如何在Swift中将Int转换为字符

你不能直接转换为整数的Character情况下,但你可以去从整数UnicodeScalar到Character,然后再返回:

let startingValue = Int(("A" as UnicodeScalar).value) // 65
for i in 0 ..< 26 {
    print(Character(UnicodeScalar(i + startingValue)))
}

试试这个

for i in 0...25
{
    let string = String(format: "%c", i+65) as String
    NSLog("%@", string)
}

到目前为止,我已经想出了这个:

for i in 0 ..< 26 {
    print(Character(UnicodeScalar(Int(UnicodeScalar("A").value) + i)))
}

如果您只是尝试将"A"生成为"Z",则可以避免数学运算,只需:

for c in UnicodeScalar("A").value...UnicodeScalar("Z").value {
    print(String(UnicodeScalar(c)))
}

为了未来的访问者,我提供问题标题的基本答案,而不是问题本身的细节.

这是一个两步的过程.转换Int为a UnicodeScalar,然后将其转换UnicodeScalar为a Character.

let myInteger: Int = 97

// convert Int to a valid UnicodeScalar
guard let myUnicodeScalar = UnicodeScalar(myInteger) else {
    return
}

// convert UnicodeScalar to Character
let myCharacter = Character(myUnicodeScalar)

// results
print(myCharacter) // a

(资源)

或者......

if let myUnicodeScalar = UnicodeScalar(97) 
    let myCharacter = Character(myUnicodeScalar)
}

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