斯威夫特迭代字符長度

我在我的第一應用的一個過程，它的目的是作爲一個密碼生成器，並告訴人們在1-1000規模有多難猜測，並有多難基於字母如何格式化，它看起來像什麼以及大腦如何記憶模式。到目前爲止，我有所有我想要在一個陣列中使用的字符，然後我有一個在循環，通過人物迭代，但我無法弄清楚如何指定密碼生成的長度，因爲目前它只是打印每個字符。於是，我問我怎樣才能使一個8個字符長的密碼生成器儘可能簡單，我至今是：斯威夫特迭代字符長度

import Foundation 


    let chars = ["a","b","c","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u"  ,"v","w","x","y","z","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","0","1","2","3","4","5","6","7","8","9"]

謝謝！

var generate: String 

    for generate in chars { 
     print(generate) 
    }

來源

2016-07-04 RufusV

儘可能簡單地用一個循環，並和隨機數在給定的索引來獲取字符：

let length = 8 
var pass = "" 
for _ in 0..<length { 
    let random = arc4random_uniform(UInt32(chars.count)) 
    pass += chars[Int(random)] 
} 

print(pass)

來源

2016-07-04 15:17:57 vadian

會不會有你知道的方式，通過所有可能的密碼進行迭代訂購？我只是想確保相同的密碼永遠不會生成兩次。儘管謝謝您的幫助！ – RufusV

@RufusV：你知道可以用8個字母數字字符創建多少個密碼嗎？答案是218340105584896，即使你創造10.000每秒然後再遍歷所有這些都需要多年... –

爲了這個目的，我建議使用散列函數 – vadian

斯威夫特迭代字符長度

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