在Apple Swift中連接NSString和Int

我已經使用arc4random_uniform（）從給定的String或Int中獲取隨機值。現在我想加入那些價值觀。在Apple Swift中連接NSString和Int

var randomNumber = arc4random_uniform(9999) + 1000 // returns random number between 1000 and 9999 e.g. - 7501 
let alphabet_array = ["A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"] 
let randomIndex = Int(arc4random_uniform(UInt32(alphabet_array.count))) 
let random_alphabet = alphabet_array[randomIndex] // returns random alphabet e.g. - E

我想是顯示7501和E在一起，就像7501E

var str="\(randomNumber)\(random_alphabet)" // error

我得到以下錯誤：

Command /Applications/Xcode6-Beta.app/Contents/Developer/Toolchains/XcodeDefault.xctoolchain/usr/bin/swift failed with exit code 254

請幫助我如何顯示此一起。

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2014-06-24 Funny

它工作正常爲了我。嘗試重新啓動Xcode並重試。 – Greg

你真的打算1000和10999之間，以獲得數這樣... –

假設你的代碼是逐字，你在第二行缺少「=」「讓alphabet_array [‘A’」否則它作爲@Nate Cook說 – Grimxn

它正常工作與正確的標點符號：

var randomNumber = arc4random_uniform(999) + 1000 
// returns random number between 1000 and 9999 e.g. - 7501 NOTE 999 not 9999 

let alphabet_array = ["A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"] 
// Note missing "=" in OP 

let randomIndex = Int(arc4random_uniform(UInt32(alphabet_array.count))) 

let random_alphabet = alphabet_array[randomIndex] // returns random alphabet e.g. - E 

var str = "\(randomNumber)\(random_alphabet)" // e.g. 1197A

來源

2014-06-24 08:04:55 Grimxn

而不是使用字符串插值的，你也可以隨機數轉換爲字符串，然後把它們相加：

let str = randomNumber.description + random_alphabet

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2014-06-24 08:08:35

在Apple Swift中連接NSString和Int

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