java- how to generate a 6 digit random hexadecimal value

Question

I have a scenario in a Android app, where a random hexadecimal value has to be generated with 6 digits. (The range of values can be hexadecimal or integer values).

What is the most efficient way to do this? Do I have to generate a random decimal number, and then convert it to hexadecimal? Or can a value be directly generated?

Numbers are numbers. Just generate them within the needed range and do the conversion when you display them. — Federico klez Culloca, Jun 18 '18 at 07:31
Possible duplicate of [this](https://stackoverflow.com/questions/11094823/java-how-to-generate-a-random-hexadecimal-value-within-specified-range-of-value) — Grenther, Jun 18 '18 at 07:35
Combine [How do I generate random integers within a specific range in Java?](https://stackoverflow.com/q/363681) with `Integer.toHexString(int)` — Pshemo, Jun 18 '18 at 07:35
6 digit hexademical number is 3 bytes. 3 bytes is 0..16777215 integer. Just generate integer in that range, and then show it as hex. — Vladyslav Matviienko, Jun 18 '18 at 07:38
Random rnd = new Random(); int n = rnd.nextInt(999999); Integer.valueOf(String.valueOf(n), 16); — Eugen, Jun 18 '18 at 07:38

gagan singh · Accepted Answer · 2018-06-18T07:48:31.637

8

    String zeros = "000000";
    Random rnd = new Random();
    String s = Integer.toString(rnd.nextInt(0X1000000), 16);
    s = zeros.substring(s.length()) + s;
    System.out.println("s = " + s);

edited Jun 18 '18 at 07:48

answered Jun 18 '18 at 07:37

gagan singh

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yes, you are right. Copy paste error from the IDE :) – gagan singh Jun 18 '18 at 07:42
Also, just to make this more useful and complete, you should prepend `0x` and assign the result to a String. Icing on the cake would be to also show how to pad. – Federico klez Culloca Jun 18 '18 at 07:43
4

Use `String.format("%06x", rnd.nextInt(0x1000000))` to get zero-padding. – Andreas Jun 18 '18 at 07:45
@Andreas way is better. – gagan singh Jun 18 '18 at 07:49
For secure application,we can go with SecureRandom class – kavie Jun 18 '18 at 10:33

score 4 · Answer 2 · answered Jun 18 '18 at 07:43

You can use hex literals in your program the same way as decimal literals. A hex literal is prefixed with 0x. Your max value is FFFFFF, so in your program you can write

int maxValue = 0xFFFFFF;

Then you need to generate random numbers in that range. Use the Random class as you normally would.

Random r = new Random();
int myValue = r.nextInt(maxValue + 1);

Note the use of maxValue + 1, because the upper bound for nextInt() is exclusive.

The final step is to print out your hex value.

System.out.printf("%06X", myValue);

score 1 · Answer 3 · edited May 01 '21 at 00:09

1

SecureRandom random = new SecureRandom();
int num = random.nextInt(0x1000000);
String formatted = String.format("%06x", num); 
System.out.println(formatted);

Code Explain

this random object use SecureRandom Class method. this class use for generate random number.
```
SecureRandom random = new SecureRandom();
```
next int num object store 6 hexadecimal digit random number
```
int num = random.nextInt(0x1000000);
```

then output num as 6 digit hexadecimal number

String formatted = String.format("%06x", num);

edited May 01 '21 at 00:09

Erhannis

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answered Jun 18 '18 at 07:56

Ravi Patel

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code only answers aren't really helpful. Please explain the code, tell us why it would offer a solution to the question asked. – jwenting Jun 18 '18 at 08:35
While this code snippet may solve the question, [including an explanation](http://meta.stackexchange.com/questions/114762/explaining-entirely-code-based-answers) really helps to improve the quality of your post. Remember that you are answering the question for readers in the future, and those people might not know the reasons for your code suggestion. – 31piy Jun 18 '18 at 08:38

java- how to generate a 6 digit random hexadecimal value

3 Answers3