@ -342,8 +342,8 @@ numbers.map({
|
||||
有很多种创建更简洁的闭包的方法。如果一个闭包的类型已知,比如作为一个回调函数,你可以忽略参数的类型和返回值。单个语句闭包会把它语句的值当做结果返回。
|
||||
|
||||
```swift
|
||||
let mappedNumbers = numbers.map({ number in 3 * number })
|
||||
print(mappedNumbers)
|
||||
let sortedNumbers = numbers.sort { $0 > $1 }
|
||||
print(sortedNumbers)
|
||||
```
|
||||
|
||||
你可以通过参数位置而不是参数名字来引用参数——这个方法在非常短的闭包中非常有用。当一个闭包作为最后一个参数传给一个函数的时候,它可以直接跟在括号后面。当一个闭包是传给函数的唯一参数,你可以完全忽略括号。
|
||||
|
||||
@ -631,12 +631,12 @@ if let constantName = someOptional {
|
||||
你可以像上面这样使用可选绑定来重写`possibleNumber`这个例子:
|
||||
|
||||
```swift
|
||||
if let actualNumber = possibleNumber.toInt() {
|
||||
print("\(possibleNumber) has an integer value of \(actualNumber)")
|
||||
if let actualNumber = Int(possibleNumber) {
|
||||
print("\'\(possibleNumber)\' has an integer value of \(actualNumber)")
|
||||
} else {
|
||||
print("\(possibleNumber) could not be converted to an integer")
|
||||
print("\'\(possibleNumber)\' could not be converted to an integer")
|
||||
}
|
||||
// 输出 "123 has an integer value of 123"
|
||||
// 输出 "'123' has an integer value of 123"
|
||||
```
|
||||
|
||||
这段代码可以被理解为:
|
||||
|
||||
Reference in New Issue
Block a user