为IEnumerable增加Combine的扩展方法

        有时我们需要合并两个集合，并同时做一些修改。下面我们实现一个扩展方法在IEnumerable<T>上，看下面的代码：

   public static class IEnumerableExtensions

    {

        /// <summary>

        /// Combines the specified seq A and B

        /// </summary>

        /// <typeparam name="TA"></typeparam>

        /// <typeparam name="TB"></typeparam>

        /// <typeparam name="T"></typeparam>

        /// <param name="seqA">The seq A.</param>

        /// <param name="seqB">The seq B.</param>

        /// <param name="func">The func.</param>

        /// <returns>IEnumerable list</returns>

        public static IEnumerable<T> Combine<TA, TB, T>(

            this IEnumerable<TA> seqA, IEnumerable<TB> seqB, Func<TA, TB, T> func)

        {

            using (var iteratorA = seqA.GetEnumerator())

            using (var iteratorB = seqB.GetEnumerator())

            {

                bool hasValueA;

                bool hasValueB;

                do

                {

                    hasValueA = iteratorA.MoveNext();

                    hasValueB = iteratorB.MoveNext();

                    if (hasValueA | hasValueB)

                    {

                        TA a = hasValueA ? iteratorA.Current : default(TA);

                        TB b = hasValueB ? iteratorB.Current : default(TB);

                        yield return func(a, b);

                    }

                } while (hasValueA || hasValueB);

            }

        }

    }

如何使用，看UnitTest:

   1:          [Test]

   2:          public  void TestIEumberableExtention()

   3:          {

   4:              int[] integers1 = new int[] { 1, 2, 3, 4, 5 };

   5:              int[] integers2 = new int[] { 10, 20, 30, 40, 50 };

   6:   

   7:              //Sums

   8:              CollectionAssert.AreEqual(new int[] { 11, 22, 33, 44, 55 }, integers1.Combine(integers2, (i, j) => i + j));

   9:   

  10:              char[] characters = new char[] { 'A', 'B', 'C', 'D', 'E', 'F' };

  11:   

  12:              //Mixed Types and Lengths

  13:              CollectionAssert.AreEqual(new [] { "A1","B2","C3","D4","E5","F0" },characters.Combine(integers1, (c, i) => string.Format("{0}{1}", c, i)));

  14:          }

希望这篇POST对您有帮助。

作者：Petter Liu
出处：http://www.cnblogs.com/wintersun/
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