replicating an array and storing it in another array

Question

I have

string[] a = {"1","2","3","4","5"};

I want to create a second array and have it store *3 what array a has.

So the second array would look like:

string[] b = {"1","2","3","4","5","1","2","3","4","5","1","2","3","4","5"};

I was thinking of using Array.Copy, but is there another way?

What is wrong with using Array.Copy? Just copy the array to the three segments of the new array. — user1726343
– user1726343, Commented May 9, 2015 at 20:25
What have you tried and where were you stuck? What problem did you reach in order to start thinking you should explore a different way of doing that? — Ondrej Tucny
– Ondrej Tucny, Commented May 9, 2015 at 20:26
There is nothing wrong with it, but sometimes there are different ways of solving the same problem, so I was asking if there is another way to do it. Sometimes there is a function in MSDN that someone knows which does exactly what you are trying to do without implementing anything else, so I thought someone might know a method that replicates an array into another without any external logic or code. — Kelbe
– Kelbe, Commented May 9, 2015 at 20:27
This question has merit, but it would be improved if you gave some desired characteristics for alternatives, rather than just seeking a list of equivalent approaches. (For example, you might want an alternative that's more flexible, more compact, more readable, or more performant.) — Dan J
– Dan J, Commented May 9, 2015 at 20:39

cmcquillan · Accepted Answer · 2015-05-09 20:29:54Z

6

Here is a fun LINQ statement which gets your desired output:

b = Enumerable.Range(0, 3).SelectMany(p => a).ToArray();

answered May 9, 2015 at 20:29

cmcquillan

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4 Comments

Pawel Maga Over a year ago

Its correct, but dirty. Sometimes it's better to write some more lines of code, but have better readability.

Dan J Over a year ago

It's a little harder to read, especially for anyone not familiar with LINQ extension methods, but this would be a decent solution for replicating an array n number of times (i.e. if the value of "3" is not fixed). I'd probably wrap it in a convenience function to make its purpose a little more clear. :)

cmcquillan Over a year ago

@PawelMaga Yep. I'd agree. However, the question was "Is there another way?". I'm simply giving an example.

Dan J Over a year ago

Also, this is neither here nor there, but I wish the API declared InclusiveRange and ExclusiveRange functions, whose parameters were the range bounds. (start, count) just seems unintuitive. :P

DWright · Accepted Answer · 2015-05-09 21:28:43Z

The below sample performs worse than use of Array.Copy--it's more just playing with ways of accomplishing what you ask. The final result is obtained with a very simple looking call of string[] newArray = a.ArrayRepeat(3);

using System;
using System.Collections.Generic;
using System.Linq;

namespace ArrayRepeat
{
    public static class ArrayHelper
    {
        public static IEnumerable<T> Repeater<T>(this T[] a, int reps)
        {
            for (int i = 0; i < reps; i++)
            {
                foreach (var item in a)
                {
                    yield return item;
                }
            }
        }

        public static T[] ArrayRepeat<T>(this T[] a, int reps)
        {
            return a.Repeater<T>(reps).ToArray();
        }
    }
    class Program
    {

        static void Main(string[] args)
        {
            string[] a = { "1", "2", "3", "4", "5" };
            string[] newArray = a.ArrayRepeat(3);

            var pressKeyToExit = Console.ReadKey();
        }
    }
}

DWright · Accepted Answer · 2015-05-09 21:51:16Z

0

Here's another approach that is not hackish, seems clean, and is maybe even useful, although it still will not perform as well as Array.Copy:

using System;
using System.Linq;

namespace RepeatArray
{
    class Program
    {
        static void Main(string[] args)
        {
            string[] a = { "1", "2", "3", "4", "5" };
            string[] newArray = Enumerable.Repeat(a, 3).SelectMany(x => x).ToArray();

            var pressKeyToExit = Console.ReadKey();
        }
    }
}

I should give props to @cmcquillan. I was trying to figure out how to get around the quirkiness of his SelectMany(p => x) which just discards p. The idea comes from him, however.

edited May 9, 2015 at 21:51

answered May 9, 2015 at 21:46

DWright

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