Question

How to assign a random number using a cryptographically secure random number generator in Swift

Answer 1

import Foundation
import Security

/// Generates a cryptographically secure random number within a given range.
func secureRandomNumber(in range: ClosedRange<Int>) -> Int? {
    // Ensure the range is valid
    guard range.lowerBound < range.upperBound else {
        print("Invalid range: \(range)")
        return nil
    }
    
    // Calculate the range size
    let rangeSize = range.upperBound - range.lowerBound + 1
    
    // Determine the number of bytes needed to represent the range
    let byteCount = MemoryLayout<UInt32>.size
    var randomBytes = [UInt8](repeating: 0, count: byteCount)
    
    // Generate random bytes
    let result = SecRandomCopyBytes(kSecRandomDefault, byteCount, &randomBytes)
    if result != errSecSuccess {
        print("Failed to generate random bytes: \(result)")
        return nil
    }
    
    // Convert the bytes to a UInt32
    let randomValue = randomBytes.withUnsafeBytes { pointer -> UInt32 in
        return pointer.load(as: UInt32.self)
    }
    
    // Map the random value to the desired range
    let randomInRange = Int(randomValue % UInt32(rangeSize)) + range.lowerBound
    return randomInRange
}

if let randomNumber = secureRandomNumber(in: 1...100) {
    print("Secure random number: \(randomNumber)")
} else {
    print("Failed to generate a secure random number.")
}



/*
run:

Secure random number: 14

*/

 

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