这也是可行的:

let arr = [1,2,3,4,5,6,7,8,9,10]
var sumedArr = arr.reduce(0, combine: {$0 + $1})
print(sumedArr)

结果是:55

@IBOutlet var valueSource: [MultipleIntBoundSource]!

private var allFieldsCount: Int {
    var sum = 0
    valueSource.forEach { sum += $0.count }
    return sum
}

用于嵌套参数

Swift 4示例

class Employee {
    var salary: Int =  0
    init (_ salary: Int){
        self.salary = salary
    }
}

let employees = [Employee(100),Employee(300),Employee(600)]
var sumSalary = employees.reduce(0, {$0 + $1.salary}) //1000

用于对象数组中元素的和

self.rankDataModelArray.flatMap{$0.count}.reduce(0, +)

对我来说，这就像使用财产

    let blueKills = match.blueTeam.participants.reduce(0, { (result, participant) -> Int in
        result + participant.kills
    })

斯威夫特3.0

我也有同样的问题，我在苹果的文档上找到了这个解决方案。

let numbers = [1, 2, 3, 4]
let addTwo: (Int, Int) -> Int = { x, y in x + y }
let numberSum = numbers.reduce(0, addTwo)
// 'numberSum' == 10

但是，在我的例子中，我有一个对象列表，然后我需要转换我的列表的值:

let numberSum = self.list.map({$0.number_here}).reduce(0, { x, y in x + y })

这对我很有用。