swift数组

2017年05月02日16:27:33

直接代码

     let cast = ["Vivien", "Marlon", "Kim", "Karl"]
     var strArray = [String]()
     for item in  cast{
          strArray += (item+",")
     }
     if  strArray.length > 0{
            strArray = strArray.subStringTo(strArray.length - 1)
     }
     print(list)
     // Prints "Vivien, Marlon, Kim, Karl"

上面代码是很多人的第一想法
下面的代码就比较简洁了 效果一样

  let cast = ["Vivien", "Marlon", "Kim", "Karl"]
  let list = cast.joined(separator: ", ")
  print(list)
  // Prints "Vivien, Marlon, Kim, Karl"

在拨盘选择数字的时候

let array  = Array(1...10)
/*
 以下两种写法一致
 */
let result = array.map { (item) -> String in
    return "\(item)"
}
let result = array.map {"\($0)"}

print(result)
//["1", "2", "3", "4", "5", "6", "7", "8", "9", "10"]

关于swift数组中对象的深浅拷贝

class IntegerReference :NSObject{
    var value = 10
    
}
var firstIntegers = [IntegerReference(), IntegerReference()]
var secondIntegers = firstIntegers


firstIntegers[0].value = 100
print(secondIntegers[0].value) //两个数组对象值都会改变
// Prints "100"

secondIntegers[0].value = 50
print(firstIntegers[0].value) //两个数组对象值都会改变
// Prints "50"

数组中拷贝对象要实现copyWithZone方法

class IntegerReference :NSObject,NSCopying{
    var value = 10
    func copy(with zone: NSZone? = nil) -> Any {
        let copy = IntegerReference()
        copy.value = self.value
        return copy
    }
}
var firstIntegers = [IntegerReference(), IntegerReference()]
var secondIntegers = firstIntegers.map{ ($0.copy() as! IntegerReference) }


firstIntegers[0].value = 100
print(secondIntegers[0].value) //改变数组一中对象的值   数组二不会被改变
// Prints "10"

secondIntegers[0].value = 50
print(firstIntegers[0].value)
// Prints "100"