Trying to implement ?.let pattern in Swift, is it even possible?

Question

I've been porting code back and forth between Swift and Kotlin. Inspired by this answer, I've been using ?.let { } ?: run {} as replacements for the many places I use if let ... { } else {} in Swift.

I've really come to like the Kotlin expression though. In my opinion it's more general than the keyword specific if let concoction. So for parity's sake--or at least for the sake of learning by trying silly experiments--I thought I'd try to implement the let construct in Swift.

Because let is a keyword in Swift, I discovered right away that I'd need to put backticks, or use a different name, like cull. That led to this uncompilable code:

extension Any {
    func cull<R>(block:(Self) -> R) -> R {
        return block(self)
    }
}

which given my apprentice level skills, probably has numerous problems. But most importantly, I'm informed that I can't extend Any (which is explained in better detail here).

So Any can't be extended. Is there any other language construct I can use to make it so that I can implement something like Kotlin's let? Basically:

<expression>.foo { <expressionResult> in ... }

As a bonus/aside question, I think even if there is a way (I do hope there is just so I can learn), it won't be the same, because the Kotlin one can be used for control flow because I can do a non local return from the Kotlin one, but not from a Swift closure. Is that also correct?

CLARIFICATION

Basically, I can implement the Kotlin let pattern for designated types:

protocol Cullable {}

extension Cullable {
    func cull<R>(_ block:(Self) -> R) -> R {
        return block(self)
    }
}

extension Int:Cullable { }

42.cull { thing in print(thing) } // this will print 42

This works as I want. But I don't want to have to extend every single non-nominal type in the system to make it work. I just want to define it globally some how. If it was possible.

It's not possible, because there is no way to have the "foo" part captured and passed into whatever function you concoct. `if let` is a language-feature, not a library feature — Alexander, Aug 05 '18 at 00:38
Oh, I see. Just write an extension on `Optional`, it's not much different. — Alexander, Aug 05 '18 at 02:33
Can you give an example of what you are actually trying to accomplish? Like give the Kotlin code you are trying to replicate in Swift? I'm a little confused because the ?.let in Kotlin is actually, in my opinion, a much worse option than what Swift has. In most cases, I'd much rather have a Swifty guard let or if let than a Kotlin ?.let. — TJ Olsen, Oct 24 '22 at 14:22

Sweeper · Accepted Answer · 2018-08-05T01:05:08.813

Actually, I think such a method already exists in Optional. Although this is only for optionals, at least it solves your problem of the lack of ?.let. The method is called map.

let optional: String? = "hello"
optional.map { print($0) } // prints hello
// note how I did not unwrap the optional. I called "map" directly on the optional, not the wrapped "String".

You can map the value to into another value, or Void, like I did above.

I don't think you can have a let method on everything.

flatMap is also available if you want to return an optional inside the closure.

And yes, after looking at what "non-local return" means in Kotlin, I don't think you can do the same in Swift. Those curly brace thingys are "closures". They are "closed", so you can't return the surrounding closure in an inner closure.

Trying to implement ?.let pattern in Swift, is it even possible?

1 Answers1