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

9

    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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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
For secure application,we can go with SecureRandom class – kavie Jun 18 '18 at 10:33

score 6 · 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 4 · Answer 3 · edited May 01 '21 at 00:09

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);

score 1 · Answer 4 · answered Nov 01 '22 at 17:15

1

More generally:

int digits = 6;
String hex = String.format("%0" + digits + "x", new BigInteger(digits * 4, new SecureRandom()));

answered Nov 01 '22 at 17:15

Jesse Glick

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java- how to generate a 6 digit random hexadecimal value

4 Answers4