CODEWITHSUNDEEP
swiftswift2
Nogurenn
Nogurenn

Reputation: 878

Why is the let variable = x syntax not working with the ternary operator?

Implicitly unwrapping optionals can be done this way.

if let foo = foo {
     print("foo")
} else {
     print("bar")
}

I'm practicing in Playground, and what I don't understand is why I can't do the one above using a ternary operator.

let foo = foo ? print("foo") : print("bar")

// or

var result:String = (let foo = foo ? print("foo") : print("bar"))

I was thinking this is perfectly normal, albeit awkward, syntax for Swift since ternary operations evaluate to expressions. My suspicion is that let foo = foo is itself an expression that Swift has to deal with, but that's coming from a Swift newbie like me.

Upvotes: 0

Views: 2000

Answers (1)

rob mayoff
rob mayoff

Reputation: 385650

if let is a special Swift form that allows you to rebind an identifier while unwrapping an Optional. A let statement by itself can't rebind an identifier declared in the same scope. Them's the rules.

Furthermore, you cannot use an Optional as if it were a BooleanType, which you're trying to do (since the conditional of ?: has to be a BooleanType).

Also note that print returns () (the empty tuple, also known as the singleton instance of the Void type). So there's no reason to rebind foo to the return value of print.

This works, though it's not good style:

foo != nil ? print("set") : print("nil")

This is good style:

if foo != nil {
    print("set")
} else {
    print("nil")
}

Or this:

print(foo != nil ? "set" : "nil")

The following is worse in this case, but is a reasonable pattern when you need to transform the wrapped value if set, and provide a default if nil:

print(foo.map { _ in "set" } ?? "nil" )

If you just want to unwrap foo for use later in the function, without increasing the brace nesting level, you can use guard:

guard let foo = foo else {
    print("nil")
    return // you must return or throw or abort here in a guard statement
}

// Here foo has been rebound and is no longer optional.
print("set to \(foo)")

Upvotes: 4

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