Secure generation of random numbers and dynamic initialization vectors
import Foundation
import CommonCrypto
let key = "1234567890123456" // 16 bytes for AES128
let iv = "abcdefghijklmnop" // 16 bytes for AES128
func encryptAES(data: Data) -> Data {
let cryptLength = size_t(kCCBlockSizeAES128 + data.count + kCCKeySizeAES128)
var cryptData = Data(count:cryptLength)
let keyLength = size_t(kCCKeySizeAES128)
let options = CCOptions(kCCOptionPKCS7Padding)
var numBytesEncrypted :size_t = 0
let cryptStatus = cryptData.withUnsafeMutableBytes {cryptBytes in
data.withUnsafeBytes {dataBytes in
CCCrypt(CCOperation(kCCEncrypt),
CCAlgorithm(kCCAlgorithmAES),
options,
key, keyLength,
iv,
dataBytes, data.count,
cryptBytes, cryptLength,
&numBytesEncrypted)
}
}
if UInt32(cryptStatus) == UInt32(kCCSuccess) {
cryptData.removeSubrange(numBytesEncrypted..
In this Swift code, the function
encryptAES(data: Data) -> Data
is used to encrypt data using AES encryption. The encryption key and initialization vector (IV) are hardcoded as
key
and
iv
respectively. This is a security vulnerability as the IV should be random and unique for each encryption process.
By using a static, hardcoded IV, it becomes possible for an attacker to predict the output of the encryption. If they have access to both the encrypted data and the IV, they can perform a reverse engineering attack to predict the random values that should remain secret.
The vulnerability lies in this line of code:
let iv = "abcdefghijklmnop"
. This is where the IV is hardcoded instead of being randomly generated. This could allow an attacker to predict the random values that should remain secret, compromising the security of the encryption.
To fix this vulnerability, the IV should be randomly generated for each encryption process, ensuring that even if an attacker has access to the encrypted data, they cannot predict the random values used in the encryption.
import Foundation
import CommonCrypto
let key = "1234567890123456" // 16 bytes for AES128
func generateRandomIV(for length: Int) -> String {
let letters = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
return String((0.. Data {
let cryptLength = size_t(kCCBlockSizeAES128 + data.count + kCCKeySizeAES128)
var cryptData = Data(count:cryptLength)
let keyLength = size_t(kCCKeySizeAES128)
let options = CCOptions(kCCOptionPKCS7Padding)
var numBytesEncrypted :size_t = 0
let iv = generateRandomIV(for: 16) // 16 bytes for AES128
let cryptStatus = cryptData.withUnsafeMutableBytes {cryptBytes in
data.withUnsafeBytes {dataBytes in
CCCrypt(CCOperation(kCCEncrypt),
CCAlgorithm(kCCAlgorithmAES),
options,
key, keyLength,
iv,
dataBytes, data.count,
cryptBytes, cryptLength,
&numBytesEncrypted)
}
}
if UInt32(cryptStatus) == UInt32(kCCSuccess) {
cryptData.removeSubrange(numBytesEncrypted..
The updated code now includes a function
generateRandomIV(for length: Int) -> String
that generates a random initialization vector (IV) for AES encryption. This function uses a secure random number generator to generate a random string of the specified length. The generated string is composed of alphanumeric characters (both lowercase and uppercase) and digits.
The
encryptAES(data: Data) -> Data
function has been updated to use this dynamically generated IV instead of a static, hardcoded value. The IV is generated for each encryption operation, ensuring its uniqueness.
The IV is generated with a length of 16 bytes, which is the block size for AES128 encryption. The generated IV is passed to the
CCCrypt
function as the IV parameter.
This approach mitigates the risk of an attacker being able to predict the IV and perform a reverse engineering attack. The IV is unique for each encryption operation and is not hardcoded, making it unpredictable.
Please note that the generated IV needs to be stored securely as it will be needed for the decryption process. The storage method is not shown in this code and should be implemented according to the security requirements of your application.