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Swift 数组用于存储相同类型的值的顺序列表。Swift 要严格检查,不允许不同类型的值在同一个数组中

声明一个数组

var someArray = [SomeType]()
var someArray = [SomeType](count: NumbeOfElements, repeatedValue: InitialValue)
var someInts = [Int](count: 3, repeatedValue: 0)//声明一个三个容量的数组,初始值为0
var someInts:[Int] = [10, 20, 30]//声明一个int类型的初始值为中括号内的数组

 

可以使用下标语法从数组中检索对应值,传递数组名后方括号内的索引对应的值,如下:

var someVar = someArray[index]

 

在这里,指数从0开始,这意味着可以使用索引0来访问第一个元素,第二元素可以通过使用索引1进行访问,其它类似。让我们来看看下面创建,初始化和访问数组的例子:

import Cocoa

var someInts = [Int](count: 3, repeatedValue: 10)//创建一个3个空间的可变数组,初始值都是10,数组类型为int类型

var someVar = someInts[0]

println( "Value of first element is \(someVar)" )
println( "Value of second element is \(someInts[1])" )
println( "Value of third element is \(someInts[2])" )

 

当上述代码被编译和执行时,它产生了以下结果:

Value of first element is 10
Value of second element is 10
Value of third element is 10

修改数组

可以使用 append() 方法或加法赋值运算符(+=)将新的项目添加到数组的末尾,在这里首先创建一个空的数组,然后添加新的元素到数组中,如下图所示:

import Cocoa

var someInts = [Int]()//初始化一int类型的空数组
someInts.append(20)//添加20
someInts.append(30)//添加30
someInts += [40]//添加40

var someVar = someInts[0]

println( "Value of first element is \(someVar)" )
println( "Value of second element is \(someInts[1])" )
println( "Value of third element is \(someInts[2])" )
Value of first element is 20
Value of second element is 30
Value of third element is 40

直接给已经存在的数组元素赋值就可以修改数组元素中的值

import Cocoa

var someInts = [Int]()

someInts.append(20)
someInts.append(30)
someInts += [40]

// Modify last element
someInts[2] = 50

var someVar = someInts[0]

println( "Value of first element is \(someVar)" )
println( "Value of second element is \(someInts[1])" )
println( "Value of third element is \(someInts[2])" )
Value of first element is 20
Value of second element is 30
Value of third element is 50

可以使用 for-in 循环迭代级数,在下面的例子是数组的整个集值,如下图所示:

import Cocoa

var someStrs = [String]()

someStrs.append("Apple")
someStrs.append("Amazon")
someStrs += ["Google"]

for item in someStrs {
   println(item)
}

 

当上述代码被编译和执行时,它产生了以下结果:

Apple
Amazon
Google

 

也可以使用 enumerate() 函数,如下面的例子所示,它返回索引及对应的值:

import Cocoa

var someStrs = [String]()

someStrs.append("Apple")
someStrs.append("Amazon")
someStrs += ["Google"]

for (index, item) in enumerate(someStrs) {
   println("Value at index = \(index) is \(item)")
}

 

当上述代码被编译和执行时,它产生了以下结果:

Value at index = 0 is Apple
Value at index = 1 is Amazon
Value at index = 2 is Google

  

两个数组相加

使用加法运算符(+),以添加的相同类型的数组,这将产生新的数组是来自两个数组值相加组合后的数组,如下:

import Cocoa

var intsA = [Int](count:2, repeatedValue: 2)
var intsB = [Int](count:3, repeatedValue: 1)

var intsC = intsA + intsB

for item in intsC {
   println(item)
}

  

当上述代码被编译和执行时,它产生了以下结果:

2
2
1
1
1

  

count 属性

可以使用只读计算 (count) 数组属性,找出下面显示出数组中元素的个数:

import Cocoa

var intsA = [Int](count:2, repeatedValue: 2)
var intsB = [Int](count:3, repeatedValue: 1)

var intsC = intsA + intsB

println("Total items in intsA = \(intsA.count)")
println("Total items in intsB = \(intsB.count)")
println("Total items in intsC = \(intsC.count)")

  

当上述代码被编译和执行时,它产生了以下结果:

Total items in intsA = 2
Total items in intsB = 3
Total items in intsC = 5

 

空属性

使用只读数组的空属性(isEmpty)找出一个数组是否为空,如下图所示:

import Cocoa

var intsA = [Int](count:2, repeatedValue: 2)
var intsB = [Int](count:3, repeatedValue: 1)
var intsC = [Int]()

println("intsA.isEmpty = \(intsA.isEmpty)")
println("intsB.isEmpty = \(intsB.isEmpty)")
println("intsC.isEmpty = \(intsC.isEmpty)")

 

当上述代码被编译和执行时,它产生了以下结果:

intsA.isEmpty = false
intsB.isEmpty = false
intsC.isEmpty = true

  

 

 

转载于:https://www.cnblogs.com/liuShanPei1024/p/5350195.html