CODEWITHSUNDEEP
c#arraysloops
Emmanuel
Emmanuel

Reputation: 19

How do I create an integer array that outputs 3 different numbers?

I want to create an int array that hold 10 numbers that range from (0-9). Then I want the array to output 3 different numbers.

I made one but it only gives me one random number. How can I fix this so it outputs three numbers instead of one.

 int[] password = { 0,1,2,3,4,5,6,7,8,9 };

 Random r = new Random();
 int inputPassword = r.Next(password.Length);
 Console.WriteLine(inputPassword);

Upvotes: 0

Views: 161

Answers (2)

ADyson
ADyson

Reputation: 62078

Unless, at some point, you are planning to change the numbers in the array to something different, then the array is completely unnecessary - other than its length property, which is fixed, you aren't actually using it.

Based on what you've stated as your goal, the code could be as simple as this:

int n = 10;
int length = 3;
Random r = new Random();
for (int i = 0; i < length; i++)
{
    int inputPassword = r.Next(n);
    Console.WriteLine(inputPassword);
}

Demo: https://dotnetfiddle.net/OKgqIE

P.S. If there are more constraints to the scenario which make this inadequate then you need to clarify them.

Upvotes: 1

Martin
Martin

Reputation: 16433

The way to achieve your aim is simple, use a loop of some kind.

In the example below I've used a for loop:

int[] password = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };
Random r = new Random();
for (int i = 0; i < 3; i++)
{
    int inputPassword = r.Next(password.Length);
    Console.WriteLine(inputPassword);
}

Incidentally, as pointed out by @ADyson, you're not actually using any of the stored values in your password array. The line where you use the Next random should be like this:

int inputPassword = password[r.Next(password.Length)];

This will fetch the value at a random index in the password array.

Upvotes: 2

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