Question

arr1 = ["h","e","l","l","o"] as! Any

for i in 0..<arr1.count
{
    let newString:String = arr1[i] as! String

    let data1:UInt8 = UInt8(newString)!

    arr[i] = data1
}

Answer 1

将一个字符数组转换为UInt8并没有多大意义，因为有些字符不能仅用8位表示，例如＆＃34;＆＃34;

这是你将字符转换为[UInt32]的方式：

let chars: [Character] = ["a", "", "c"]
let result = chars.map { UnicodeScalar($0.description)?.value }.filter{ $0 != nil }

或者，您可以使用此方法将字符转换为UInt8，但过滤掉那些无法用8位表示的字符：

let chars: [Character] = ["a", "", "c"]
let result = chars.map { UnicodeScalar($0.description) }.filter { $0 != nil && $0!.isASCII }.map { UInt8($0!.value) }

Answer 2

可以通过String代表Character转换UInt8来实现您可以使用String.UTF8View，这是一种集合类型，现在您可以使用UInt8轻松转换为map。如

let charArray: [Character] = ["h","e","l","l","o"]
let arrayUInt8 = String(charArray).utf8.map { (val) -> UInt8 in
    return val
}

print(charArray)
//["h", "e", "l", "l", "o"]

print(arrayUInt8)
//[104, 101, 108, 108, 111]

如何在swift 3中将char数组转换为uint8数组？

2 个答案: