Generate unique number with number of digits between 10 and 16

Question

Generate unique number with number of digits between 10 and 16

Noah Aas 985

Hello,

I have a client server application.

I have to send a message identification number. What options do I have to generate a unique number?

The number of digits should be between 10 and 16. It must be ensured that no duplicate numbers are generated during runtime.

Many thanks for the good and helpful tips.

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Answer 1

Marcin Policht 50,735 MVP Volunteer Moderator

Here are a few options to consider:

Atomic counter (single process) A simple, thread-safe counter starting from the current time in milliseconds.

private static long _messageIdCounter = DateTimeOffset.UtcNow.ToUnixTimeMilliseconds();

public static long GetNextMessageId()
{
    return Interlocked.Increment(ref _messageIdCounter);
}

Produces 13-digit numbers.
Unique within the current process runtime.
Not safe across multiple processes or restarts.

Timestamp with random suffix (safe for distributed systems) Combines current time with a random number to reduce the chance of collisions.

public static long GenerateMessageId()
{
    long timestamp = DateTimeOffset.UtcNow.ToUnixTimeMilliseconds(); // 13 digits
    int random = RandomNumberGenerator.GetInt32(0, 100); // 2-digit random number
    return long.Parse($"{timestamp}{random:D2}");
}

Results in 15-digit numbers.
Very low probability of collision, even across systems.

GUID to numeric conversion Use a GUID and convert it to a long, then use modulo to limit the number to a maximum of 16 digits.

public static long GenerateNumericFromGuid()
{
    Guid guid = Guid.NewGuid();
    byte[] bytes = guid.ToByteArray();
    long raw = BitConverter.ToInt64(bytes, 0);
    
    // Ensure it's positive and within 16 digits
    return Math.Abs(raw % 1_000_000_000_000_0000); // max 16 digits
}

While GUIDs have very high uniqueness, using modulo slightly reduces that guarantee.
Still safe for most practical applications.

If the above response helps answer your question, remember to "Accept Answer" so that others in the community facing similar issues can easily find the solution. Your contribution is highly appreciated.

hth

Marcin

Noah Aas 985 Reputation points

2025-04-14T12:56:16.5033333+00:00

Unfortunately this does not work with the GetInt32, see picture.

Do you have another tip? No definition CS0117

The other options are good. Thanks.
Noah Aas 985 Reputation points

2025-04-21T06:58:31.2366667+00:00

GUID to numeric conversion Use a GUID

Do you have any other tips? Otherwise, I'll take solution or 2
Marcin Policht 50,735 Reputation points MVP Volunteer Moderator

2025-04-21T12:17:52.1+00:00
The error you're getting is likely because RandomNumberGenerator.GetInt32() is only available in .NET Core 3.0+ and .NET 5+, but not in older versions of .NET Framework. If you're targeting .NET Framework or an older .NET Standard version, you won’t have access to this method.

You can replace the RandomNumberGenerator.GetInt32(0, 100) line with your own implementation like this:

public static long GenerateMessageId() { long timestamp = DateTimeOffset.UtcNow.ToUnixTimeMilliseconds(); // 13 digits // Generate a 2-digit random number securely byte[] rngBytes = new byte[4]; RandomNumberGenerator.Create().GetBytes(rngBytes); int random = BitConverter.ToInt32(rngBytes, 0) % 100; // Ensure it's positive and 2-digit random = Math.Abs(random % 100); return long.Parse($"{timestamp}{random:D2}"); }

Alternatively, you can use any of the methods suggested earlier that already are working for you.

If the above response helps answer your question, remember to "Accept Answer" so that others in the community facing similar issues can easily find the solution. Your contribution is highly appreciated.

hth

Marcin
Noah Aas 985 Reputation points

2025-04-21T14:16:00.1366667+00:00

Thanks for the feedback.

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Generate unique number with number of digits between 10 and 16

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