苹果Mac Pro垃圾桶 最低配的ME253CH
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Thank you very much.
Since you have learned Swift (编程语言) more than a couple of years already, what suggestions and guidance will you give me for learning the language?
My first and current reading is [The Swift Programming Language (Swift 3.1)] with the help of the IBM Swift Sandbox interpreter. What else will you suggest?
后续网站会更新一些视频教程，希望能够对你有用。
后续网站会更新一些视频教程，希望能够对你有用。
I tried the sample code in [Subscript Options] of [The Swift Programming Language (Swift 3.1)], I had compilation errors.
Can you help debugging ?
struct Matrix {
let rows: Int, columns: Int
var grid: [Double]
init(rows: Int, columns: Int) {
self.rows = rows
self.columns = columns
grid = Array(repeating: 0.0, count: rows * columns)
}
func indexIsValid(row: Int, column: Int) -> Bool {
return row >= 0 && row < rows && column >= 0 && column < columns
}
subscript(row: Int, column: Int) -> Double {
get {
assert(indexIsValid(row: row, column: column), "Index out of range")
return grid[(row * columns) + column]
}
set {
assert(indexIsValid(row: row, column: column), "Index out of range")
grid[(row * columns) + column] = newValue
}
}
}
var matrix = Matrix(rows: 2, columns: 2)
matrix[0, 1] = 1.5
matrix[1, 0] = 3.2
func indexIsValidForRow(row: Int, column: Int) -> Bool {
return row >= 0 && row < rows && column >= 0 && column < columns
}
let someValue = matrix[2, 2]
// this triggers an assert, because [2, 2] is outside of the matrix bounds
@Daniel 之前用Xcode7.1编译出错，今天装了个Xcode8.3，indexIsValidForRow函数放到Matrix结构体中就可以了。修改的代码如下：
struct Matrix {
let rows: Int, columns: Int
var grid: [Double]
init(rows: Int, columns: Int) {
self.rows = rows
self.columns = columns
grid = Array(repeating: 0.0, count: rows * columns)
}
func indexIsValid(row: Int, column: Int) -> Bool {
return row >= 0 && row < rows && column >= 0 && column < columns
}
subscript(row: Int, column: Int) -> Double {
get {
assert(indexIsValid(row: row, column: column), "Index out of range")
return grid[(row * columns) + column]
}
set {
assert(indexIsValid(row: row, column: column), "Index out of range")
grid[(row * columns) + column] = newValue
}
}
func indexIsValidForRow(row: Int, column: Int) -> Bool {
return row >= 0 && row < rows && column >= 0 && column < columns
}
}
var matrix = Matrix(rows: 2, columns: 2)
matrix[0, 1] = 1.5
matrix[1, 0] = 3.2
let someValue = matrix[1, 1]
// let someValue = matrix[2, 2]
// this triggers an assert, because [2, 2] is outside of the matrix bounds
You are great!
Recently I bought 3 electronic books, but I will wait to read them until I purchase a new Mac Pro next year when it is available.
Also I need to spend about RMB50,000 to fix my teeth this year.
dkychang@qq.com
Functions are a first-class type. This means that a function can return another function as its value:
func makeIncrementer() -> ((Int) -> Int) {
func addOne(number: Int) -> Int {
return 1 + number
}
return addOne
}
var increment = makeIncrementer()
print(increment(7))
Swift is not easy, I don't understand the above code. Do you?
//返回addOne这个函数
func makeIncrementer() -> ((Int) -> Int) {
func addOne(number: Int) -> Int {
return 1 + number
}
return addOne
}
//increment存放makeIncrementer()返回的函数，即increment就是addOne()
var increment = makeIncrementer()
//实际就是调用了addOne()，输出结果为8
print(increment(7))
I still don't understand.
Why increment can take an argument?
makeIncrementer () takes NO argument.
因为increment是addOne，addOne是带参数的；makeIncrementer()定义时就时没有参数。
你不要把increment当成makeIncrementer()，increment只是makeIncrementer()返回的一个函数。在Java中可以理解为increment是makeIncrementer()返回的一个对象，区别就是在于在Java中函数不是class，而Swift中函数也算是class，所以在Swift中可以把一个函数作为返回值，也可以作为参数。
func anyCommonElements<T: Sequence, U: Sequence>(_ lhs: T, _ rhs: U) -> Bool
where T.Iterator.Element: Equatable, T.Iterator.Element == U.Iterator.Element {
for lhsItem in lhs {
for rhsItem in rhs {
if lhsItem == rhsItem {
return true
}
}
}
return false
}
anyCommonElements([1, 2, 3], [3]) Another piece of code that I don't understand, maybe you can explain?
//这个函数是用于判断两个数组中是否有相同元素
//<T: Sequence, U: Sequence> 这里用了泛型，指定了参数T和U的类型必须实现Sequence协议，即参数必须是数组或集合等可遍历的
func anyCommonElements<T: Sequence, U: Sequence>(_ lhs: T, _ rhs: U) -> Bool
//where子句是对泛型对参数T和U做了约束，调用函数时传递的参数达到以下2个条件才行
//T.Iterator.Element: Equatable 参数T必须实现Equatable协议，实现Equatable协议的类型才可以使用 == 和 != 进行比较
//T.Iterator.Element == U.Iterator.Element 这句要求T和U的类型必须是相同的，例：要么都是字符，要么都是数字。
where T.Iterator.Element: Equatable, T.Iterator.Element == U.Iterator.Element {
//这里就是循环对比内容了，一旦发现有相同的元素则返回true
for lhsItem in lhs {
for rhsItem in rhs {
if lhsItem == rhsItem {
return true
}
}
}
return false
}
//这里结果是true，因为两个数组中都有3
anyCommonElements([1, 2, 3], [3, 4])
//上面调用时参数时用的数字，下面例子使用的字符，其它实现Equatable协议的类型也可以使用。
//这里结果是false
anyCommonElements(["1", "2"], ["3", "7"])
//总结：泛型可以使同一个函数，支持多种类型，而不需要为不同类型分别写一个函数。
关于泛型可以参考：泛型 - Swift 3.0 教程
I know the results of the examples because I tried the IBM compiler. But the terms (Sequence, Iterator and Equatable) have never been explained in the document [The Swift Programming Language (Swift 3.1)]?
Have you read other documents?