给定带有 Array
的 struct
import Foundation
struct Card {
var flag: String = ""
}
var cards = Array<Card>()
cards.append(Card())
以下操作不会修改原始数组元素 // A copy is created.
var cardCopy = cards[0]
// Will NOT modify cards[0]
cardCopy.flag = "modify0"
print(cards[0].flag)
以下操作将修改原始数组元素 // We can modify cards[0] by
cards[0].flag = "modify"
print(cards[0].flag)
然而,它在某种意义上并不高效,我们每次都需要执行索引访问。想象 cards[0].flag0 = "modify"
cards[0].flag1 = "modify"
cards[0].flag2 = "modify"
cards[0].flag3 = "modify"
...
有没有办法,我们可以创建对结构数组元素的引用? 这样我们就可以写// How to create a reference to cards[0]?
var cardReference = ...
cardReference.flag0 = "modify"
cardReference.flag1 = "modify"
cardReference.flag2 = "modify"
cardReference.flag3 = "modify"
...
一种可能性是用 struct
替换 class
。但是,在此之前,我想探索其他替代方案。 最佳答案
您可以使用函数进行更改并通过引用传递 Card
结构,如下所示:
func update(card: inout Card) {
card.flag0 = "modify"
card.flag1 = "modify"
card.flag2 = "modify"
card.flag3 = "modify"
}
var cards = Array<Card>()
cards.append(Card())
update(card: &cards[0])
或者通过在 Card
类型中使用变异函数并作为闭包传递您的更改,甚至更好:struct Card {
var flag0: String = ""
var flag1: String = ""
var flag2: String = ""
var flag3: String = ""
mutating func update(block: (inout Card) -> Void) {
block(&self)
}
}
var cards = Array<Card>()
cards.append(Card())
cards[0].update {
$0.flag0 = "modify"
$0.flag1 = "modify"
$0.flag2 = "modify"
$0.flag3 = "modify"
}
更新 :为了使第二种方法更具可重用性,您可以定义如下协议(protocol):protocol Updatable {
mutating func update(block: (inout Self) -> Void)
}
extension Updatable {
mutating func update(block: (inout Self) -> Void) {
block(&self)
}
}
并使 Card
结构符合它:struct Card: Updatable {
var flag0: String = ""
var flag1: String = ""
var flag2: String = ""
var flag3: String = ""
}
然后你可以像上面一样使用它。关于ios - 是否有可能创建对结构数组元素的引用?,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/63485670/