I wish to use guard-let to assign a variable to an expression, but I want to modify the expression before assigning. If the expression is nil, then the else block should be entered, otherwise the variable should be assigned to f(expression). Here is an example of what I would like to do:
let arr: [Int] = []
// Do stuff, maybe add elements to arr
guard let x = abs(arr.first) else { return } // Syntax error
// If arr was nonempty, then we want x = abs(arr.first!)
But Swift does not allow this syntax because abs requires a non-optional argument, and arr.first is optional. So is there any way to evaluate arr.first, and then if it is not nil to assign abs(arr.first!) to x? I know that I could do this with if-let or by using two variables (one from the guard-let and then one that gets assigned to the absolute value of that variable). But guard-let seems like the tool for the job, if only there were some way to accomplish this.
let arr:[Int] = [-1,1,3,-9]
guard let x = arr.first.flatMap({ $0 < 0 ? -$0: $0 }) else { return }
// ...
or (UPDATE based on dfri's notes)
// ....
let arr:[Int] = [-1,1,3,-9]
guard let x = arr.first.map(abs) else { return }
Optional(Some<Int>) -> Int -> Optional<abs(Some<Int)> -> Int ... meh
You could do a dirty guard let ..., let ... else fix as follows (forcing the binded certainly-not-nil value of x to become an optional which you subsequently immediately unwrap and bind to xAbs)
func foo() {
let arr: [Int] = [-1, 2, -3, 4]
guard let x = arr.first,
let xAbs = Optional(abs(xAbs)) else { return }
print(xAbs, xAbs.dynamicType)
}
foo() // 1 Int
This doesn't look very pretty however, and I would, personally, prefer adding an Int extension and make use of optional chaining, as I will cover next.
Instead: use extensions and optional chaining
Unless you explicitly need to store x as well as xAbs, an alternative and more Swifty approach is to use optional chaining in combination with a simple extension to Int:
extension Int {
var absValue: Int { return abs(self) }
}
func foo() {
let arr: [Int] = [-1, 2, -3, 4]
guard let xAbs = arr.first?.absValue else { return }
print(xAbs, xAbs.dynamicType)
}
foo() // 1 Int
Since arr.first is an optional Int variable, you can implement whatever method/computed property you wish onto self as an extension to Int, and access that method/property using optional chaining arr.first?.someMethod()/arr.first?.someProperty (as .absValue above).
Or, simply modify your arr.first (unwrapped) value after the guard let ... else block
I see no reason, however (other than the technical discussion) not to introduce an additional immutable holding the absolute value of x. This will also increase code readability, at least w.r.t. to the dirty guard let ..., let ... else fix above.
// ...
guard let x = arr.first else { return }
let xAbs = abs(x)
Or, if you find it acceptable for your xAbs property to be mutable, out of a theoretical perspective your could remove the middle-man immutable by using a guard var ... block rather than guard let ...
guard var xAbs = arr.first else { return }
xAbs = abs(xAbs)
This should probably only be used, however, if xAbs is to be mutated again (i.e., use immutables whenever you really don't need mutables, and never the other way around).
I think the cleanest and simplest solution would be like this:
guard let first = arr.first else { return }
let x = abs(first)
Now the calculation abs(first) is only reached if arr.first != nil.
What you want can be achieved using case let.
let arr: [Int] = [1,2,3,4]
guard let first = arr.first, case let absolute = abs(first) else { return }
// use `absolute`
Related
I am using an SQLite library in which queries return optional values as well as can throw errors. I would like to conditionally unwrap the value, or receive nil if it returns an error. I'm not totally sure how to word this, this code will explain, this is what it looks like:
func getSomething() throws -> Value? {
//example function from library, returns optional or throws errors
}
func myFunctionToGetSpecificDate() -> Date? {
if let specificValue = db!.getSomething() {
let returnedValue = specificValue!
// it says I need to force unwrap specificValue,
// shouldn't it be unwrapped already?
let specificDate = Date.init(timeIntervalSinceReferenceDate: TimeInterval(returnedValue))
return time
} else {
return nil
}
}
Is there a way to avoid having to force unwrap there? Prior to updating to Swift3, I wasn't forced to force unwrap here.
The following is the actual code. Just trying to get the latest timestamp from all entries:
func getLastDateWithData() -> Date? {
if let max = try? db!.scalar(eventTable.select(timestamp.max)){
let time = Date.init(timeIntervalSinceReferenceDate: TimeInterval(max!))
// will max ever be nil here? I don't want to force unwrap!
return time
} else {
return nil
}
}
Update: As of Swift 5, try? applied to an optional expression does not add another level of optionality, so that a “simple” optional binding is sufficient. It succeeds if the function did not throw an error and did not return nil. val is then bound to the unwrapped result:
if let val = try? getSomething() {
// ...
}
(Previous answer for Swift ≤ 4:) If a function throws and returns an optional
func getSomething() throws -> Value? { ... }
then try? getSomething() returns a "double optional" of the
type Value?? and you have to unwrap twice:
if let optval = try? getSomething(), let val = optval {
}
Here the first binding let optval = ... succeeds if the function did
not throw, and the second binding let val = optval succeeds
if the return value is not nil.
This can be shortened with case let pattern matching to
if case let val?? = try? getSomething() {
}
where val?? is a shortcut for .some(.some(val)).
I like Martin's answer but wanted to show another option:
if let value = (try? getSomething()) ?? nil {
}
This has the advantage of working outside of if, guard, or switch statements. The type specifier Any? isn't necessary but just included to show that it returns an optional:
let value: Any? = (try? getSomething()) ?? nil
I have a class with an optional member :
class A {
var i: Int? = nil
}
Then I have an array of objects of type A. Some objects in the array have a value for i, some others don't.
I want to iterate over objects in the array that have a value for i while unwrapping the optional at the same time. I didn't find a way to do both at the same time (I don't even know if it's possible), forcing me to write a if let construct inside the loop.
For example :
// a1, a2 have a value for i
let arr: [A] = [a1, a2, a3]
for obj in arr where obj.i != nil {
// I want to avoid if let, or force unwrapping here
if let unwrapped = obj.i {
print(i)
}
// let unwrapped = obj.i! ...
}
Is it possible in Swift ?
1.Maybe you can use flatMap to get value i, then print it
arr.flatMap{ $0.i }.forEach{ print($0) }
2.or Trying simple guard statement
arr.forEach { element in
guard let i = element.i else { return }
print(i)
}
I don't think that's possible.
Even if you have a where clause in your loop the type of obj is still of type A and as such i still remains optional.
To see why this is so think about the fact that you can change the value of i on object obj inside the loop, so the compiler is not sure that the value of i is valid until you unwrapp it.
You can try something like this
for obj in arr where obj.i != nil {
guard let i = obj.i else { continue }
print( i )
}
but if you start using guard you also skip the where clause
for obj in arr {
guard let i = obj.i else { continue }
print( i )
}
You can use case let syntax, but not without the help of map, and the result isn't the most readable:
for case let .some(unwrapped) in arr.map(\.i) {
print(unwrapped)
}
It's more useful if you're e.g. casting the outer object, e.g.:
for case let object as String in arrayOfAny {
if object.hasPrefix("tw") {
print("Starts with 'tw'")
}
}
instead of:
for object in arrayOfAny where object is String {
if object.hasPrefix("tw") { // Error: Value of type 'Any' has no member 'hasPrefix'
print("Starts with 'tw'")
}
}
I am using an SQLite library in which queries return optional values as well as can throw errors. I would like to conditionally unwrap the value, or receive nil if it returns an error. I'm not totally sure how to word this, this code will explain, this is what it looks like:
func getSomething() throws -> Value? {
//example function from library, returns optional or throws errors
}
func myFunctionToGetSpecificDate() -> Date? {
if let specificValue = db!.getSomething() {
let returnedValue = specificValue!
// it says I need to force unwrap specificValue,
// shouldn't it be unwrapped already?
let specificDate = Date.init(timeIntervalSinceReferenceDate: TimeInterval(returnedValue))
return time
} else {
return nil
}
}
Is there a way to avoid having to force unwrap there? Prior to updating to Swift3, I wasn't forced to force unwrap here.
The following is the actual code. Just trying to get the latest timestamp from all entries:
func getLastDateWithData() -> Date? {
if let max = try? db!.scalar(eventTable.select(timestamp.max)){
let time = Date.init(timeIntervalSinceReferenceDate: TimeInterval(max!))
// will max ever be nil here? I don't want to force unwrap!
return time
} else {
return nil
}
}
Update: As of Swift 5, try? applied to an optional expression does not add another level of optionality, so that a “simple” optional binding is sufficient. It succeeds if the function did not throw an error and did not return nil. val is then bound to the unwrapped result:
if let val = try? getSomething() {
// ...
}
(Previous answer for Swift ≤ 4:) If a function throws and returns an optional
func getSomething() throws -> Value? { ... }
then try? getSomething() returns a "double optional" of the
type Value?? and you have to unwrap twice:
if let optval = try? getSomething(), let val = optval {
}
Here the first binding let optval = ... succeeds if the function did
not throw, and the second binding let val = optval succeeds
if the return value is not nil.
This can be shortened with case let pattern matching to
if case let val?? = try? getSomething() {
}
where val?? is a shortcut for .some(.some(val)).
I like Martin's answer but wanted to show another option:
if let value = (try? getSomething()) ?? nil {
}
This has the advantage of working outside of if, guard, or switch statements. The type specifier Any? isn't necessary but just included to show that it returns an optional:
let value: Any? = (try? getSomething()) ?? nil
I have a quick question that is confusing me a little bit. I made a simple average function that takes an array of optional Ints. I check to make sure the array does not contain a nil value but when I use reduce I have to force unwrap one of the two elements in the closure. Why is it that I only force unwrap the second one (in my case $1!)
func average2(array: [Int?]) -> Double? {
let N = Double(array.count)
guard N > 0 && !array.contains({$0 == nil}) else {
return nil
}
let sum = Double(array.reduce(0) {$0+$1!})
let average = sum / N
return average
}
I know it is simple but I would like to understand it properly.
The first parameter of reduce is the sum, which is 0 in the beginning. The second one is the current element of your array which is an optional Int and therefore has to be unwrapped.
Your invocation of reduce does this:
var sum = 0 // Your starting value (an Int)
for elem in array {
sum = sum + elem! // This is the $0 + $1!
}
EDIT: I couldn't get a more functional approach than this to work:
func average(array: [Int?]) -> Double? {
guard !array.isEmpty else { return nil }
let nonNilArray = array.flatMap{ $0 }
guard nonNilArray.count == array.count else { return nil }
return Double(nonNilArray.reduce(0, combine: +)) / Double(nonNilArray.count)
}
You can also discard the second guard if you want something like average([1, 2, nil]) to return 1.5 instead of nil
As far as I know the recommended way to use optionals (Int in this example) is the following:
var one:Int?
if var maybe = one {
println(maybe)
}
Is it possible to use a shorter way to do something like the following?
var one:Int?
var two:Int?
var three:Int?
var result1 = one + two + three // error because not using !
var result2 = one! + two! + three! // error because they all are nil
Update
To be more clear about what I'm trying to do: I have the following optionals
var one:Int?
var two:Int?
var three:Int?
I don't know if either one or two or three are nil or not. If they are nil, I wan't them to be ignored in the addition. If they have a value, I wan't them to be added.
If I have to use the recommended way I know, it would look something like this: (unnested)
var result = 0
if var maybe = one {
result += maybe
}
if var maybe = two {
result += maybe
}
if var maybe = three {
result += maybe
}
Is there a shorter way to do this?
Quick note - if let is preferred for optional binding - let should always be used where possible.
Perhaps Optionals aren't a good choice for this situation. Why not make them standard Ints with a default value of 0? Then any manipulation becomes trivial and you can worry about handling None values at the point of assignment, rather than when you're working on the values?
However, if you really want to do this then a tidier option is to put the Optionals into an Array and use reduce on it:
let sum = [one,two,three,four,five].reduce(0) {
if ($1) {
return $0 + $1!
}
return $0
}
That's exactly the point of optionals — they may be nil or non-nil, but unwrapping them when they're nil is an error. There are two types of optionals:
T? or Optional<T>
var maybeOne: Int?
// ...
// Check if you're not sure
if let one = maybeOne {
// maybeOne was not nil, now it's unwrapped
println(5 + one)
}
// Explicitly unwrap if you know it's not nil
println(5 + one!)
T! or ImplicitlyUnwrappedOptional<T>
var hopefullyOne: Int!
// ...
// Check if you're not sure
if hopefullyOne {
// hopefullyOne was not nil
println(5 + hopefullyOne)
}
// Just use it if you know it's not nil (implicitly unwrapped)
println(5 + hopefullyOne)
If you need to check multiple optionals at once here there are a few things you might try:
if maybeOne && maybeTwo {
println(maybeOne! + maybeTwo!)
}
if hopefullyOne && hopefullyTwo {
println(hopefullyOne + hopefullyTwo)
}
let opts = [maybeOne, maybeTwo]
var total = 0
for opt in opts {
if opt { total += opt! }
}
(It seems you can't use the let optional binding syntax with more than one optional at once, at least for now...)
Or for extra fun, something more generic and Swifty:
// Remove the nils from a sequence of Optionals
func sift<T, S: Sequence where S.GeneratorType.Element == Optional<T>>(xs: S) -> GeneratorOf<T> {
var gen = xs.generate()
return GeneratorOf<T> {
var next: T??
do {
next = gen.next()
if !next { return nil } // Stop at the end of the original sequence
} while !(next!) // Skip to the next non-nil value
return next!
}
}
let opts: [Int?] = [1, 3, nil, 4, 7]
reduce(sift(opts), 0) { $0 + $1 } // 1+3+4+7 = 15