Random float in C#

Question

I need to generate a random number in C# for a program I'm writing. Let's say I need to generate a number between -3 and 3. I know I can generate an integer with

System.Random random = new System.Random(); 
float rndNumber = random.Next(-3, 4);

However, this only gives whole numbers (ex. -2, 1, 2, 0). This doesn't generate any decimals. Although you can generate decimals like this

var rand = new Random();
var rndNumber = new decimal(rand.NextDouble());

This gives decimals (ex. 0.47, 0.93, 0.58). But how would I generate numbers that aren't just a whole number? For example 1.04, 1.98, or -0.73 but no going under or over -3 and 3?

fyi `decimal` is not a float. – Daniel A. White Aug 29 '20 at 18:51 — Daniel A. White, Aug 29 '20 at 18:51

score 2 · Accepted Answer · answered Aug 29 '20 at 18:57

2

 var rand = new Random(); 
    var rndNumber = new decimal(rand.NextDouble());
rndNumber = rndNumber * 6 - 3;

NextDouble() returns a double between 0 and 1 multiply this by the range you want 6 then shift it by the offset you.want -3

answered Aug 29 '20 at 18:57

Rich

528
6
11

Works great! Is there a way to only generate the first 3 decimals? – Erik Norberg Aug 29 '20 at 19:13
You can use Math.Round pass the value in and the number of decimal places you want .... Math.Round(rndNumber, 3) – Rich Aug 29 '20 at 19:26

Random float in C#

1 Answers1