C#•10mo ago

Random Generator method

could someone help me with this method? if the list contains a number we should generate a new random number that is not in the list

c#
public int GeefUniekGetal(int maximum)
{
    const int minimum = 1;
    int result = 0;
    
    if (getallen.Count != maximum)
    {
        if (!getallen.Contains(result))
        {
            result = new Random().Next(minimum, maximum);
            getallen.Add(result);
        } else
        {
            int newResult = new Random().Next(minimum, maximum);
            getallen.Add(newResult);
        }
    } else
    {
        result = 0;
    }
    return result;
}

c#
public int GeefUniekGetal(int maximum)
{
    const int minimum = 1;
    int result = 0;
    
    if (getallen.Count != maximum)
    {
        if (!getallen.Contains(result))
        {
            result = new Random().Next(minimum, maximum);
            getallen.Add(result);
        } else
        {
            int newResult = new Random().Next(minimum, maximum);
            getallen.Add(newResult);
        }
    } else
    {
        result = 0;
    }
    return result;
}

28 Replies

CTJ•10mo ago

its not ideal but you might want to create loop and loop till you find number thats not in list and then break out and of cousrse add a limiting case so you dont get stuck in the loop forever

Yawnder•10mo ago

Also, you can use Random.Shared, rather than creating a new one. Also, make sure you doublecheck the inclusivity/exclusivity of the bounds, otherwise you'll have a bad surprise. Also, no need to set the result to 0 in the else, it's already set to zero.

Angius•10mo ago

Just FYI, if you're creating new instances of

Random()~each time, chances are you'll be getting a *lot* of repeated values
Either use a single instance of it and keep it in a static private field
Or use

Random.Shared.Next()` instead

TheUnquietOP•10mo ago

could you explain what this method actually does?

Angius•10mo ago

Shared is a static property on the Random class that's already initialized with a new Random() So you can easily reuse it without instantiating it yourself

TheUnquietOP•10mo ago

I'll try this now! should I loop through the empty list first or should I loop through something else? I cant find this property

Angius•10mo ago

What version of .NET are you using?

TheUnquietOP•10mo ago

where could I check that?

Angius•10mo ago

In the csproj file for example Or project properties

TheUnquietOP•10mo ago

8 I believe

TheUnquietOP•10mo ago

TheUnquietOP•10mo ago

is this correct?

Angius•10mo ago

Random.Shared.Next() should work, then

TheUnquietOP•10mo ago

it wont run now

Angius•10mo ago

This seems to be a whole another issue entirely The project being on .NET 8, but you not having .NET 8 SDK installed

TheUnquietOP•10mo ago

oh I see, I need to update I cloned it from git and on my laptop I did update but not on my pc thanks for explaining

CTJ•10mo ago

Well I imagined it something like Enter loop Generate random number check if number is in list if yes add and break if loop is repeated more than say 100 times break and do something or alternatively you can build checker to see if every possible number was used yet oooooor you could make a list1 with all numbers from 0 to maxint then remove ones that are already on list2 and then pick a random number from that list1 and insert it to your output list2 that way you'd avoid loops altogether

TheUnquietOP•10mo ago

still wont budge

Angius•10mo ago

It tells you why It's not

new Random().Shared.Next();

new Random().Shared.Next();

it's not

var rnd = new Random();
rnd.Shared.Next();

var rnd = new Random();
rnd.Shared.Next();

It's

Random.Shared.Next();

Random.Shared.Next();

Exactly as I said a good three times now

TheUnquietOP•10mo ago

like this? int result = new Random.Shared.Next(minimum, maximum);

Angius•10mo ago

No Read my code again Whhere do you see new?

Random.Shared.Next();
Random.Shared.Next();
Random.Shared.Next();
Random.Shared.Next();
Random.Shared.Next();
Random.Shared.Next();
Random.Shared.Next();
Random.Shared.Next();
Random.Shared.Next();
Random.Shared.Next();
Random.Shared.Next();
Random.Shared.Next();
Random.Shared.Next();
Random.Shared.Next();
Random.Shared.Next();
Random.Shared.Next();
Random.Shared.Next();

Random.Shared.Next();
Random.Shared.Next();
Random.Shared.Next();
Random.Shared.Next();
Random.Shared.Next();
Random.Shared.Next();
Random.Shared.Next();
Random.Shared.Next();
Random.Shared.Next();
Random.Shared.Next();
Random.Shared.Next();
Random.Shared.Next();
Random.Shared.Next();
Random.Shared.Next();
Random.Shared.Next();
Random.Shared.Next();
Random.Shared.Next();

TheUnquietOP•10mo ago

like so? int result = Random.Shared.Next(minimum, maximum);

Angius•10mo ago

Yes

TheUnquietOP•10mo ago

that wont run : 'RandomGeneratorOpdracht.exe' (CoreCLR: DefaultDomain): Loaded 'C:\Program Files\dotnet\shared\Microsoft.NETCore.App\8.0.2\System.Private.CoreLib.dll'. Skipped loading symbols. Module is optimized and the debugger option 'Just My Code' is enabled. 'RandomGeneratorOpdracht.exe' (CoreCLR: clrhost): Loaded 'C:\Users\Gebruiker\Source\Repos\TheUnquiet\RandomGenerator\RandomGenerator\bin\Debug\net8.0\RandomGeneratorOpdracht.dll'. Symbols loaded. 'RandomGeneratorOpdracht.exe' (CoreCLR: clrhost): Loaded 'C:\Program Files\dotnet\shared\Microsoft.NETCore.App\8.0.2\System.Runtime.dll'. Skipped loading symbols. Module is optimized and the debugger option 'Just My Code' is enabled. 'RandomGeneratorOpdracht.exe' (CoreCLR: clrhost): Loaded 'C:\Program Files\dotnet\shared\Microsoft.NETCore.App\8.0.2\System.Console.dll'. Skipped loading symbols. Module is optimized and the debugger option 'Just My Code' is enabled. 'RandomGeneratorOpdracht.exe' (CoreCLR: clrhost): Loaded 'C:\Program Files\dotnet\shared\Microsoft.NETCore.App\8.0.2\System.Collections.dll'. Skipped loading symbols. Module is optimized and the debugger option 'Just My Code' is enabled. 'RandomGeneratorOpdracht.exe' (CoreCLR: clrhost): Loaded 'C:\Program Files\dotnet\shared\Microsoft.NETCore.App\8.0.2\System.Threading.dll'. Skipped loading symbols. Module is optimized and the debugger option 'Just My Code' is enabled. 'RandomGeneratorOpdracht.exe' (CoreCLR: clrhost): Loaded 'C:\Program Files\dotnet\shared\Microsoft.NETCore.App\8.0.2\System.Text.Encoding.Extensions.dll'. Skipped loading symbols. Module is optimized and the debugger option 'Just My Code' is enabled. 'RandomGeneratorOpdracht.exe' (CoreCLR: clrhost): Loaded 'C:\Program Files\dotnet\shared\Microsoft.NETCore.App\8.0.2\System.Runtime.InteropServices.dll'. Skipped loading symbols. Module is optimized and the debugger option 'Just My Code' is enabled. The program '[10972] RandomGeneratorOpdracht.exe' has exited with code 0 (0x0). nvm thanks!

Angius•10mo ago

No idea what those errors could be tbh Ah, they're just messages telling you VS is not debugging framework code Which is expected and good You can turn off those messages, I see: https://stackoverflow.com/a/55866842/6042255

TheUnquietOP•10mo ago

I'll look into that if the number is in the list I need to generate a new number that is not equal to the previous generated number

CTJ•10mo ago

yes.

static void Main(string[] args)
{
    List<int> MyNumbers = new();

    while (true)
    {
        MyNumbers.Add(GenerateNewNumber(MyNumbers, 6));
        Console.WriteLine(string.Join(", ", MyNumbers));
    }

}


static int GenerateNewNumber(List<int> myList, int MaxInt, int MinInt = 0)
{
    List<int> AllNumbers = Enumerable.Range(MinInt, MaxInt).ToList();
    AllNumbers.RemoveAll(myList.Contains);

    if (AllNumbers.Count == 0)
        throw new Exception("No new numbers can be generated in this range.");
        //return 0;
    
    return AllNumbers[new Random().Next(AllNumbers.Count())];
}

static void Main(string[] args)
{
    List<int> MyNumbers = new();

    while (true)
    {
        MyNumbers.Add(GenerateNewNumber(MyNumbers, 6));
        Console.WriteLine(string.Join(", ", MyNumbers));
    }

}


static int GenerateNewNumber(List<int> myList, int MaxInt, int MinInt = 0)
{
    List<int> AllNumbers = Enumerable.Range(MinInt, MaxInt).ToList();
    AllNumbers.RemoveAll(myList.Contains);

    if (AllNumbers.Count == 0)
        throw new Exception("No new numbers can be generated in this range.");
        //return 0;
    
    return AllNumbers[new Random().Next(AllNumbers.Count())];
}

Console output:

4
4, 2
4, 2, 1
4, 2, 1, 5
4, 2, 1, 5, 0
4, 2, 1, 5, 0, 3

4
4, 2
4, 2, 1
4, 2, 1, 5
4, 2, 1, 5, 0
4, 2, 1, 5, 0, 3

and program ends with exception being thrown

ZacharyPatten•10mo ago

I'm a little late to this thread, but I have multiple articles I've documented that seem relevant to this. In particular: https://github.com/ZacharyPatten/ZacharyPatten/blob/main/Articles/2019-12-18.md "Generating Unique Random Data" If it is relevant and you have any questions, feel free to "@" me 🙂 @TheUnquiet

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