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"
