在Swift 3.0中生成随机字节
我想在Swift 3.0中使用SecRandomCopyBytes
生成随机字节。 这是我在Swift 2.2中做的
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,我试图这样做,因为我知道unsafemutablebytes的概念现在是不同的,但它不允许我返回。 如果我注释掉返回部分,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
你很近,但是return
里面的闭包,而不是从外层函数返回。 因此只有SecRandomCopyBytes()
应该在闭包中被调用,并且结果被传回。
func generateRandomBytes() -> String? { var keyData = Data(count: 32) let result = keyData.withUnsafeMutableBytes { (mutableBytes: UnsafeMutablePointer<UInt8>) -> Int32 in SecRandomCopyBytes(kSecRandomDefault, keyData.count, mutableBytes) } if result == errSecSuccess { return keyData.base64EncodedString() } else { print("Problem generating random bytes") return nil } }
对于“单expression式闭包”,闭包types可以自动推断出来,所以这可以缩短为
func generateRandomBytes() -> String? { var keyData = Data(count: 32) let result = keyData.withUnsafeMutableBytes { SecRandomCopyBytes(kSecRandomDefault, keyData.count, $0) } if result == errSecSuccess { return keyData.base64EncodedString() } else { print("Problem generating random bytes") return nil } }