How do I generate a random uint with a maximum?

Question

How in C# can I generate a pseudorandom uint with some maximum? (A minimum is not needed.) There seem to be plenty of questions asking for a fully random one, but nothing with an upper limit.

Clarification: This upper limit may be greater than int.MaxValue, so just casting Random.Next(int) does not help.

[`(uint)random.Next(10000)`](https://msdn.microsoft.com/en-us/library/zd1bc8e5(v=vs.110).aspx)? — Tim Schmelter, Jul 16 '15 at 10:20
Please try to do and show some research (_which_ questions did you find?) and show your code and how that doesn't work. If you can't: see [duplicate](http://stackoverflow.com/questions/9025002/unable-to-include-upper-bound-when-generating-random-integer-numbers). — CodeCaster, Jul 16 '15 at 10:22
@TimSchmelter Right, but that won't let me make a `uint` greater that `int.MaxValue`, defeating the point, as this is a possibility. — AlphaModder, Jul 16 '15 at 10:22
@CodeCaster That's not a duplicate, I specifically need to be able to handle an upper bound greater than `int.MaxValue`. — AlphaModder, Jul 16 '15 at 10:23
@AlphaMCubed: you haven't stated that in your question. What do you want to happen if the value is greater than `int.Maxvalue`? — Tim Schmelter, Jul 16 '15 at 10:24
You did not include that constraint in your initial question. Update your question with all relevant information. — CodeCaster, Jul 16 '15 at 10:24
@CodeCaster Have done so now. It seemed implicit, but I see the confusion. — AlphaModder, Jul 16 '15 at 10:25

CodeCaster · Accepted Answer · 2015-07-16T10:55:30.367

7

You can simply generate a number between Int32.MinValue and In32.MaxValue and shift up by half your range:

var random = new Random();
int number = random.Next(Int32.MinValue, Int32.MaxValue);
uint uintNumber = (uint)(number + (uint)Int32.MaxValue);

edited Jul 16 '15 at 10:55

answered Jul 16 '15 at 10:26

CodeCaster

147,647
23
218
272

1

I can't believe I didn't think of this... But wouldn't you use `random.Next(Int32.MinValue, Int32.MinValue + maximum)`? (Considering the first int could be larger than `Int32.MaxValue / 2`, causing an overflow? – AlphaModder Jul 16 '15 at 10:40
I think the math is a tad off. Thanks for accepting though. – CodeCaster Jul 16 '15 at 11:13
Sure thing. Seems to be the best solution. – AlphaModder Jul 16 '15 at 11:16
This seems not to work correctly. The only working solution I found was https://stackoverflow.com/a/17080161/1365061 – Alex Feb 01 '21 at 09:21
@Alex can you explain why it seems that way? – CodeCaster Feb 01 '21 at 10:39

Andriy Horen · Answer 2 · 2015-07-16T11:08:47.913

3

Idea is to generate random 4-byte array then it doesn't matter whether you convert into Int32 or UInt32 it's up to you:

var buffer = new byte[sizeof(uint)];
new Random().NextBytes(buffer);
uint result = BitConverter.ToUInt32(buffer, 0);

result = (result % (max - min)) + min;

edited Jul 16 '15 at 11:08

answered Jul 16 '15 at 10:36

Andriy Horen

2,861
4
18
38

score 0 · Answer 3 · answered Jul 16 '15 at 10:35

this trick would do it

    static void Main(string[] args)
    {
        uint MaxValue = 5555;
        uint value;
        var rnd = new Random();
        var bytes = new byte[4];

        do
        {
            rnd.NextBytes(bytes);
            value = BitConverter.ToUInt32(bytes, 0);
        } while (value > MaxValue);

    }

xXliolauXx · Answer 4 · 2015-07-16T15:40:48.213

-3

First, create a new object of the random class:

Random r = new Random();

Then, get your numbers out of it:

int myRandomNumber = r.Next(myMaximum + 1);

So r.Next(10) would return a number 1-9 inclusive.

Finally, cast it into uint: myRandomUInt = (UInt)myRandomNumber;

edited Jul 16 '15 at 15:40

answered Jul 16 '15 at 10:27

xXliolauXx

1,273
1
11
25

How do I generate a random uint with a maximum?

4 Answers4