在Swift 3.0中生成随机字节

Question

我想SecRandomCopyBytes在Swift 3.0中使用生成随机字节.这是我在Swift 2.2中的表现

private static func generateRandomBytes() -> String? {
    let data = NSMutableData(length: Int(32))

    let result = SecRandomCopyBytes(kSecRandomDefault, 32, UnsafeMutablePointer<UInt8>(data!.mutableBytes))
    if result == errSecSuccess {
        return data!.base64EncodedString(options: NSData.Base64EncodingOptions(rawValue: 0))
    } else {
        print("Problem generating random bytes")
        return nil
    }
}

在Swift 3中,我尝试这样做,因为我知道unsafemutablebytes的概念现在不同了,但它不允许我返回.如果我评论退货部分,它仍然说Generic Parameter ResultType could not be inferred

fileprivate static func generateRandomBytes() -> String? {
    var keyData = Data(count: 32)
    _ = keyData.withUnsafeMutableBytes {mutableBytes in
        let result = SecRandomCopyBytes(kSecRandomDefault, keyData.count, mutableBytes)
        if result == errSecSuccess {
            return keyData.base64EncodedString(options: NSData.Base64EncodingOptions(rawValue: 0))
        } else {
            print("Problem generating random bytes")
            return nil
        }
    }
    return nil
}

有谁知道如何解决这一问题？

谢谢

Answer 1

你很接近,但是return在闭包内部从闭包返回,而不是从外部函数返回.因此只SecRandomCopyBytes()应在闭包中调用,并将结果传回.

func generateRandomBytes() -> String? {

    var keyData = Data(count: 32)
    let result = keyData.withUnsafeMutableBytes {
        (mutableBytes: UnsafeMutablePointer<UInt8>) -> Int32 in
        SecRandomCopyBytes(kSecRandomDefault, 32, mutableBytes)
    }
    if result == errSecSuccess {
        return keyData.base64EncodedString()
    } else {
        print("Problem generating random bytes")
        return nil
    }
}

对于"单表达式闭包",闭包类型可以自动推断,因此可以缩短为

func generateRandomBytes() -> String? {

    var keyData = Data(count: 32)
    let result = keyData.withUnsafeMutableBytes {
        SecRandomCopyBytes(kSecRandomDefault, 32, $0)
    }
    if result == errSecSuccess {
        return keyData.base64EncodedString()
    } else {
        print("Problem generating random bytes")
        return nil
    }
}

Answer 2

这是使用 Swift 5 实现函数的最简单和“最快捷”的方法：

func generateRandomBytes() -> String? {
    var bytes = [UInt8](repeating: 0, count: 32)
    let result = SecRandomCopyBytes(kSecRandomDefault, bytes.count, &bytes)

    guard result == errSecSuccess else {
        print("Problem generating random bytes")
        return nil
    }

    return Data(bytes).base64EncodedString()
}

通常，当函数的控制流取决于表达式的成功或失败或非 nil 值的存在时，Swift 中的最佳实践是使用保护语句而不是 if/else 语句。

Answer 3

根据Apple文档，它看起来类似于：

public func randomData(ofLength length: Int) throws -> Data {
    var bytes = [UInt8](repeating: 0, count: length)
    let status = SecRandomCopyBytes(kSecRandomDefault, length, &bytes)
    if status == errSecSuccess {
        return Data(bytes: bytes)
    }
    // throw an error
}

或作为其他初始化程序：

public extension Data {
    public init(randomOfLength length: Int) throws {
        var bytes = [UInt8](repeating: 0, count: length)
        let status = SecRandomCopyBytes(kSecRandomDefault, length, &bytes)
        if status == errSecSuccess {
            self.init(bytes: bytes)
        } else {
            // throw an error
        }
    }
}