Well...here is the code
func howMany() -> Int {
return 10
}
func Call() -> Void {
guard case let output = howMany(), output > 5 else { return }
}
Call()
I do not really understand how the guard case works. This looks pretty much like a pattern matching condition where we compare whether the result of howMany() is equal to output, if it is, assign the value to output and then compare it to the literal value 5. However, when I deleted the line of output > 5, the compiler said, "the guard condition is always true, body is unreachable."
According to the pattern, if we translate it into a switch statement, it looks pretty much like this
switch howMany() {
case let output where output > 5:
break;
}
The question is if it could be directly translated into the switch statement, then there should not be a warning of "the guard condition is always true, body is unreachable" when we delete the where condition.
I hope somebody could shed some light on this.
Consider:
func foo() {
guard case let output = howMany(), output > 5 else {
print("less than or equal to 5")
return
}
print("\(output) is greater than 5")
}
That is roughly equivalent to:
func bar() {
switch howMany() {
case let output where output > 5:
print("\(output) is greater than 5")
default:
print("less than or equal to 5")
return
}
}
If you remove that > 5 criteria:
func foo() {
guard case let output = howMany() else {
print("less than or equal to 5")
return
}
print("\(output) is greater than 5")
}
You get your warning:
'guard' condition is always true, body is unreachable
That warning is correct, because the body is unreachable.
And if you do the equivalent in the switch example:
func bar() {
switch howMany() {
case let output:
print("\(output) is greater than 5")
default:
print("less than or equal to 5")
return
}
}
If you do that, you will receive an analogous warning:
default will never be executed
And, again, it makes sense, because default will not be reached.
Now, consider your example switch with no default clause:
func bar() {
switch howMany() {
case let output:
print("output:", output)
}
}
You don't receive a warning here only because there is no default clause (the analog of the "body" of the guard statement).
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.
This is what I've tried and can't figure out where the error is coming from. Is there something missing? Syntax error? I tried doing similar with if-else in the function and also getting errors.
var steps = 0
func incrementSteps() -> Int {
steps += 1
print(steps)
return steps
}
incrementSteps()
let goal = 10000
func progressUpdate() -> Int {
let updated = progressUpdate()/goal
switch updated {
case (0.0..<0.1):
print("Great start")
case (0.11..<0.5):
print("Almost halfway")
case (0.51..<0.9):
print("Almost complete")
default:
print("Beat goal")
}
}
progressUpdate()
You need to specify updated as Double. And cast it back to Int when returning(if you require Int for your requirement).
Note: Also, you need to modify calling the progressUpdate function within progressUpdate definition which causes a recursion. If you want to do so you might want to give condition to break the loop.
func progressUpdate() -> Int {
let updated = Double(steps/goal)
switch updated {
case (0.0..<0.1):
print("Great start")
case (0.11..<0.5):
print("Almost halfway")
case (0.51..<0.9):
print("Almost complete")
default:
print("Beat goal")
}
return Int(updated)
}
I have a number that I get from JSON, this number represents an age. Users give me a range of two ages and may code is supposed to check if this number I'm getting from JSON is in the range.
here is my code and it gives me error
Type of Expression is ambiguous without more context?
let age = "40"
if Int(AgeFrom) ... Int(AgeTO) ~= Int(age) {
print("yes")
}
Update
if let value: AnyObject = response.result.value as AnyObject? {
var ages = String
let json = JSON(value)
for (key, subJson) in json {
ages.append(subJson["age"].string!)
}
guard let min = Int(self.DropDownFrom.selectedItem!) else { return }
guard let max = Int(self.DropDownTo.selectedItem!) else { return }
for fitage in ages {
switch ages
{
case (min...max):
print ("Age is in range")
default:
print ("Nope, not this time")
}
}
Still gives me an error.
You need to unwrap the optionals because the Int(:String) method might not have a valid answer.
Best way to do this is kind of thing is with guard
guard let min = Int(AgeFrom) else { return }
guard let max = Int(AgeTo) else { return }
And from there you can go with the simple if statement:
if (min <= age && age <= max)
{
print ("Age is in range")
}
or get really fancy and use the switch statement pattern matching syntax (which I much prefer)
switch age
{
case (min...max):
print ("Age is in range")
default:
print ("Nope, not this time")
}
if - simplicity and readability
It is a basic thing in programming, checking if an optional is between two other optional values with an if:
if Int(AgeFrom)! <= Int(age)! && Int(AgeTO)! >= Int(age)! {
print("It is in the range!")
}
switch - multiple cases handling
However, I recommend using a switch for case handling:
switch(Int(AgeFrom)! <= Int(age)!, Int(AgeTO)! >= Int(age)!){
case (true,true): print("Yes, it fits the range")
case (false,true): print("Too young!")
case (true,false): print("Too old!")
}
The second solution is far better for multiple cases of the age value, especially when it's outside the range.
Hope it helps!
You can also use optional binding:
if let ageFrom = Int(ageFrom),
let ageTo = Int(ageTo),
ageFrom...ageTo ~= age
{
print("yes")
} else {
print("no")
}
You have to unwrap the optionals:
if Int(AgeFrom)!...Int(AgeTO)! ~= Int(age)! {
print("yes")
}
of course that is the unsafe way of unwrapping, since it will crash if the conversion of AgeFrom, AgeTO or age fail.
I am using the nice guard statement from Swift 3.0 (in Xcode 8.0) and have the following function:
func parse(suffix s: String) throws -> Instruction {
guard let count = Int(s) where count >= 0 else {
throw InstructionParsingError(message: s + " should be a positive integer")
}
return CallInstruction(argCount: count)
}
My issue is that the swift compiler complains twice about the line containing my guard statement:
CallInstruction.swift:42:29: Boolean condition requires 'where' to separate it from variable binding
CallInstruction.swift:42:30: Expected ',' joining parts of a multi-clause condition
I tried
replacing the where with a , then the second error disappears but the first one is still there.
replacing the where with , where but then this line can't even be parsed
replacing the count in the where by Int(s) but have the same errors.
How should I change my code so that it compiles? (With a nice single guard statement I mean, of course I could have multiple guards, or ifs or switch but from what I read about the guard statement I should be able to have a clean readable line).
For solving of this issue i recommend you to use model Swift syntax in guard statement that replace where with ,.
func parse(suffix s: String) {
guard let count = Int(s), count >= 0 else {
return
}
}
Also it is possible to use if let statement :
func parseTwo(suffix s: String) {
if let count = Int(s), count >= 0 {
// done
print(count)
} else {
return
}
}