arrays - 迭代多个结构数组的最简单方法？

Question

希望通过多个结构数组递增，使用它们的变量之一作为每个循环的数量。谢谢!

struct Example {
    var partOne: Int
    var partTwo: Int
    var partThree: Int
}

var one = Example(partOne: 10, partTwo: 11, partThree: 12)
var two = Example(partOne: 10, partTwo: 11, partThree: 12)

var arrayOfExamples = [one, two]

for i in 0...arrayOfExamples[0].partThree {
    print(i)
}

//once i = 12, then

for i in 0...arrayOfExamples[1].partThree {
    print(i)
}

Answer 1

只需使用嵌套循环，外层循环遍历 arrayOfExamples 的项目:

for item in arrayOfExamples {
    for i in 0...item.partThree {
        print(i)
    }
}

---

通过使用 KeyPath，您可以编写一个函数来迭代具有调用者指定的属性的值:

func iterateOverKeyPath(array: [Example], keyPath: KeyPath<Example, Int>) {
    for item in array {
        for i in 0...item[keyPath: keyPath] {
            print(i)
        }
    }
}

// iterate using partThree property
iterateOverKeyPath(array: arrayOfExamples, keyPath: \Example.partThree)

// now do the same for partTwo
iterateOverKeyPath(array: arrayOfExamples, keyPath: \Example.partTwo)

---

而且 Example struct 并没有什么特别之处，所以我们可以让这个泛型适用于任何类型:

func iterateOverKeyPath<T>(array: [T], keyPath: KeyPath<T, Int>) {
    for item in array {
        for i in 0...item[keyPath: keyPath] {
            print(i)
        }
    }
}