名单上的所有位置数据覆盖 [英] List overwriting data on all positions

问题描述

我创建了一个类3个字段：

I created a class with 3 fields:

class Gente
    {
        int _ID;
        string _nome, _sexo;


        public Gente()
        {
            _ID = 0;
            _nome = "";
            _sexo = "";
        }

        public int ID
        {
            set { _ID = value; }
            get { return _ID; }
        }

        public string Nome
        {
            set { _nome = value; }
            get { return _nome; }
        }

        public string Sexo
        {
            set { _sexo = value; }
            get { return _sexo; }
        }
    }

然后，我宣布从类和列表。对象从该类这样我可以能够添加到列表

Then I declared a List from that class and an object from that class so that I can be able to add to the List.

List<Gente> Alunos = new List<Gente>();        

        Gente professor = new Gente();

我第一次做Alunos.Add（教授）;到列表中它正确地将信息发送到0位置。

The first time I do the Alunos.Add(professor); to the list it correctly sends the information to the 0 position.

但是，当我做第二次它会覆盖新数据位置0的一切，除了添加新数据到一个新的位置，1。

But when I do it second time it overwrites everything on position 0 with the new data besides adding the new data to a new position, 1.

推荐答案

什么是您的代码，这两个对象添加到列表中的样子？我的猜测是：

What does your code to add the two objects to the list look like? My guess is:

Gente professor = new Gente();
professor.Nome = "Fu";
Alunos.Add(professor);

professor.Nome = "Bar";
Alunos.Add(professor);

这是不正确。当您添加引用类型列表，你只需要添加到原来的对象的引用。在这种情况下，你要添加两个引用同一个对象两次。

Which is incorrect. When you add reference types to a list, you're only adding a reference to the original object. In this case you're adding two references to the same object twice.

因此，当你修改的第二个对象...你还修改了第一位。

Therefore, when you modify the second object...you're also modifying the first.

一个简单的一行修复将重新初始化教授第二次使用前：

A simple one-liner fix would be to re-initialize professor before its second use:

Gente professor = new Gente();
professor.Nome = "Fu";
Alunos.Add(professor);

professor = new Gente();
professor.Nome = "Bar";
Alunos.Add(professor);

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