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I'm trying to make a function that can generate a random number and specify how many bits the random number can be.

More specifically, I am trying to generate a number that could either be 8, 10, or 12 bits long. This bit amount is specified by another variable.

I then want to represent this number as a hex in string or as an signed and unsigned integer in decimal. I think i can just use the various Integer.toString methods for that part.

How would I go about generating specific bit-length random numbers in Java?

So far, I have this function skeleton in place:

public static String generateQuestion(int numBits, int taskType){
    String questionStr = "";
    String questionVal = ""; // This is the string hexadecimal version of the randomly generated number
    // TODO: Code to generate random value based on numBits
    

    switch(taskType){
        case 0: // Hex to Decimal
            questionStr = "What are the signed and unsigned values for the " + numBits + "-bit value " + questionVal + "?";
            return questionStr; // Return the constructed string
        case 1: // Signed Decimal to Hex
        case 2: // Unsigned Decimal to Hex
        default:
            throw new IllegalStateException("Unexpected Question Type: " + taskType); // In case the task type is something other than the 3 possible question types
    }
}
Mark Rotteveel
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Prithvi Boinpally
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  • Do you need cryptographically secure numbers or merely good distribution? – Martheen Oct 16 '20 at 01:59
  • I don't need secure numbers. I just need something very simple that has a decent range and randomness. As for the link, I saw it before posting this but I don't think it deals with the specific bit requirement I have. It generates number within a range, but I don't know how I can set the range to match bit length. – Prithvi Boinpally Oct 16 '20 at 02:04
  • Perhaps could create a max_var for each specified bit length? – Prithvi Boinpally Oct 16 '20 at 02:04
  • To generate a number with 8 bits, use 2^8, 10, 2^10, 12, 2^12, isn't that literally the definition? – Martheen Oct 16 '20 at 02:05
  • Ah yes that could work. I was wondering if that only applied to base-10 but I can just generate it as a decimal and convert to hex later. – Prithvi Boinpally Oct 16 '20 at 02:07

1 Answers1

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Make a new instance of java.util.Random, preferably only one for your entire app.

Then, you can invoke .nextInt(x) on it, where x is the upper bound (exclusive). So, if you want a 4 bit random number, you'd call .nextInt(16), or if you find it more readable (it turns into the exact same bytecode), .nextInt(0x10), which generates a random number between 0 (inclusive) and 16 (exclusive), so, from 0000 to 1111.

To find the '16', it's just 1 << (bits-1). e.g. rnd.nextInt(1 << (12-1)) will give you a 12 bit random number.

Numbers just are. An int has no notion of hex or decimal or binary. That's a thing you choose when you print it. String.format("%x", 10) will print A for example.

rzwitserloot
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