构造函数调用的可选链接？(Optional chaining for constructor calls?)

构造函数调用的可选链接？(Optional chaining for constructor calls?)

我想选择施放NSNumber? 到Int? ，但Int的初始化方法只接受init(NSNumber) ，所以我无法传递NSNumber? 。

有没有办法让我压缩这段代码，以便它使用像可选链接这样的东西？

// number: NSNumber? let integer = number == nil ? nil : Int(number!)

I want to optionally cast an NSNumber? to an Int?, but the initialiser method for Int only takes init(NSNumber), so I can't pass an NSNumber?.

Is there a way for me to compact this code so that it uses something like optional chaining?

// number: NSNumber? let integer = number == nil ? nil : Int(number!)

最满意答案

Int构造函数不接受可选参数。 你可以将结构“包装”到map() ：

let number : NSNumber? = ... let n = number.map { Int($0) } // `n` is an `Int?`

但是在这里使用NSNumber的integerValue属性和可选链接更容易：

let n = number?.integerValue // `n` is an `Int?`

要不就

let n = number as? Int // `n` is an `Int?`

因为Swift会自动在NSNumber和Int之间“桥接”。

The Int constructors don't take optional arguments. You could "wrap" the construction into map():

let number : NSNumber? = ... let n = number.map { Int($0) } // `n` is an `Int?`

But here it is easier to use the integerValue property of NSNumber with optional chaining:

let n = number?.integerValue // `n` is an `Int?`

or just

let n = number as? Int // `n` is an `Int?`

since Swift "bridges" between NSNumber and Int automatically.