在关闭中破坏元组的元组 [英] Destructuring tuple of tuple in closure

问题描述

我可以很容易地破坏一个元组的元组：

I can destructure a tuple of tuple easily:

let tt = (2, (3, 4))

let (a, (b, c)) = tt
b // => 3

我想在声明关闭时做同样的事情，例如，我以为我可以写：

I'd like to do the same when declaring a closure, for example I thought I could write:

[tt].map { (a, (b, c)) in
    // Use b
}

Xcode抱怨未命名参数必须用空名称写。

Xcode complains with "Unnamed parameters must be written with the empty name".

让我工作的唯一方法是：

Only way I got it to "work" was:

[tt].map { (a, tuple: (b: Int, c: Int)) in
    // Use tuple.b
}

我想避免两个缺点：

• 我需要使用 tuple.b 代替 b


• 我需要指定类型 b 和 c

• I need to use tuple.b instead of b
• I need to specify the types of b and c

顺便说一句，我的用例是我想对索引进行 reduce ，所以我尝试使用 array.enumerate（）。reduce

BTW, my use case is that I want to do a reduce with index so I'm trying using array.enumerate().reduce

推荐答案

使用附加的赋值行，可以将数组中的值赋给（a ，（b，c））解构元组：

With an additional assignment line, you can assign the values in the array to (a, (b, c)) to deconstruct the tuple:

let tt1 = (2, (3, 4))
let tt2 = (5, (6, 7))

[tt1, tt2].map { tt in
    let (a, (b, c)) = tt
    print(b)
}

输出：

3
6

---

或者：

---

Alternatively:

[tt1, tt2].map {
    let (a, (b, c)) = $0
    print(b)
}

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