8 Digit Random Number Generation in C#3.5

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Hi All,

I have a requirement to generate the 8 digit random number in C#3.5.I do not want to use the syste.random class since the random number can predictable.

So I used Cryptographic random number genearation but this gives me always random bytes only.Iam expecting an 8 digit random number from these random bytes.Is there any possibility getting only possitive random bytes and 8 digit random number from these random bytes.

Thanks,
Uday

Posted 2-Feb-11 15:39pm

aruidsp

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4 solutions

Solution 4

Maybe you could use part of this article where it generates a random encryption key:

Create and Share (with a client app) a Random Encryption Key[^]

Posted 3-Feb-11 21:45pm

#realJSOP

Solution 2

You can use it

long randNum = 0;
Random random = new Random();
for (int i = 1; i <= 8; i++)
{
    long t = 0;
    if (i == 1)
        t = random.Next(1, 9);
    else
        t = random.Next(0,9);
    randNum = randNum * 10 + t;
}
Console.WriteLine(randNum);

Posted 2-Feb-11 16:23pm

shakil0304003

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aruidsp 2-Feb-11 23:29pm

Thanks for code but I don't want use the random class.

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Sergey Alexandrovich Kryukov · Accepted Answer · 2011-02-02T15:56:00

Solution 1

First of all, I don't think System.Random gives predictable results. If you say so, would you please demonstrate your prediction? However, I'm not sure if the result of random generation is not correlated and obey required distribution. Even if it not perfect, it does not mean predictability.

Now, let's assume you have random byte. Let's say your method is byte Random().

C#

int RandomInt() {
    byte numBytes = sizeof(int);
    int value = 0;
    for (int index = 0; index < numBytes; index ++)
       value |= Random() << index * 8;
    return value;
}

Same thing with any other integer type. All you need is to shift each byte by the number of bits need to put it in all byte positions in sequence and OR with previous result. As number of bytes is calculated from the target type, it will work no matter what type you want to use as a target but... to write a generic (in a simple way) is nearly impossible.

—SA

Posted 2-Feb-11 15:56pm

Sergey Alexandrovich Kryukov

Updated 2-Feb-11 18:08pm

v3

Comments

wizardzz 2-Feb-11 21:58pm

I agree, I'm not sure why Random is not suitable without a better explanation from the OP

Sergey Alexandrovich Kryukov 2-Feb-11 22:07pm

Thank you.
Chances are, these concerns are because of rumors "they say something's wrong with random". Another reason: seeding is often used incorrectly (I saw some example here in Questions and Answers), and then the patient blames the generator. In real life, quality of the generator can be very important, for example, in numeric simulations where statistics is derived from a numeric experiment. In other cases, it does not make big difference: solutions of equations using Monte-Carlo method for example: high precision of distribution is not so critical; it is also not critical in... security -- not at all. However, predictability would be fatal for security. In practice, this is not an issue.
--SA

aruidsp 2-Feb-11 23:28pm

Hi SA,

Thanks a lot for the code.

The field value never assigned any value.
is it as below

index |= Random() << index * 8;

value = index;
return value;

Thanks,
Uday

Sergey Alexandrovich Kryukov 3-Feb-11 0:09am

Sorry, I just fixed a bug, please see the update (the modification you provided did not fix it yet).
--SA

aruidsp 3-Feb-11 9:26am

Hi SA,

Thanks for the update.
I have some issue with the code.
here is my code.

byte Random()
{
RNGCryptoServiceProvider random = new RNGCryptoServiceProvider();
byte[] randomBytes = new byte[4];
random.GetBytes(randomBytes);
return randomBytes;
}

public int randomint()
{
byte numBytes = sizeof(int);
int value = 0;
for (int index = 0; index < numBytes; index++)
value |= Random() << index * 8;

return value;
}

Now the randomBytes can not convert to byte.

Thanks.

Sergey Alexandrovich Kryukov 4-Feb-11 3:14am

OK, this is a different question. How can I'm supposed to help you? At least show me the interface of this class: RNGCryptoServiceProvider. [Scratch] Or, no, I have it....
Please wait...
--SA

Espen Harlinn 4-Feb-11 11:42am

5+ times 2 - good answer, both of them :)

Sergey Alexandrovich Kryukov · Accepted Answer · 2011-02-03T21:27:00

Solution 3

Oh! Please concentrate, look at the interface and types properly!

To answer follow-up question:

C#

int RandomInt(RNGCryptoServiceProvider random) {
    byte numBytes = sizeof(int);
    byte randomBytes = byte[numBytes];
    random.GetBytes(randomBytes);
    int value = 0;
    for (int index = 0; index < numBytes; index ++)
       value |= randomBytes[index] << index * 8;
    return value;
}

Didn't you see the difference between byte[] and byte?! How could they convert?
Don't implement my Random, it was just for example, make random all numBytes (==4) bytes at once (shown above).

—SA

Posted 3-Feb-11 21:27pm

Sergey Alexandrovich Kryukov

Updated 4-Feb-11 9:45am

v3

Comments

aruidsp 4-Feb-11 21:34pm

Hi SA,

Thanks.

But it returns more than 8 digit random number.
I assigned numbytes = 4 and run my code, it returns more than 8 digits.
But I need always an 8 digit random number only.

Thanks.

Sergey Alexandrovich Kryukov 4-Feb-11 21:46pm

What code "returns"? I eliminated all functions returning arrays above.
Don't say "digits"! The integer types are not composed of digits. May be here is a big misconception. Now I'm not sure you understand computer presentation of integers and mix it up with string presentation ("digits" only exist in string presentation, int32 and uint32 are always 4 bytes).
If you're not going to understand that, nobody will help you.

Now, please concentrate and read carefully. The code above is correct and returns random integer byte-by-byte (not digit by digit). If you need that (8 digits), you should explain why very accurately. You should also define what a digit is, it depend (it could be decimal, hexadecimal, etc.)

The above code does not have a constant which defines the size of data. It is calculated by the type of parameter. If you change type to any other integer type, the number of bytes will be 1, 2, 4, or 8.

--SA