Generating an array of letters in the alphabet

Question

Is there an easy way to generate an array containing the letters of the alphabet in C#? It\'s not too hard to do it by hand, but I was wondering if there was a built in way

Answer 1

Unfortunately there is no ready-to-use way.

You can use; char[] characters = "ABCDEFGHIJKLMNOPQRSTUVWXYZ".ToCharArray();

Answer 2

char alphaStart = Char.Parse("A");
char alphaEnd = Char.Parse("Z");
for(char i = alphaStart; i <= alphaEnd; i++) {
    string anchorLetter = i.ToString();
}

Answer 3

var alphabets = Enumerable.Range('A', 26).Select((num) => ((char)num).ToString()).ToList();

Answer 4

//generate a list of alphabet using csharp
//this recurcive function will return you
//a string with position of passed int
//say if pass 0 will return A ,1-B,2-C,.....,26-AA,27-AB,....,701-ZZ,702-AAA,703-AAB,...

static string CharacterIncrement(int colCount)
{
    int TempCount = 0;
    string returnCharCount = string.Empty;

    if (colCount <= 25)
    {
        TempCount = colCount;
        char CharCount = Convert.ToChar((Convert.ToInt32('A') + TempCount));
        returnCharCount += CharCount;
        return returnCharCount;
    }
    else
    {
        var rev = 0;

        while (colCount >= 26)
        {
            colCount = colCount - 26;
            rev++;
        }

        returnCharCount += CharacterIncrement(rev-1);
        returnCharCount += CharacterIncrement(colCount);
        return returnCharCount;
    }
}

//--------this loop call this function---------//
int i = 0;
while (i <>
    {
        string CharCount = string.Empty;
        CharCount = CharacterIncrement(i);

        i++;
    }

Answer 5

C# 3.0 :

char[] az = Enumerable.Range('a', 'z' - 'a' + 1).Select(i => (Char)i).ToArray();
foreach (var c in az)
{
    Console.WriteLine(c);
}

yes it does work even if the only overload of Enumerable.Range accepts int parameters ;-)

Answer 6

I don't think there is a built in way, but I think the easiest would be

  char[] alpha = "ABCDEFGHIJKLMNOPQRSTUVWXYZ".ToCharArray();