Is there a LINQ way to iterate through an array of strings?

Question

I have an array of strings like this:

test[1] = "AA";
test[2] = "BB";

I like to do things in good ways. Now I need to iterate through the array so it looks like this:

1. "AA"
2. "BB"
etc ..

I think I can do this with a for loop and index but I am wondering if I can also do it with LINQ.

Without knowing what you actually want to do its hard to suggest a solution. — Oliver Weichhold
– Oliver Weichhold, Commented Jul 13, 2011 at 7:32
Nice answer from @polishchuk. Why would you jump into LINQ for something as simple as this? Using LINQ will generate a new IEnumerable - so you will have two instances of your list floating around, which becomes problematic if you are dealing with a very large array. LINQ is great, but it isn't the solution to every little iteration problem, there is still a place for a simple for loop. — slugster
– slugster, Commented Jul 13, 2011 at 7:52
If I didn't use LINQ then can I do it really easy with another way? — MIMI
– MIMI, Commented Jul 13, 2011 at 7:55
Simple for loop: for (int i = 0; i < arr.Length; i++) arr[i] = string.Format("{0}. {1}", i+1, arr[i]);. It's all good if you are simply learning how to use LINQ, but sometimes there is nothing wrong with the basics... — slugster
– slugster, Commented Jul 13, 2011 at 8:13

Kirill Polishchuk · Accepted Answer · 2017-08-27 23:00:23Z

20

Prior to C# 6.0:

var result = test.Select((s, i) => string.Format("{0}. {1}", i + 1, s));

Starting from C# 6.0 you can use interpolated strings:

var result = test.Select((s, i) => $"{i + 1}. {s}");

edited Aug 27, 2017 at 23:00

answered Jul 13, 2011 at 7:34

Kirill Polishchuk

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3 Comments

MIMI Over a year ago

I get a message "does not contain a definition for 'Select' and no extension method 'Select' accepting a first argument of type". If I don't use LINQ how I could I do this?

Daniel Hilgarth Over a year ago

You are missing using System.Linq;

Filburt Over a year ago

@polishchuk Nice. The only thing I miss is ++i instead of i + 1 ;-)

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