問(wèn)題描述

我只想將 character 轉(zhuǎn)換為 Int.

I just want to convert a character into an Int.

這應(yīng)該很簡(jiǎn)單.但我還沒(méi)有發(fā)現(xiàn)以前的答案有幫助.總是有一些錯(cuò)誤.也許是因?yàn)槲艺?Swift 2.0 中嘗試它.

This should be simple. But I haven't found the previous answers helpful. There is always some error. Perhaps it is because I'm trying it in Swift 2.0.

for i in (unsolved.characters) {
  fileLines += String(i).toInt()
  print(i)
}

推薦答案

在 Swift 2.0 中，toInt() 等已被替換為初始化器.(在這種情況下，Int(someString).)

In Swift 2.0, toInt(), etc., have been replaced with initializers. (In this case, Int(someString).)

因?yàn)椴皇撬械淖址伎梢赞D(zhuǎn)換成int，所以這個(gè)初始化器是failable的，也就是說(shuō)它返回一個(gè)可選的int(Int?)而不僅僅是一個(gè) Int.最好的辦法是使用 if let 解開這個(gè)可選項(xiàng).

Because not all strings can be converted to ints, this initializer is failable, which means it returns an optional int (Int?) instead of just an Int. The best thing to do is unwrap this optional using if let.

我不確定你到底想要什么，但這段代碼在 Swift 2 中工作，并完成了我認(rèn)為你正在嘗試做的事情:

I'm not sure exactly what you're going for, but this code works in Swift 2, and accomplishes what I think you're trying to do:

let unsolved = "123abc"

var fileLines = [Int]()

for i in unsolved.characters {
    let someString = String(i)
    if let someInt = Int(someString) {
        fileLines += [someInt]
    }
    print(i)
}

或者，對(duì)于更快捷的解決方案:

Or, for a Swiftier solution:

let unsolved = "123abc"

let fileLines = unsolved.characters.filter({ Int(String($0)) != nil }).map({ Int(String($0))! })

// fileLines = [1, 2, 3]

您可以使用 flatMap 進(jìn)一步縮短它:

You can shorten this more with flatMap:

let fileLines = unsolved.characters.flatMap { Int(String($0)) }

flatMap 返回一個(gè) Array，其中包含將 transform 映射到 self 的非零結(jié)果"……所以當(dāng) Int(String($0)) 為 nil 時(shí)，結(jié)果被丟棄.

flatMap returns "an Array containing the non-nil results of mapping transform over self"… so when Int(String($0)) is nil, the result is discarded.

這篇關(guān)于在 Swift 2.0 中將字符轉(zhuǎn)換為 Int的文章就介紹到這了，希望我們推薦的答案對(duì)大家有所幫助，也希望大家多多支持html5模板網(wǎng)！