I'm trying to change a tuple in an array , however ,when I try
emo = (type:emo.type,strength:increaseStrength(emo.strength))
it gives me error
"cannot assign to 'let' value 'emo'
here is my code :
var emotions : [(type : String, strength: Int)] = [("happy",0),("scared",0),("tender",0),("excited",0),("sad",0)]
func increaseStrength(i:Int)->Int {
switch i {
case 0: return 1
case 1: return 2
case 2: return 3
case 3: return 0
default :return 0
}
}
#IBAction func HappyBA(sender: AnyObject) {
for emo in emotions {
if (emo.type == "happy" ){
emo = (type:emo.type,strength:increaseStrength(emo.strength))
}
}
println(emotions)
}
If there are better way to do the assignment please tell me I am so appreciated ! Thanks..
There is no point assigning to emo even if you could do it. This is not the same as replacing the corresponding object in the array - which is what you do want to do. emo is a copy; even if you were to set a property of it, it wouldn't affect the one back in the array. And certainly setting the variable would not magically read back into the array!
Here's one solution. Instead of cycling thru emotions in your for-loop, cycle thru enumerate(emotions). Now you have a tuple of an index number along with an emotion. If this is the right emotion type, write into the array via the index number.
for (ix,emo) in enumerate(emotions) {
if emo.type == "happy" {
emotions[ix] = (type:emo.type,strength:increaseStrength(emo.strength))
}
}
Or you could use map.
emotions = emotions.map {
emo in
if emo.type == "happy" {
return (type:emo.type,strength:increaseStrength(emo.strength))
} else {
return emo
}
}
Related
How do i get the result of of i to return from the function here? I can return it from inside the if { } but then I get an error there’s no global return (for the function). My goal is to write a function that accepts an Integer and returns the square root. My next step is to throw an error if it's not an Int or between 1 and 10,000, but please assume simple numbers for this question.
let userInputInteger = 9
func squareRootCheckpoint4(userInputInteger: Int) -> Int {
for i in 1...100 {
let userInputInteger = i * i
if i == userInputInteger {
break
}
}
return i
}
update:
To be able to return it, you will have to declared a variable outside the for-loop, then based on the logic assign the "i" counter to that outside variable.
func squareRootCheckpoint4(userInputInteger: Int) -> Int {
var result = 0 // this is to keep the result
for i in 1...100 {
let powerOfTwoResult = i * i
if powerOfTwoResult == userInputInteger {
result = i // assign result variable based on the logic
break
}
}
return result // return result
}
The error shown is because if the for-loop finishes without any hit on the if statement, the function will not be able to return any Int. That's why the compiler produces that "no global return" error.
One way to do it if you don't want to return a default value is to throw an error. You could throw an error from a function (https://docs.swift.org/swift-book/LanguageGuide/ErrorHandling.html). For example for your goal of returning square root of an integer and throw error if the input integer is not between 1 and 10000 (if I understand correctly), you could do something like this
enum InputError: Error {
case invalidInteger
}
func squareRootCheckpoint4(userInputInteger: Int) throws -> Int {
if userInputInteger < 1 || userInputInteger > 10000 {
throw InputError.invalidInteger
}
// calculate square root
return resultOfSquareroot
}
// and to handle the error, you could encapsulate the squareRootCheckpoint4 function in a do-catch statement
do {
let squareRootResult = try squareRootCheckpoint4(userInputInteger: 4)
} catch InputError.invalidInteger {
// handle your error here
}
Alternatively, you could do a validation on the input integer separately before calling the squareRootCheckpoint4 function. For example
func validate(_ input: Int) -> Bool {
if userInputInteger < 1 || userInputInteger > 10000 {
return false
}
return true
}
func squareRootCheckpoint4(userInputInteger: Int) -> Int {
// calculate square root
return resultOfSquareroot
}
var input: Int = 9
if validate(input) {
let squareRootResult = squareRootCheckpoint4(input)
} else {
// handle when input is invalid
}
A couple of things are swapped around, and something (anything) needs to be returned from the function. The framing of the question only allows solutions that don't work if the input is not an integer with a square root.
List of Specifics:
The function argument name userInputInteger should not have been used to instantiate a new object. That's confusing. I did it because I'm learning and thought it was necessary.
Instead, i times i (or i squared) should have been assigned to a new object.
The if statement should be equal to this new object, instead of i.
Functions should return something. When i is returned in the if statement, it's not available anywhere outside the for loop. The print statements in the code example help explain what is available in these different scopes, as well as what's not.
The for loop stops when i is returned. That happens when the guess (i) is equal to the input (userInputInteger).
The for loop will continue through 100 if the input doesn't have a square root.
The function should return the correct number that the if statement was trying to guess, but you have to tell it to (with a return statement) and do it in the right spot. This is the "global return" error. However, because this approach in the question is setup poorly without error handling, only an Integer with a sqare root will return correctly. And in this case, the function's global return doesn't matter; Swift just requires something be returned, regardless of whether or not it's used.
Code Example - Direct Answer:
import Cocoa
let userInputInteger = 25
let doesNotMatter = 0
print("sqrt(\(userInputInteger)) is \(sqrt(25))") //correct answer for reference
func squareRootCheckpoint4(userInputInteger2: Int) -> Int {
var j: Int
for i in 1...100 {
print("Starting Loop \(i)")
j = i * i
if j == userInputInteger2 {
print("i if-loop \(i)")
print("j if-loop \(j)")
return i
}
print("i for-loop \(i)")
print("j for-loop \(j)")
}
//print("i func-end \(i)") //not in scope
//print("j func-end \(j)") //not in scope
//return i //not in scope
//return j //not in scope
print("Nothing prints here")
// but something must be returned here
return doesNotMatter
}
print("Input was \(userInputInteger)")
print("The Square Root of \(userInputInteger) is \(squareRootCheckpoint4(userInputInteger2: userInputInteger))")
Code Example - Improved with Error Handling
import Cocoa
enum ErrorMsg : Error {
case outOfBoundsError, noRootError
}
func checkSquareRoot (Input userInputInteger2: Int) throws -> Int {
if userInputInteger < 1 || userInputInteger > 10_000 {
throw ErrorMsg.outOfBoundsError
}
var j : Int
for i in 1...100 {
print("Starting Loop \(i)")
j = i * i
if j == userInputInteger2 {
print("i if-loop \(i)")
print("j if-loop \(j)")
return i
}
print("i for-loop \(i)")
print("j for-loop \(j)")
}
throw ErrorMsg.noRootError
}
let userInputInteger = 25
print("Input was \(userInputInteger)")
do {
let result = try checkSquareRoot(Input: userInputInteger)
print("The Square Root of \(userInputInteger) is \(result)")
} catch ErrorMsg.outOfBoundsError {
print("out of bounds: the userInputInteger is less than 1 or greater than 10,000")
} catch ErrorMsg.noRootError {
print("no root")
}
print("Swift built-in function for reference: sqrt(\(userInputInteger)) is \(sqrt(25))") //correct answer for reference
while true {
print ( """
1. Log in
2. Create new user
3. Quit
""")
if let numberString = readLine(), let number = Int(numberString) {
print("you entered \(number)")
break
} else {
print("Try again")
}
}
It is necessary to use the number variable in the future to compare it.
How can it be taken out of sight?
When I try to make a global var something goes wrong:
var numberString: String?
var number: Int?
while true {
print ( """
1. Log in
2. Create new user
3. Quit
""")
if numberString = readLine(), number = Int(numberString) {
print("you entered \(number)")
break
} else {
print("Try again")
}
}
Considering that you are creating a command prompt and that choice has no meaning outside your loop there is no need to make it global. You only need to switch the user selection and decide what to do from there. Note that if you try to break from inside the switch you won't exit your loop as I showed to you in your last question. To allow the compiler to know that you want to break the while loop instead of the switch you need to label your while loop statement, this way you can specify what you want to break when you have a switch inside a loop. Try like this:
func getValue() -> Int? {
guard let line = readLine(), let value = Int(line) else {
return nil
}
return value
}
question: while true {
print("""
1. Log in
2. Create new user
3. Quit
""")
guard let value = getValue() else {
continue
}
switch value {
case 1:
print("you have selected number one")
case 2:
print("you have selected number two")
case 3:
print("Good bye")
break question
default:
print("Try again")
}
}
There is not much information to go off of here, but I think I can help. In the first script, you are creating a new variable and setting the value for that variable:
if let numberString = readLine(), let number = Int(numberString) { ... }
In your second script, you have to global variables, but at no point have you provided them a value. What you need to do is provide both global variables (numberString and number) a value before comparing them. As such, you need to use the == sign to compare different variables/ types. For instance:
var numberString: String?
var number: Int?
while true {
print("""
1. Log in
2. Create new user
3. Quit
""")
numberString = readLine()
number = Int(numberString!)
}
You can now compare and or print your global variables...
Final Code
var numberString: String?
var number: Int?
while true {
print("""
1. Log in
2. Create new user
3. Quit
""")
numberString = readLine()
number = Int(numberString!)
if number == 1 {
print("you have selected number one")
} else if number == 2 {
print("you have selected number two")
} else if number == 3 {
print("you have selected number three")
} else {
print("Try again")
}
}
Due to the fact that both variables numberString and number are now global variables, you can access them in the future if you need to compare them again.
Does anybody know of a shortcut to test whether three numbers are the same? I know this works:
if number1 == number2 && number2 == number3 {
}
But I would like something cleaner, such as;
if number1 == number2 == number3 {
}
It's quite important as I'm comparing a lot of different values.
You could use a set
if Set([number1, number2, number3]).count == 1 {
...
though I'd argue it isn't as transparent as multiple if clauses
You can use the power of tuples and the Transitive Property of Equality.
if (number1, number2) == (number2, number3) {
}
The clause of this IF is true only when number1 is equals to number2 AND number2 is equals to number3. It means that the 3 values must be equals.
You can add them in an array and use sets:
var someSet = NSSet(array: [2,2,2])
if someSet.count == 1 {
print("Same values")
}
Don't know of anything other than a Set, I'd suggest wrapping it in a function to make your intent clear. Something along these lines:
func allItemsEqual<T>(items:[T]) -> Bool {
guard items.count > 1 else { fatalError("Must have at least two objects to check for equality") }
return Set(items).count == 1
}
func allItemsEqual(items:T...) -> Bool {
return equal(items)
}
if allItemsEqual(2,3,2) {
// false
}
if allItemsEqual(2, 2, 2) {
// true
}
Beyond that, maybe you could get fancy with operator overloading?
Try this:
func areEqual<T: NumericType>(numbers: T...) -> Bool {
let num = numbers[0]
for number in numbers {
if number != num {
return false
}
}
return true
}
Where NumericType is defined in this post: What protocol should be adopted by a Type for a generic function to take any number type as an argument in Swift?
This will allow you to use the function for all number types
You just pass any number of numbers like:
//returns true
if areEqual(1, 1, 1) {
print("equal")
}
I have an if statement that checks to see if an array element matches a local variable.
if pinArray.contains(where: {$0.title == restaurantName})
How would I create a variable of this element?
I attempted
let thePin = pinArray.contains(where: {$0.title == restaurantName})
but this comes with "could not cast boolean to MKAnnotation".
I also tried variations of
let pins = [pinArray.indexPath.row]
let pinn = pins(where: pin.title == restaurantName) (or close to it)
mapp.selectAnnotation(thePin as! MKAnnotation, animated: true)
to no avail. What basic step am I missing?
contains(where:) returns a Bool indicating whether a match was found or not. It does not return the matched value.
So thePin is a Bool which you then attempt to force-cast to a MKAnnotation which of course crashes.
If you want the matching value, change your code to:
if let thePin = pinArray.first(where: { $0.title == restaurantName }) {
do {
mapp.selectionAnnotation(thePin, animated: true)
} catch {
}
} else {
// no match in the array
}
No need for contains at all. No need to cast (assuming pinArray is an array of MKAnnotation).
I am downloading information from a Firebase database and it is being inputted via a for loop into:
static var Reports = [String:[String:String]]()
I need to figure out a way to search the inside values for a certain string
I have messed around with this but can't seem to get it inside the inside dictionary (If that makes sense)
for values in Reports.count {
if let item = Reports["favorite drink"] {
print(item)
}
}
I need to have a search string then a number of times the value appears like so:
func findString(dict Dictionary) -> Int {
var ReportsLevel1 = 0
(for loop I'm guessing)
search here for string
return ReportsLevel1
}
Tip: the outside dictionary keys are not set in stone, they depend on what time and date the report was submitted
To find out the numberOfTimes in which "yourSearchString" appears you can do as follows
var numberOfTimes = 0
for internalDictionary in reports.values
{
for value in internalDictionary.values
{
if (value == "yourSearchString") { numberOfTimes += 1 }
}
}
or
let numberOfTimes = reports.flatMap { internalDictsArray in internalDictsArray.value.filter { $0.value == "yourSearchString" } }.count