Create a Secure Random number in iOS using Swift?

Question

public func createSecureRandomKey(numberOfBits: Int) -> Any {
    let attributes: [String: Any] =
        [kSecAttrKeyType as String:CFString.self,
         kSecAttrKeySizeInBits as String:numberOfBits]

    var error: Unmanaged<CFError>?
    guard let privateKey = SecKeyCreateRandomKey(attributes as CFDictionary, &error) else {
        return ""
    }
    return privateKey
}

I am trying to create Secure random number like above way, but returning nothing, Could any one please help me. Thanks.

have you tried printing error?

user234461
– user234461

2018-06-07 17:16:02 +00:00
Commented Jun 7, 2018 at 17:16 — user234461
– user234461, Commented Jun 7, 2018 at 17:16
Yes.. But No error displyed, always going to return ""

Satish Balaga
– Satish Balaga

2018-06-07 17:34:05 +00:00
Commented Jun 7, 2018 at 17:34 — Satish Balaga
– Satish Balaga, Commented Jun 7, 2018 at 17:34
Add print(error) inside the guard.

rmaddy
– rmaddy

2018-06-07 21:21:04 +00:00
Commented Jun 7, 2018 at 21:21 — rmaddy
– rmaddy, Commented Jun 7, 2018 at 21:21

Biba · Accepted Answer · 2020-02-03 12:33:03Z

2

It looks like you are using the wrong function. With your function you are generating a new key. But as your title says you want to generate secure random numbers. For this there is a function called: SecRandomCopyBytes(::_:)

Here is a code snippet taken from the official apple documentation how to use it:

var bytes = [Int8](repeating: 0, count: 10)
let status = SecRandomCopyBytes(kSecRandomDefault, bytes.count, &bytes)

if status == errSecSuccess { // Always test the status.
    print(bytes)
    // Prints something different every time you run.
}

Source: Apple doc

answered Feb 3, 2020 at 12:33

Biba

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