WebSep 28, 2016 · private int letterNumber (string letter) { int sum = 0; char c = letter [0]; for (int i = 0; i < letter.Length; i++) { c = (char)letter [i]; sum += char.ToUpper (c) - 64; } return sum; } Also, A = 1 + D = 4 + E = 4, => 9 :) Share Improve this answer Follow edited Sep 28, 2016 at 8:55 answered Sep 28, 2016 at 8:46 Aleksandar Matic WebNov 25, 2016 · var alphabet = Enumerable.Range (0, 26).Select (i => Convert.ToChar ('A' + i)); Share Improve this answer Follow answered May 3, 2016 at 10:25 shankar.siva 189 2 14 Add a comment 5 Enumerable.Range (65, 26).Select (a => new { A = (char) (a) }).ToList ().ForEach (c => Console.WriteLine (c.A)); Share Improve this answer Follow
Fastest function to generate Excel column letters in C#
WebJun 23, 2011 · You can use string builder to achieve this. int length = value.Length; var lastString = value [length - 1]; if (lastString == 'Z') { if ( (length - 2) >= 0) ++value [length - 2]; else value.Append ('A'); value.Replace ("Z", "A"); } else ++value [length - 1]; Share Improve this answer Follow edited Apr 27, 2016 at 6:06 WebApr 12, 2024 · A Computer Science portal for geeks. It contains well written, well thought and well explained computer science and programming articles, quizzes and practice/competitive programming/company interview Questions. iris county spca
int - Convert alphabetic string into Integer in C# - Stack Overflow
WebJan 14, 2013 · public string GetCode(int number) { int start = (int)'A' - 1; if (number chars = new List(); while (nxt != 0) { int rem = nxt % 26; if (rem == 0) rem = 26; chars.Add( (char) (rem + start)); nxt = nxt / 26; if (rem == 26) nxt = nxt - 1; } for (int i = chars.Count - 1; i >= 0; i--) { str.Append( (char) (chars[i])); } return str.ToString(); } … WebNov 26, 2024 · A Computer Science portal for geeks. It contains well written, well thought and well explained computer science and programming articles, quizzes and practice/competitive programming/company interview Questions. WebFeb 22, 2016 · One way would simply be to assign a - l to values, as variables, such as: int a = 1; int b = 2; int c = 3; ... Another way would be with an Dictionary Dictionary Alpha = new Dictionary () { {'a', 1}, {'b', 2}, {'c', 3} ... }; You could use an enumeration public enum Alphabet { A = 1, B = 2, C = 3, ... porky disease