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Random.NextBytes() Method in C#
The Random.NextBytes() method in C# is used to fill the elements of a specified array of bytes with random numbers.
Syntax
The syntax is as follows −
public virtual void NextBytes (byte[] buffer);
Above the buffer is the array of bytes.
Example
Let us now see an example −
using System;
public class Demo {
public static void Main(){
Random r = new Random();
Random r2 = new Random();
Random r3 = new Random();
Byte[] arr = new Byte[5];
r3.NextBytes(arr);
Console.WriteLine("Random numbers.....");
for (int i = 1; i <= 5; i++)
Console.WriteLine(r.Next());
Console.WriteLine("
Random numbers from 1 to 10.....");
for (int i = 1; i <= 5; i++)
Console.WriteLine(r2.Next(10));
Console.WriteLine("
Random numbers in the byte array...");
for (int i = 0; i < arr.GetUpperBound(0); i++)
Console.WriteLine(arr[i]);
}
}Output
This will produce the following output −
Random numbers..... 2081486546 329484380 1639318640 1499756340 2122408387 Random numbers from 1 to 10..... 9 1 7 6 9 Random numbers in the byte array... 210 92 112 52
Example
Let us now see another example −
using System;
public class Demo {
public static void Main(){
Random r = new Random();
Byte[] arr = new Byte[2];
r.NextBytes(arr);
Console.WriteLine("Random numbers in the byte array...");
for (int i = 0; i < 2; i++)
Console.WriteLine(arr[i]);
}
}Output
This will produce the following output −
Random numbers in the byte array... 173 11
Updated on: 03-Dec-2019
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