好的,让我们从这个非常简单的按钮点击方法开始
private void button1_Click(object sender, EventArgs e)
{
int counter = 1;
List<int> items = new int[] { 1, 2, 3 }.ToList();
List<int>.Enumerator enm = items.GetEnumerator();
// 1
if (!enm.MoveNext())
throw new Exception("Unexpected end of list");
if (enm.Current != counter)
throw new Exception(String.Format("Expect {0} but actual {1}", counter, enm.Current));
counter++;
// 2
if (!enm.MoveNext())
throw new Exception("Unexpected end of list");
if (enm.Current != counter)
throw new Exception(String.Format("Expect {0} but actual {1}", counter, enm.Current));
counter++;
//3
if (!enm.MoveNext())
throw new Exception("Unexpected end of list");
if (enm.Current != counter)
throw new Exception(String.Format("Expect {0} but actual {1}", counter, enm.Current));
counter++;
if (enm.MoveNext())
throw new Exception("Unexpected continuation of list");
}
这个方法什么也不做,因为每个断言都优雅地传递。事情很好,直到我相信我应该引入一种方法来消除冗余
static void AssertNext(ref int counter, List<int>.Enumerator e)
{
if (!e.MoveNext())
throw new Exception("Unexpected end of list");
if (e.Current != counter)
throw new Exception(String.Format("Expect {0} but actual {1}", counter, e.Current));
counter++;
}
private void button2_Click(object sender, EventArgs e)
{
var counter = 1;
var items = new int[] { 1, 2, 3 }.ToList();
var enm = items.GetEnumerator();
AssertNext(ref counter, enm);
AssertNext(ref counter, enm);
AssertNext(ref counter, enm);
if (enm.MoveNext()) throw new Exception("Unexpected continuation of list");
}
尽管如此,这种重构很简单(至少对我来说)。它打破了程序! 在第二次调用AssertNext时,似乎枚举器已经重置为起始点并导致断言失败。
我无法理解发生了什么。我真的觉得这个谜题初学者。
我想念的是什么?
答案 0 :(得分:5)
我想象它与List.Enumerator是一个结构有关。你将它传递给一个方法,操纵它,然后返回。原始实例可能不会发生操作。
答案 1 :(得分:3)
List<T>.Enumerator
是一个值类型,意味着它被复制到方法的本地范围,被更改,然后在离开方法时被销毁。尝试通过引用传递它。