闭包如何捕获来自以前调用的值？

问题描述

typealias IntMaker = (Void)->Int

func makeCounter() ->IntMaker{
    var n = 0 // Line A

    func adder()->Integer{
        n = n + 1
        return n 
    }
    return adder
}

let counter1 = makeCounter()

counter1() // returns 1
counter1() // returns 2
counter1() // returns 3

每次调用 / code>？这意味着 var n = 0 应该每次调用...

Isn't 'Line A' called each time we call counter1() ? meaning that var n = 0 should be called every time...

为什么计数器返回不同的值？

Why is that the counter returns different values? Shouldn't they always be returning '1' ?

推荐答案

你已经调用 makeCounter / code>一次。这将创建您的新闭包，并将其分配给 counter1 。这个闭包关闭了可变的 var n ，只要这个闭包存在，它就会被捕获。

You've called makeCounter() once. That creates your new closure, and assigns it to counter1. This closure closes over the mutable var n, and will remain captured as long as this closure exists.

code> counter1（）将执行它，但它保留相同的捕获 n ，并将其改变。这个特定的加法器总是捕获这个相同的 n ，只要它存在..

Calling counter1() will execute it, but it retains the same captured n, and mutates it. This particular "adder" will ALWAYS capture this same n, so long as it exists..

您需要创建新的闭包，捕获 n 的新实例：

To get the behavior you're suggesting, you need to make new closures which capture new instances of n:

let counter1 = makeCounter()

counter1() // returns 1
counter1() // returns 2
counter1() // returns 3

var counter2 = makeCounter()
counter2() // returns 1
counter2 = makeCounter()
counter2() // returns 1
counter2 = makeCounter()
counter2() // returns 1

现在 counter1 和 counter2 每个都有自己的单独实例 n 。

Now both counter1 and counter2 each have their own separate instances of n.

这篇关于闭包如何捕获来自以前调用的值？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！