static IEnumerable<string> Split(string str, double chunkSize)
{
    return Enumerable.Range(0, (int) Math.Ceiling(str.Length/chunkSize))
       .Select(i => new string(str
           .Skip(i * (int)chunkSize)
           .Take((int)chunkSize)
           .ToArray()));
}

另一种方法是:

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

public class Program
{
    public static void Main()
    {

        var x = "Hello World";
        foreach(var i in x.ChunkString(2)) Console.WriteLine(i);
    }
}

public static class Ext{
    public static IEnumerable<string> ChunkString(this string val, int chunkSize){
        return val.Select((x,i) => new {Index = i, Value = x})
                  .GroupBy(x => x.Index/chunkSize, x => x.Value)
                  .Select(x => string.Join("",x));
    }
}

使用IX库中的Buffer扩展

    static IEnumerable<string> Split( this string str, int chunkSize )
    {
        return str.Buffer(chunkSize).Select(l => String.Concat(l));
    }

为什么不是循环?这里有一些东西可以很好地做到这一点:

        string str = "111122223333444455";
        int chunkSize = 4;
        int stringLength = str.Length;
        for (int i = 0; i < stringLength ; i += chunkSize)
        {
            if (i + chunkSize > stringLength) chunkSize = stringLength  - i;
            Console.WriteLine(str.Substring(i, chunkSize));

        }
        Console.ReadLine();

我不知道你会如何处理字符串不是因子4的情况，但不是说你的想法是不可能的，只是想知道它的动机，如果一个简单的for循环做得很好?显然，上面的内容可以被清除，甚至可以作为扩展方法加入。

或者正如评论中提到的，你知道它是/4

str = "1111222233334444";
for (int i = 0; i < stringLength; i += chunkSize) 
  {Console.WriteLine(str.Substring(i, chunkSize));} 

        public static List<string> DevideByStringLength(string text, int chunkSize)
    {
        double a = (double)text.Length / chunkSize;
        var numberOfChunks = Math.Ceiling(a);

        Console.WriteLine($"{text.Length} | {numberOfChunks}");

        List<string> chunkList = new List<string>();
        for (int i = 0; i < numberOfChunks; i++)
        {
            string subString = string.Empty;
            if (i == (numberOfChunks - 1))
            {
                subString = text.Substring(chunkSize * i, text.Length - chunkSize * i);
                chunkList.Add(subString);
                continue;
            }
            subString = text.Substring(chunkSize * i, chunkSize);
            chunkList.Add(subString);
        }
        return chunkList;
    }

修改(现在它接受任何非空字符串和任何正chunkSize) Konstantin Spirin的解决方案:

public static IEnumerable<String> Split(String value, int chunkSize) {
  if (null == value)
    throw new ArgumentNullException("value");
  else if (chunkSize <= 0)
    throw new ArgumentOutOfRangeException("chunkSize", "Chunk size should be positive");

  return Enumerable
    .Range(0, value.Length / chunkSize + ((value.Length % chunkSize) == 0 ? 0 : 1))
    .Select(index => (index + 1) * chunkSize < value.Length 
      ? value.Substring(index * chunkSize, chunkSize)
      : value.Substring(index * chunkSize));
}

测试:

  String source = @"ABCDEF";

  // "ABCD,EF"
  String test1 = String.Join(",", Split(source, 4));
  // "AB,CD,EF"
  String test2 = String.Join(",", Split(source, 2));
  // "ABCDEF"
  String test3 = String.Join(",", Split(source, 123));

它基于@dove解决方案，但作为扩展方法实现。

好处:

扩展方法 涵盖角落案例 分割字符串与任何字符:数字，字母，其他符号

Code

public static class EnumerableEx
{    
    public static IEnumerable<string> SplitBy(this string str, int chunkLength)
    {
        if (String.IsNullOrEmpty(str)) throw new ArgumentException();
        if (chunkLength < 1) throw new ArgumentException();

        for (int i = 0; i < str.Length; i += chunkLength)
        {
            if (chunkLength + i > str.Length)
                chunkLength = str.Length - i;

            yield return str.Substring(i, chunkLength);
        }
    }
}

使用

var result = "bobjoecat".SplitBy(3); // bob, joe, cat

为简洁起见，删除了单元测试(请参阅以前的修订版)