List<string> SplitString(int chunk, string input)
{
    List<string> list = new List<string>();
    int cycles = input.Length / chunk;

    if (input.Length % chunk != 0)
        cycles++;

    for (int i = 0; i < cycles; i++)
    {
        try
        {
            list.Add(input.Substring(i * chunk, chunk));
        }
        catch
        {
            list.Add(input.Substring(i * chunk));
        }
    }
    return list;
}

最近我不得不在工作中写一些东西来完成这个任务，所以我想我会把我对这个问题的解决方案贴出来。作为一个额外的好处，这个解决方案的功能提供了一种方法，以相反的方向分割字符串，并且它正确地处理上面Marvin Pinto提到的unicode字符。所以，就是这样:

using System;
using Extensions;

namespace TestCSharp
{
    class Program
    {
        static void Main(string[] args)
        {    
            string asciiStr = "This is a string.";
            string unicodeStr = "これは文字列です。";

            string[] array1 = asciiStr.Split(4);
            string[] array2 = asciiStr.Split(-4);

            string[] array3 = asciiStr.Split(7);
            string[] array4 = asciiStr.Split(-7);

            string[] array5 = unicodeStr.Split(5);
            string[] array6 = unicodeStr.Split(-5);
        }
    }
}

namespace Extensions
{
    public static class StringExtensions
    {
        /// <summary>Returns a string array that contains the substrings in this string that are seperated a given fixed length.</summary>
        /// <param name="s">This string object.</param>
        /// <param name="length">Size of each substring.
        ///     <para>CASE: length &gt; 0 , RESULT: String is split from left to right.</para>
        ///     <para>CASE: length == 0 , RESULT: String is returned as the only entry in the array.</para>
        ///     <para>CASE: length &lt; 0 , RESULT: String is split from right to left.</para>
        /// </param>
        /// <returns>String array that has been split into substrings of equal length.</returns>
        /// <example>
        ///     <code>
        ///         string s = "1234567890";
        ///         string[] a = s.Split(4); // a == { "1234", "5678", "90" }
        ///     </code>
        /// </example>            
        public static string[] Split(this string s, int length)
        {
            System.Globalization.StringInfo str = new System.Globalization.StringInfo(s);

            int lengthAbs = Math.Abs(length);

            if (str == null || str.LengthInTextElements == 0 || lengthAbs == 0 || str.LengthInTextElements <= lengthAbs)
                return new string[] { str.ToString() };

            string[] array = new string[(str.LengthInTextElements % lengthAbs == 0 ? str.LengthInTextElements / lengthAbs: (str.LengthInTextElements / lengthAbs) + 1)];

            if (length > 0)
                for (int iStr = 0, iArray = 0; iStr < str.LengthInTextElements && iArray < array.Length; iStr += lengthAbs, iArray++)
                    array[iArray] = str.SubstringByTextElements(iStr, (str.LengthInTextElements - iStr < lengthAbs ? str.LengthInTextElements - iStr : lengthAbs));
            else // if (length < 0)
                for (int iStr = str.LengthInTextElements - 1, iArray = array.Length - 1; iStr >= 0 && iArray >= 0; iStr -= lengthAbs, iArray--)
                    array[iArray] = str.SubstringByTextElements((iStr - lengthAbs < 0 ? 0 : iStr - lengthAbs + 1), (iStr - lengthAbs < 0 ? iStr + 1 : lengthAbs));

            return array;
        }
    }
}

此外，这里还有一个运行此代码结果的图像链接:http://i.imgur.com/16Iih.png

我个人更喜欢我的解决方案:-)

它处理:

是块大小的倍数的字符串长度。 不是块大小倍数的字符串长度。 小于块大小的字符串长度。 NULL和空字符串(抛出异常)。 块大小小于1(引发异常)。

它是作为一个扩展方法实现的，它预先计算将要生成的块的数量。它检查最后一个块，因为如果文本长度不是一个倍数，它需要更短。干净、简短、容易理解……和工作!

    public static string[] Split(this string value, int chunkSize)
    {
        if (string.IsNullOrEmpty(value)) throw new ArgumentException("The string cannot be null.");
        if (chunkSize < 1) throw new ArgumentException("The chunk size should be equal or greater than one.");

        int remainder;
        int divResult = Math.DivRem(value.Length, chunkSize, out remainder);

        int numberOfChunks = remainder > 0 ? divResult + 1 : divResult;
        var result = new string[numberOfChunks];

        int i = 0;
        while (i < numberOfChunks - 1)
        {
            result[i] = value.Substring(i * chunkSize, chunkSize);
            i++;
        }

        int lastChunkSize = remainder > 0 ? remainder : chunkSize;
        result[i] = value.Substring(i * chunkSize, lastChunkSize);

        return result;
    }

它不是很漂亮，也不是很快，但它是有效的，它是一行程序，它是LINQy:

List<string> a = text.Select((c, i) => new { Char = c, Index = i }).GroupBy(o => o.Index / 4).Select(g => new String(g.Select(o => o.Char).ToArray())).ToList();

static IEnumerable<string> Split(string str, int chunkSize)
{
    return Enumerable.Range(0, str.Length / chunkSize)
        .Select(i => str.Substring(i * chunkSize, chunkSize));
}

请注意，可能需要额外的代码来优雅地处理边缘情况(null或空输入字符串，chunkSize == 0，输入字符串长度不能被chunkSize整除，等等)。最初的问题没有为这些边缘情况指定任何需求，在现实生活中，需求可能会有所不同，因此它们超出了这个答案的范围。

        List<string> chunks = new List<string>();
        var longString = new string('s', 3000);
        var chunkLength = 1273;
        var ratio = (double)longString.Length / chunkLength;
        var countOfChunks = Convert.ToByte(Math.Round(ratio, MidpointRounding.ToPositiveInfinity));

        for (byte i = 0; i < countOfChunks; i++)
        {
            var remainingLength = longString.Length - chunkLength * i;
            if (chunkLength > remainingLength)
                chunks.Add(longString.Substring(i * chunkLength, remainingLength));
            else
                chunks.Add(longString.Substring(i * chunkLength, chunkLength));
        }