Question

我想在Swift 3.0中使用SecRandomCopyBytes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。如果我注释掉了返回部分，它仍会显示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个问题？

由于

Answer 1

你很接近，但是关闭内的return返回从封闭，而不是从外部功能。因此，只应调用SecRandomCopyBytes() 关闭，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”，闭包类型可以推断出来自动，所以这可以缩短为

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
    }
}

Swift 5更新：

func generateRandomBytes() -> String? {

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

Answer 2

根据Apple Documentation，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
        }
    }
}

Answer 3

这是使用Swift 5实现功能的最简单，最快捷的方式：

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，在Swift中，最好的做法是在函数的控制流取决于表达式的成功或失败或是否存在非null值时，使用保护语句而不是if / else语句。

在Swift中使用SecRandomCopyBytes

3 个答案: