Creating long randoms

Question

So I have this

Random random1 = new Random();
int intrandom1 = random1.Next();

I want to put a long after the .Next. How do I do that? It only accepts ints.

@BeniBela : I dont know about any language where method/function/procedure begins with capital letter :( — Anoop Vaidya, Dec 28 '12 at 18:17
@AnoopVaidya: I know one: Pascal. But Pascal does not have a new operator. And it is case insensitive anyways — BeniBela, Dec 28 '12 at 18:19
possible duplicate of [Random number in long range, is this the way?](http://stackoverflow.com/questions/6651554/random-number-in-long-range-is-this-the-way) — Sean Bright, Dec 28 '12 at 18:21
Here comes someone from same area... now all problem solved :) — Anoop Vaidya, Dec 28 '12 at 18:21
@BeniBela In Java it's `nextInt()` and `nextLong()` not next or Next. — Peter Lawrey, Dec 28 '12 at 18:22

score 3 · Answer 1 · answered Dec 28 '12 at 18:13

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First idea: A long as 64 bit integer, is the combination of 2 32 bit integers, so you can use:

((long)random1.Next() << 32)  | random1.Next()

Or perhaps

((long)random1.Next() <<< 32)  | random1.Next()

if you use java (?) and need a unsigned shift

edit: does not look like Java. Java has random1.nextLong() for that. Perhaps C#? I do not know that

answered Dec 28 '12 at 18:13

BeniBela

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Java already has [`Random.nextLong()`](http://docs.oracle.com/javase/6/docs/api/java/util/Random.html#nextLong()) -which in facts works as you explain. +1 for the lateral thinking, anyway. – SJuan76 Dec 28 '12 at 18:16
Java doesn't have a Next() method and nextInt() it is signed so every second number would have all 1s for the top bits. – Peter Lawrey Dec 28 '12 at 18:20
Are you sure this will result in an even distribution along the number range? I'm inclined to think that it will bias the results a bit, but I'm not a good enough mathematician to prove it. – Servy Dec 28 '12 at 18:36

Mihai8 · Answer 2 · 2012-12-28T19:14:03.557

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It can be generated arbitrary long number by using a simple linked list. Just imagine that every node from list can store a number random generated, and a function can be read that list like an unitary number. Using an algorithm like that you can obtain an arbitrary long random number.

edited Dec 28 '12 at 19:14

answered Dec 28 '12 at 18:19

Mihai8

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Creating long randoms

2 Answers2