Swift语言：如何调用SecRandomCopyBytes函数

Question

从 Objective-C，我可以做到这一点：

NSMutableData *data = [NSMutableData dataWithLength:length];
int result = SecRandomCopyBytes(kSecRandomDefault, length, data.mutableBytes);

试图在Swift中进行此操作时，我有以下代码：

let data = NSMutableData(length: Int(length))
let result = SecRandomCopyBytes(kSecRandomDefault, length, data.mutableBytes)

但我遇到了这个编译器错误：

'Void' is not identical to 'UInt8'

由于类型不匹配，数据的mutableBytes参数被拒绝，但我无法强制转换该参数（并且我假设这样做是安全的）。

- Daniel

Answer 1

这似乎是有效的：

let data = NSMutableData(length: Int(length))
let result = SecRandomCopyBytes(kSecRandomDefault, length, UnsafeMutablePointer<UInt8>(data.mutableBytes))

- Daniel

Answer 2

Swift 5

let count: Int = <byteCount>
var data = Data(count: count)
let result = data.withUnsafeMutableBytes {    
    SecRandomCopyBytes(kSecRandomDefault, count, $0.baseAddress!) 
}

Swift 4:

var data = Data(count: <count>)
let result = data.withUnsafeMutableBytes { mutableBytes in
    SecRandomCopyBytes(kSecRandomDefault, data.count, mutableBytes)
}

- Ondrej Stocek

---

你可以使用 nil 代替 kSecRandomDefault。 - Sulthan

Swift 5 会产生一个错误 Overlapping accesses to 'data', but modification requires exclusive access; consider copying to a local variable。解决方案是用你传递的 <count> 替换 data.count。 - Andy Ibanez

感谢您的提醒，@AndyIbanez。我已经改进了Swift 5的答案。 - Ondrej Stocek

Answer 3

Swift 4版本：

let count = 16
var data = Data(count: count)
_ = data.withUnsafeMutableBytes {
    SecRandomCopyBytes(kSecRandomDefault, count, $0)
}

- sundance