List<>中Find的用法小结

i've been looking for help on how to find objects in generics with list.find() method .... and ... take a look what i have found.
in the follow example, i created a simple class:

public class person
{
       private int _id;
       private string _name;

       public int id {  get{ return _id;} set{ _id = value;}}
       public int name {  get{ return _name;} set{ _name= value;}}

       public person(int id, string name)
       {
             _id = id;
             _name = name;
       }
}

in the example, there's a simple class with two private attributes. now we're going to create a typed list of this object and take advantage of the find() method

public void createandsearchlist()
{
      //create and fill the collection
      list<person> mylist = new list<person>();
      mylist.add(new person(1, "andreysanches"));
      mylist.add(new person(2, "alexandretarifa"));
      mylist.add(new person(3, "emersonfacunte"));

     //find a specific object
     person mylocatedobject = mylist.find(delegate(person p) {return p.id == 1; });
}

备注：在list和array集合中搜索元素经常使用该方法，主要技术是泛型委托

list用法注意：如果增加一个对象，必须重新new一个对象，看下面的例子：

person p=new pewson();
for(int i=0;i<5;i++)
{
p.id=i;
p.name="xxxx";
list.add(p);
}

for(int i=0;i<5;i++)
{
person p=new person();
p.id=i;
p.name="xxxx";
list.add(p);
}

上面有区别吗？在输出list的值是有区别了，第一个list里面存放的都是一样的，结果和最后一个一样，都是同一个人对象，第二个list达到预期的效果，存储不同的人对象。这是为什么？原因很简单，一个list对象中存储的都是对一个共有的对象进行引用，所以以最后改变的值为准。

对象的排序：本文主要阐述对存储datatable的list进行按tablename排序

1、新建一个sort类

public class scenesort:icomparer<datatable>
    {
       public int compare(datatable obj1, datatable obj2)
       {
           int tablenamelength1;
           int tablenamelength2;
           if (obj1 == null)
           {
               if (obj2 == null)
                   return 0;
               else
                   return -1;
           }
           else
           {
               if (obj2 == null)
                   return 1;
               else
               {
                   tablenamelength1=obj1.tablename.length;
                   tablenamelength2=obj2.tablename.length;
                   if (convert.toint32(obj1.tablename.substring(2,tablenamelength1-2))>convert.toint32(obj2.tablename.substring(2,tablenamelength2-2)))
                       return 1;                            //大于返回1
                   else if (convert.toint32(obj1.tablename.substring(2,tablenamelength1-2))<convert.toint32(obj2.tablename.substring(2,tablenamelength2-2)))
                       return -1;                           //小于返回-1
                   else
                       return 0;                            //相等返回0
               }
           }

       }

2 排序：
list<datatable> ldt = new list<datatable>();
scenesort ss=new scenesort ();
tablelist.sort(ss);即可对list进行排序；