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Optionally bind elements before appending to Arrays #170

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Mar 3, 2016
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Optionally bind elements before appending to Arrays #170

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8 changes: 4 additions & 4 deletions Sources/Utility/ArrayExtensions.swift
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Original file line number Diff line number Diff line change
Expand Up @@ -20,10 +20,10 @@ extension Array {
var t = [T]()
var u = [U]()
for e in self {
if e is T {
t.append(e as! T)
} else {
u.append(e as! U)
if let e = e as? T {
t.append(e)
} else if let e = e as? U {
u.append(e)
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how about

if case let e as T = e {
    t.append(e)
} else if case let e as U = e {
    u.append(e)
}

Does the same thing but avoids the ? in as

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sure - that works!

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as for me if case let is to heavy

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actually, that does not work at all. Still asks to force-downcast. Agree with @kostiakoval.

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Works for me

extension Array {
    public func partition<T, U>() -> ([T], [U]) {
        var t = [T]()
        var u = [U]()
        for e in self {
            if case let e as T = e {
                t.append(e)
            } else if case let e as U = e {
                u.append(e)
            }
        }
        return (t, u)
    }
}

let a = [1, 2, 3, "2", "3"]
let partition: ([Int], [String]) = a.partition()

agreed though that if case let looks heavier at first but I have grown to like it lol

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wouldn't say better or worse just different syntactically I guess?

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@aciidb0mb3r's solution is syntactic sugar (https://developer.apple.com/library/ios/documentation/Swift/Conceptual/Swift_Programming_Language/Patterns.html#//apple_ref/swift/grammar/optional-pattern). Same solution, different way of implementing, which I would deem non-traditional.

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Both solutions works fine and there is no big difference. if let e = e as? T solutions is safe, very clear and it's very often used in Swift community.
I feel there is no need for further discussion.

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👍

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Thanks! 👍

}
}
return (t, u)
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