https://stackoverflow.com/a/33209762/887210
Or just use the optional as an empty and you can match all scenarios.
func compare_colours(a1:[String]?, a2:[String]?) -> Bool {
return a1 ?? [""] == a2 ?? [""]
}
In the code below, I'm confused as to where the return statement in the code returns to? When executed, it works as expected, but does it return to:
if userIsInTheMiddleOfTyping == true
or does it return to:
if let digit = sender.currentTitle
Below is the full chunk of code where this applies.
class ViewController: UIViewController {
private var userIsInTheMiddleOfTyping = false
private var decimalUsed = false
#IBAction private func touchDigit(sender: UIButton)
{
if let digit = sender.currentTitle {
if userIsInTheMiddleOfTyping == true {
if digit == "." && decimalUsed == true {
return //where does this return to?
} else if digit == "." && decimalUsed == false {
decimalUsed = true
}
let textCurrentlyInDisplay = display.text!
display.text = textCurrentlyInDisplay + digit
} else {
display.text = digit
}
userIsInTheMiddleOfTyping = true
}
}
A return always returns out of the function, so in this case it returns to the line of code that calls touchDigit(...)
Basically here, the return just stops the execution of the touchDigit function.
(Which means that none of the code following the return will be run)
The return simply stops the code. You can put it in functions if you would like. For example:
If I want to continue running some code only if a certain statement is true, then you can return the function to stop it if it is false.
func something(a: Int, b: Int) {
if a != b {
return//Stops the code
}
//Some more code -- if a is not equal to b, this will not be called
}
Remember, this only works with void functions. It can work with others as well, but that is slightly different. You must return something along with it. Another example:
func somethingElse(a: Int, b: Int) -> Bool{
if a != b {
return false //stops the code, but also returns a value
}
return true //Will only get called if a == b
}
In this function, it is return a Boolean. If a != b, then return false is written because that returns false while also stoping the code.
For more on returns, you can visit Apple's documentation on functions.
I like many of the features in Swift, but using manipulating strings are still a big pain in the ass.
func checkPalindrome(word: String) -> Bool {
print(word)
if word == "" {
return true
} else {
if word.characters.first == word.characters.last {
return checkPalindrome(word.substringWithRange(word.startIndex.successor() ..< word.endIndex.predecessor()))
} else {
return false
}
}
}
This code fails miserably whenever the string's length is an odd number. Of course I could make it so the first line of the block would be if word.characters.count < 2, but is there a way in Swift to get substrings and check easily?
Update
I like many of the suggestions, but I guess the original question could be misleading a little, since it's a question about String more than getting the right results for the function.
For instance, in Python, checkPalindrome(word[1:-1]) would work fine for the recursive definition, whereas Swift code is much less graceful since it needs other bells and whistles.
return word == String(word.reversed())
func isPalindrome(myString:String) -> Bool {
let reverseString = String(myString.characters.reversed())
if(myString != "" && myString == reverseString) {
return true
} else {
return false
}
}
print(isPalindrome("madam"))
I have used the below extension to find whether the number is Palindrome or Not.
extension String {
var isPalindrome: Bool {
return self == String(self.reversed())
}
}
Sometimes having a front end for a recursion can simplify life. I sometimes do this when the arguments which are most convenient to use are not what I want in the user interface.
Would the following meet your needs?
func checkPalindrome(str: String) -> Bool {
func recursiveTest(var charSet: String.CharacterView) -> Bool {
if charSet.count < 2 {
return true
} else {
if charSet.popFirst() != charSet.popLast() {
return false
} else {
return recursiveTest(charSet)
}
}
}
return recursiveTest(str.characters)
}
just add on more condition in if
func checkPalindrome(word: String) -> Bool {
print(word)
if (word == "" || word.characters.count == 1){
return true
}
else {
if word.characters.first == word.characters.last {
return checkPalindrome(word.substringWithRange(word.startIndex.successor() ..< word.endIndex.predecessor()))
} else {
return false
}
}
}
extension StringProtocol where Self: RangeReplaceableCollection {
var letters: Self { filter(\.isLetter) }
var isPalindrome: Bool {
let letters = self.letters
return String(letters.reversed()).caseInsensitiveCompare(letters) == .orderedSame
}
}
"Dammit I'm Mad".isPalindrome // true
"Socorram-me subi no onibus em marrocos".isPalindrome // true
You can also break your string into an array of characters and iterate through them until its half comparing one by one with its counterpart:
func checkPalindrome(_ word: String) -> Bool {
let chars = Array(word.letters.lowercased())
for index in 0..<chars.count/2 {
if chars[index] != chars[chars.count - 1 - index] {
return false
}
}
return true
}
And the recursive version fixing the range issue where can't form a range with endIndex < startIndex:
func checkPalindrome<T: StringProtocol>(_ word: T) -> Bool {
let word = word.lowercased()
.components(separatedBy: .punctuationCharacters).joined()
.components(separatedBy: .whitespacesAndNewlines).joined()
if word == "" || word.count == 1 {
return true
} else {
if word.first == word.last {
let start = word.index(word.startIndex,offsetBy: 1, limitedBy: word.endIndex) ?? word.startIndex
let end = word.index(word.endIndex,offsetBy: -1, limitedBy: word.startIndex) ?? word.endIndex
return checkPalindrome(word[start..<end])
} else {
return false
}
}
}
checkPalindrome("Dammit I'm Mad")
I think if you make an extension to String like this one then it will make your life easier:
extension String {
var length: Int { return characters.count }
subscript(index: Int) -> Character {
return self[startIndex.advancedBy(index)]
}
subscript(range: Range<Int>) -> String {
return self[Range<Index>(start: startIndex.advancedBy(range.startIndex), end: startIndex.advancedBy(range.endIndex))]
}
}
With it in place, you can change your function to this:
func checkPalindrome(word: String) -> Bool {
if word.length < 2 {
return true
}
if word.characters.first != word.characters.last {
return false
}
return checkPalindrome(word[1..<word.length - 1])
}
Quick test:
print(checkPalindrome("aba")) // Prints "true"
print(checkPalindrome("abc")) // Prints "false"
extension String {
func trimmingFirstAndLastCharacters() -> String {
guard let startIndex = index(self.startIndex, offsetBy: 1, limitedBy: self.endIndex) else {
return self
}
guard let endIndex = index(self.endIndex, offsetBy: -1, limitedBy: self.startIndex) else {
return self
}
guard endIndex >= startIndex else {
return self
}
return String(self[startIndex..<endIndex])
}
var isPalindrome: Bool {
guard count > 1 else {
return true
}
return first == last && trimmingFirstAndLastCharacters().isPalindrome
}
}
We first declare a function that removes first and last characters from a string.
Next we declare a computer property which will contain the actual recursive code that checks if a string is palindrome.
If string's size is less than or equal 1 we immediately return true (strings composed by one character like "a" or the empty string "" are considered palindrome), otherwise we check if first and last characters of the string are the same and we recursively call isPalindrome on the current string deprived of the first and last characters.
Convert the string into an Array. When the loop is executed get the first index and compare with the last index.
func palindrome(string: String)-> Bool{
let char = Array(string)
for i in 0..<char.count / 2 {
if char[i] != char[char.count - 1 - i] {
return false
}
}
return true
}
This solution is not recursive, but it is a O(n) pure index based solution without filtering anything and without creating new objects. Non-letter characters are ignored as well.
It uses two indexes and walks outside in from both sides.
I admit that the extension type and property name is stolen from Leo, I apologize. 😉
extension StringProtocol where Self: RangeReplaceableCollection {
var isPalindrome : Bool {
if isEmpty { return false }
if index(after: startIndex) == endIndex { return true }
var forward = startIndex
var backward = endIndex
while forward < backward {
repeat { formIndex(before: &backward) } while !self[backward].isLetter
if self[forward].lowercased() != self[backward].lowercased() { return false }
repeat { formIndex(after: &forward) } while !self[forward].isLetter
}
return true
}
}
Wasn't really thinking of this, but I think I came up with a pretty cool extension, and thought I'd share.
extension String {
var subString: (Int?) -> (Int?) -> String {
return { (start) in
{ (end) in
let startIndex = start ?? 0 < 0 ? self.endIndex.advancedBy(start!) : self.startIndex.advancedBy(start ?? 0)
let endIndex = end ?? self.characters.count < 0 ? self.endIndex.advancedBy(end!) : self.startIndex.advancedBy(end ?? self.characters.count)
return startIndex > endIndex ? "" : self.substringWithRange(startIndex ..< endIndex)
}
}
}
}
let test = ["Eye", "Pop", "Noon", "Level", "Radar", "Kayak", "Rotator", "Redivider", "Detartrated", "Tattarrattat", "Aibohphobia", "Eve", "Bob", "Otto", "Anna", "Hannah", "Evil olive", "Mirror rim", "Stack cats", "Doom mood", "Rise to vote sir", "Step on no pets", "Never odd or even", "A nut for a jar of tuna", "No lemon, no melon", "Some men interpret nine memos", "Gateman sees name, garageman sees nametag"]
func checkPalindrome(word: String) -> Bool {
if word.isEmpty { return true }
else {
if word.subString(nil)(1) == word.subString(-1)(nil) {
return checkPalindrome(word.subString(1)(-1))
} else {
return false
}
}
}
for item in test.map({ $0.lowercaseString.stringByReplacingOccurrencesOfString(",", withString: "").stringByReplacingOccurrencesOfString(" ", withString: "") }) {
if !checkPalindrome(item) {
print(item)
}
}
A simple solution in Swift:
func isPalindrome(word: String) -> Bool {
// If no string found, return false
if word.count == 0 { return false }
var index = 0
var characters = Array(word) // make array of characters
while index < characters.count / 2 { // repeat loop only for half length of given string
if characters[index] != characters[(characters.count - 1) - index] {
return false
}
index += 1
}
return true
}
func checkPalindrome(_ inputString: String) -> Bool {
if inputString.count % 2 == 0 {
return false
} else if inputString.count == 1 {
return true
} else {
var stringCount = inputString.count
while stringCount != 1 {
if inputString.first == inputString.last {
stringCount -= 2
} else {
continue
}
}
if stringCount == 1 {
return true
} else {
return false
}
}
}