C# Generate random long number from double number

Question

I am trying to convert from python to C#

python:

rand = random.randint(0, 2 ** 50)

and here's what I did in C#

var value = Math.Pow(2, 50);
var result = random.Next(0, Convert.ToInt32(value));

I'm getting this error

Unhandled exception. System.OverflowException: Value was either too large or too small for an Int32.
   at System.Convert.ToInt32(Double value)

I wanted to generate from range 0 to value

Do you need the value to be an `int` *and* also range between 0 and 2^50? Can it be a `long`? — harold, Sep 03 '21 at 11:58
An `int` (`Int32`) can only hold a value up to near 2^31, so: yes, it will overflow a lot of the time if you ask for a value in the 2^50 range; a `long` can handle 2^63 - perhaps more useful? — Marc Gravell, Sep 03 '21 at 11:58
instead of `Math.Pow(2, 50);` use `1L << 50` instead. In python use `1 << 50` — phuclv, Sep 03 '21 at 12:01
Then you may be interested in the answers to [Random number in long range, is this the way?](https://stackoverflow.com/q/6651554/555045) — harold, Sep 03 '21 at 12:03
You're going to need something like this: `long value = ((long)random.Next() << 18) ^ (long)random.Next();` — Enigmativity, Sep 03 '21 at 12:08

H H · Answer 1 · 2021-09-03T12:02:05.367

3

In C#, an int is 32 bits. You have 2^50, that needs 50 bits. It will always be too large.

You could use Convert.ToInt64(value), that would fit but Random only returns an Int32. So it is kind of pointless.

I would suggest using 2^30. You're not going to get any more range. That 50 looks kind of arbitrary anyway.

var value = Math.Pow(2, 30);

On second thought, just use

var result = random.Next();  // returns 0 .. Int32.MaxValue-1

and remove everything with value.

edited Sep 03 '21 at 12:02

answered Sep 03 '21 at 11:58

H H

But the required I need is 50 – mark12345 Sep 03 '21 at 15:55
Then look up the linked question, with NextBytes(). Als, be sure you need 50 bits _random_ , not 50 bits _unique_. – H H Sep 03 '21 at 16:22

1 Answers1