How to generate Random numbers using SecureRandom

Question

I have the below code which uses java.util.Random, now I want to change to java.security.SecureRandom. How do I do the same operations with SecurRandom?

int rand = 0;
for (int i = 0; i < 8; i++) {
    rand = (int) (Math.random() * 3);
    switch (rand) {
        case 0:
            c = '0' + (int) (Math.random() * 10);
            break;
        case 1:
            c = 'a' + (int) (Math.random() * 26);
            break;
        case 2:
            c = 'A' + (int) (Math.random() * 26);
            break;
    }
}

Take a took at [Generate secure random number uniformly over a range in Java](https://stackoverflow.com/questions/28742702/generate-secure-random-number-uniformly-over-a-range-in-java) — Mathjoh, Apr 23 '18 at 10:56

score 3 · Answer 1 · answered Apr 23 '18 at 10:55

3

SecureRandom random = SecureRandom.getInstance("SHA1PRNG");
int rand = 0;

for (int i = 0; i < 8; i++) {
    rand = random.nextInt(3);
    switch (rand) {
    case 0:
        c = '0' + random.nextInt(10);
        break;
    case 1:
        c = 'a' + random.nextInt(26);
        break;
    case 2:
        c = 'A' + random.nextInt(26);
        break;
    }

answered Apr 23 '18 at 10:55

Maurice Perry

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I didn't mean to say it was you. But left the comment in case the downvoter sees this comment on top. – Shanu Gupta Apr 23 '18 at 12:22

score 2 · Accepted Answer · edited Apr 23 '18 at 11:09

2

secureRandomObject.nextDouble() is essentially equivalent to Math.random()

So following code will work...

SecureRandom secureRandom = new SecureRandom();
int rand = 0;
for (int i = 0; i < 8; i++) {
    rand = (int) (secureRandom.nextDouble() * 3);
    switch (rand) {
        case 0:
            c = '0' + (int) (secureRandom.nextDouble() * 10);
            break;
        case 1:
            c = 'a' + (int) (secureRandom.nextDouble() * 26);
            break;
        case 2:
            c = 'A' + (int) (secureRandom.nextDouble() * 26);
            break;
    }
}

edited Apr 23 '18 at 11:09

Elarbi Mohamed Aymen

1,617
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answered Apr 23 '18 at 10:56

Shanu Gupta

3,699
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1

Why use `nextDouble` when you want an `int` in the end? – Thilo Apr 23 '18 at 10:57
@Thilo Kindly check http://grepcode.com/file/repository.grepcode.com/java/root/jdk/openjdk/6-b27/java/lang/Math.java#Math.random%28%29 – Shanu Gupta Apr 23 '18 at 10:58
Thats because `Math.random()` essentialy calls `nextDouble`. – Shanu Gupta Apr 23 '18 at 10:59
1

For this specific case `int` would work. But its not a exact replacement of `Math.random()` – Shanu Gupta Apr 23 '18 at 11:01
A `downvote` seems pretty harsh on this one. An explanation would be appreciated. – Shanu Gupta Apr 23 '18 at 11:11

score -1 · Answer 3 · answered Apr 23 '18 at 11:09

Do you wish to achieve something like this :

    int rand=0,c=0;;
SecureRandom secrnd=new SecureRandom();
for (int i = 0; i < 8; i++) {
    rand = (int) (secrnd.nextInt(10)% 3);
    switch (rand) {
    case 0:
        c = '0' + (int) (secrnd.nextInt(10) % 10);
        break;
    case 1:
        c = 'a' + (int) (secrnd.nextInt(10) % 26);
        break;
    case 2:
        c = 'A' + (int) (secrnd.nextInt(10) % 26);
        break;
    }

    System.out.println(c);
}

Please clarify your question , so that i can understand what you want to achieve with the above code.

How to generate Random numbers using SecureRandom

3 Answers3