ios - 快速迭代字符长度

Question

我正在制作我的第一个应用程序，它的目的是作为一个密码生成器，并以 1-1000 的范围告诉人们猜测有多难，以及记住有多难关于字母的格式、外观以及大脑如何记住模式。到目前为止，我拥有想要在数组中使用的所有字符，然后我有一个 for in 循环来迭代这些字符，但我无法弄清楚如何指定要生成的密码的长度，因为目前它只是打印每个字符。所以，我问如何才能尽可能简单地制作一个 8 个字符长的密码生成器，到目前为止我所拥有的是:

    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"]

谢谢!

    var generate: String

    for generate in chars {
        print(generate)
    }

Answer 1

尽可能简单地使用循环和随机数来获取给定索引处的字符:

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

print(pass)