How to get random values from array in C# [duplicate]

Answer 1

I use the random.next(0, array.length), but this give random number of the length and i need the random array numbers.

Use the return value from random.next(0, array.length) as index to get value from the array

 Random random = new Random();
 int start2 = random.Next(0, caminohormiga.Length);
 Console.Write(caminohormiga[start2]);

Answer 2

I noticed in the comments you wanted no repeats, so you want the numbers to be 'shuffled' similar to a deck of cards.

I would use a List<> for the source items, grab them at random and push them to a Stack<> to create the deck of numbers.

Here is an example:

private static Stack<T> CreateShuffledDeck<T>(IEnumerable<T> values)
{
  var rand = new Random();

  var list = new List<T>(values);
  var stack = new Stack<T>();

  while(list.Count > 0)
  {
    // Get the next item at random.
    var index = rand.Next(0, list.Count);
    var item = list[index];

    // Remove the item from the list and push it to the top of the deck.
    list.RemoveAt(index);
    stack.Push(item);
  }

  return stack;
}

So then:

var numbers = new int[] {0, 1, 4, 6, 8, 2};
var deck = CreateShuffledDeck(numbers);

while(deck.Count > 0)
{
  var number = deck.Pop();
  Console.WriteLine(number.ToString());
}

Answer 3

You just need to use the random number as a reference to the array:

var arr1 = new[]{1,2,3,4,5,6}
var rndMember = arr1[random.Next(arr1.Length)];

Answer 4

Console.Write(caminohormiga[start2]);

Answer 5

Try like this

int start2 = caminohormiga[ran.Next(0, caminohormiga.Length)];

instead of

int start2 = random.Next(0, caminohormiga.Length);

Answer 6

To shuffle

int[] numbers = new [] {0, 1, 4, 6, 8, 2};
int[] shuffled = numbers.OrderBy(n => Guid.NewGuid()).ToArray();