swift - UnsafeMutablePointer 的奇怪行为

Question

我构建了一个小结构，其中包含伪 C 字符串 UnsafeMutablePointer。现在我想添加一种方法来获取字符串的长度，但有些行为非常奇怪:

struct Char {
    let value: CChar

    init(_ char: CChar) {
        self.value = char
    }
}

extension Char: CustomStringConvertible {
    var description: String {
        return String(UnicodeScalar(UInt8(bitPattern: self.value)))
    }
}

extension Char: Equatable {
    public static func ==(lhs: Char, rhs: Char) -> Bool {
        return lhs.value == rhs.value
    }
}

struct CharPointer {
    let pointer: UnsafeMutablePointer<Char>

    init(_ string: String) {
        let chars = string.cString(using: .utf8)!.map { Char($0) }
        self.pointer = UnsafeMutablePointer(mutating: chars)
    }

    subscript(_ offset: Int) -> Char {
        get { return self.pointer[offset] }
        set { self.pointer[offset] = newValue }
    }

    var length: Int {
        var ptr = self.pointer

        while ptr.pointee != Char(0) {
            ptr += 1
        }

        return ptr - self.pointer
    }
}

let pointer = CharPointer("Hello World!")

print(pointer[0])

// print(pointer.length)

如果我不使用 length 属性，一切都会正常工作，并且 print 语句会打印“H”。但是，如果我尝试访问 pointer.length，它会打印一个换行符，尽管我在 print 语句之后使用了该属性，并且长度为 0。这是一个错误还是我犯了一个错误？

Answer 1

感谢@Hamish，我找到了一个解决方案，我只需使用数组中的指针初始化第二个指针:

init(_ string: String) {
    let chars = string.cString(using: .utf8)!.map { Char($0) }
    let pointer = UnsafeMutablePointer(mutating: chars)
    let pointer2 = UnsafeMutablePointer<Char>.allocate(capacity: chars.count)
    pointer2.initialize(from: pointer, count: chars.count)

    self.pointer = pointer2
}