我正在制作一个变异结构，其派生值的计算成本非常高。所以我想做的是懒惰地计算这个派生值并存储结果，直到结构再次发生变异，此时派生值不再有效，需要重新计算。

(失败)选项 1:生成的属性

如果派生值是生成的属性(如下所示)，则始终返回正确的值，但始终会重新计算。

(失败)选项 2:延迟加载属性

如果它是惰性属性，则计算只进行一次……永远。因此，一旦结构发生变异，派生值就会错误，并且不会重新计算。此外，如果我从结构中分配常量值，我将无法访问该属性。

在 Swift 1.2 中是否有任何可能的解决方案，或者我是否需要提交雷达？

struct Struct {
    var value: Int

    // Option 1: Generated property
    var derivedValue: Int {
        println("Doing expensive calculation")
        return self.value * 2
    }

    // Option 2: Lazy property
    lazy var derivedValue: Int = {
        println("Doing expensive calculation")
        return self.value * 2
    }()

    init(value: Int) {
        self.value = value
    }

    mutating func mutate() {
        value = random()
    }
}

var test = Struct(value: 2)
test.derivedValue
test.derivedValue // If not lazy, expensive calculation is done again here
test.mutate()
test.derivedValue // If lazy, this has wrong value

let test2 = test
test2.derivedValue // Compiler error if using lazy implementation

最佳答案

使用嵌入式类绕过了改变结构的限制。这使您可以使用按值类型，该类型在需要时才运行昂贵的计算，但之后仍会记住结果。

下面的示例 Number 结构以您描述的方式计算并记住其 square 属性。数学本身效率低得离谱，但它是说明解决方案的简单方法。

struct Number {

    // Store a cache in a nested class.
    // The struct only contains a reference to the class, not the class itself,
    // so the struct cannot prevent the class from mutating.
    private class Cache {
        var square: Int?
        var multiples: [Int: Int] = [:]
    }
    private var cache = Cache()

    // Empty the cache whenever the struct mutates.
    var value: Int {
        willSet {
            cache = Cache()
        }
    }

    // Prevent Swift from generating an unwanted default initializer.
    // (i.e. init(cache: Number.Cache, value: Int))
    init(value: Int) {
        self.value = value
    }

    var square: Int {
        // If the computed variable has been cached...
        if let result = cache.square {

            // ...return it.
            print("I’m glad I don’t have to do that again.")
            return result
        } else {

            // Otherwise perform the expensive calculation...
            print("This is taking forever!")
            var result = 0
            for var i = 1; i <= value; ++i {
                result += value
            }

            // ...store the result to the cache...
            cache.square = result

            // ...and return it.
                return result
        }
    }

    // A more complex example that caches the varying results
    // of performing an expensive operation on an input parameter.
    func multiple(coefficient: Int) -> Int {
        if let result = cache.multiples[coefficient] {
            return result
        } else {

            var result = 0
            for var i = 1; i <= coefficient; ++i {
                result += value
            }

            cache.multiples[coefficient] = result
                return result
        }
    }
}

这是它的表现:

// The expensive calculation only happens once...
var number = Number(value: 1000)
let a = number.square // “This is taking forever!”
let b = number.square // “I’m glad I don’t have to do that again.”
let c = number.square // “I’m glad I don’t have to do that again.”

// Unless there has been a mutation since last time.
number.value = 10000
let d = number.square // “This is taking forever!”
let e = number.square // “I’m glad I don’t have to do that again.”

// The cache even persists across copies...
var anotherNumber = number
let f = anotherNumber.square // “I’m glad I don’t have to do that again.”

// ... until they mutate.
anotherNumber.value = 100
let g = anotherNumber.square // “This is taking forever!”

作为一个更现实的例子，我在日期结构上使用了这种技术，以确保尽可能少地运行用于在日历系统之间转换的重要计算。

关于swift - 如何正确地在 Swift 中的变异结构上创建惰性派生属性？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/28901871/