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 to do this.
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possible duplicate of [Localization: How to map culture info to a script name or Unicode character range?](http://stackoverflow.com/questions/252662/localization-how-to-map-culture-info-to-a-script-name-or-unicode-character-rang) – Johan Dec 22 '11 at 12:15
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FYI, [this question has been asked before](http://stackoverflow.com/questions/252662/localization-how-to-map-culture-info-to-a-script-name-or-unicode-character-range) with internationalization in mind. – Serge Wautier Nov 24 '08 at 15:43
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The reverse can be found here http://stackoverflow.com/a/667902/2343 – Sameer Alibhai Mar 22 '12 at 20:03
14 Answers
I don't think there is a built in way, but I think the easiest would be
char[] alpha = "ABCDEFGHIJKLMNOPQRSTUVWXYZ".ToCharArray();

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8It also works on other alphabets, as long as you place the string in a Resource :) – Olmo Nov 24 '08 at 21:56
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1This is the best way if you want to ensure that your code works on machines with different languages. Otherwise if you generate your alphabet dynamically (as other answers below) you can end with a different alphabet on different machines. – Luis Feb 10 '14 at 19:24
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1How does this work with different languages? Will this somehow magically contain chines letters, or letters form other languages? – Paweł Audionysos Aug 06 '19 at 03:44
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1This answer almost seems like a joke answer. The OP specifically said he could do it by hand but wants an automated way. And it ignores culture, language, etc. – Emperor Eto Jan 26 '21 at 17:08
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 ;-)

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'z' - 'a' + 1 = It just looks really clumsy, but I can't see a way around it :( – CasperT Apr 13 '09 at 12:15
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`String.Concat(Enumerable.Range('a', 'z' - 'a' + 1).Select(c => ((char)c).ToString().ToUpperInvariant()));` returns `ABCDEFGHIJKLMNOPQRSTUVWXYZ`; – abatishchev Jul 19 '10 at 09:41
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17
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2This won't work for the languages that have letters whose ASCII codes fall outside of the range 'a' - 'z'. You cannot use this method in such cases. Nice solution otherwise. – Michal B. Feb 27 '12 at 16:24
for (char letter = 'A'; letter <= 'Z'; letter++)
{
Debug.WriteLine(letter);
}

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char[] alphabet = Enumerable.Range('A', 26).Select(x => (char)x).ToArray();

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I wrote this to get the MS excel column code (A,B,C, ..., Z, AA, AB, ..., ZZ, AAA, AAB, ...) based on a 1-based index. (Of course, switching to zero-based is simply leaving off the column--;
at the start.)
public static String getColumnNameFromIndex(int column)
{
column--;
String col = Convert.ToString((char)('A' + (column % 26)));
while (column >= 26)
{
column = (column / 26) -1;
col = Convert.ToString((char)('A' + (column % 26))) + col;
}
return col;
}

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2Oh hey, just replacing `Convert.ToString` with `String.valueOf` makes it work in Java :) – Nyerguds Dec 08 '11 at 11:51
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1
Assuming you mean the letters of the English alphabet...
for ( int i = 0; i < 26; i++ )
{
Console.WriteLine( Convert.ToChar( i + 65 ) );
}
Console.WriteLine( "Press any key to continue." );
Console.ReadKey();

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9It's better to use (int)'A' instead of hardcoding 65. It'll make the code more readable and less prone to errors. – Mehrdad Afshari Nov 24 '08 at 15:42
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However, in the leading encoding where 'A' != 65, (EBCDIC), A to Z aren't sequencial. – James Curran Nov 24 '08 at 16:51
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5
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@MehrdadAfshari Internally used encoding is irrelevant. What _is_ relevant is that the values of type Char are Unicode code points, and _those_ are A-Z sequential. – Nyerguds Nov 24 '16 at 22:22
You could do something like this, based on the ascii values of the characters:
char[26] alphabet;
for(int i = 0; i <26; i++)
{
alphabet[i] = (char)(i+65); //65 is the offset for capital A in the ascaii table
}
(See the table here.) You are just casting from the int value of the character to the character value - but, that only works for ascii characters not different languages etc.
EDIT: As suggested by Mehrdad in the comment to a similar solution, it's better to do this:
alphabet[i] = (char)(i+(int)('A'));
This casts the A character to it's int value and then increments based on this, so it's not hardcoded.
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This code has a minor error. i = 0 to < 27 includes 27 letters (array element 0, then elements 1 to 26). – Brian Nov 24 '08 at 15:38
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You can make it even better: alphabet[i] = (char)(i + 'A'); Same result – CasperT Apr 13 '09 at 13:03
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2
Surprised no one has suggested a yield solution:
public static IEnumerable<char> Alphabet()
{
for (char letter = 'A'; letter <= 'Z'; letter++)
{
yield return letter;
}
}
Example:
foreach (var c in Alphabet())
{
Console.Write(c);
}

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Note also, the string has a operator[] which returns a Char, and is an IEnumerable<char>
, so for most purposes, you can use a string as a char[]. Hence:
string alpha = "ABCDEFGHIJKLMNOPQRSTUVQXYZ";
for (int i =0; i < 26; ++i)
{
Console.WriteLine(alpha[i]);
}
foreach(char c in alpha)
{
Console.WriteLine(c);
}

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

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

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It seems you have not learned how to use the correct quotes for chars. There is no reason to parse that from string; just use `char alphaStart = 'A';`, with single quotes instead of double quotes. – Nyerguds Oct 30 '13 at 09:48
//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++;
}
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fixed formatting, but code seems to be incomplete (take a look at the `if` and `while` statements) – Oliver Jul 16 '10 at 12:50
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1Woah, recursion for something that simple? You can do that with a simple While, and dividing by 26. – Nyerguds Mar 11 '11 at 10:35
4 ways get English alphabet in Console:
public void ShowEnglishAlphabet()
{
var firstLetter = 'a';
var endLetter = 'z';
for (var letter = firstLetter; letter <= endLetter; letter++)
{
Console.WriteLine($"{letter}-{letter.ToString().ToUpper()}");
}
}
public void ShowEnglishAlphabetFromUnicodeTableDecNumber()
{
var firstLetter = 97;
var endLetter = 122;
for (var letterNumberUnicodeTable = firstLetter;
letterNumberUnicodeTable <= endLetter; letterNumberUnicodeTable++)
{
Console.WriteLine($"{(char)letterNumberUnicodeTable}-
{((char)letterNumberUnicodeTable).ToString().ToUpper()}");
}
}
public void ShowEnglishAlphabetUnicodeTableEscapeSequence()
{
var firstLetter = '\u0061';
var endLetter = '\u007A';
for (var letterNumberUnicodeTable = firstLetter;
letterNumberUnicodeTable <= endLetter; letterNumberUnicodeTable++)
{
Console.WriteLine($"{letterNumberUnicodeTable}-
{letterNumberUnicodeTable.ToString().ToUpper()}");
}
}
public void ShowEnglishAlphabetUnicodeTableLinq()
{
var alphabets = Enumerable.Range('a', 26).Select(letter =>
((char)letter).ToString()).ToList();
foreach (var letter in alphabets)
{
Console.WriteLine($"{letter}-{letter.ToUpper()}");
}
}

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Even if this is a solution, please provide more information about it and what it does. – George Z. Sep 28 '19 at 21:02
Unfortunately there is no ready-to-use way.
You can use; char[] characters = "ABCDEFGHIJKLMNOPQRSTUVWXYZ".ToCharArray();

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