1

我花了一些调试来解决这个问题(或者我认为如此)。我会让你的代码松散,看看你想出什么。有一个简单的Contact类:

  1. 一些自动属性,
  2. 一个参数化的构造函数,它总是增加Contact.ID属性并根据它获得的参数设置其他属性
  3. 一个无参数构造函数,它总是使用默认值调用参数化构造函数。

先看代码;它的输出和问题遵循代码:

 using System;

 class Program 
 {
   private static void Main(string[] args) 
   {
      Contact[] contacts_array = {

         //Contact 0
         new Contact(),

         //Contact 1
         new Contact {
            Name = "contactName1",
            Age = 40,
            Email = "emailaddress1@email.com"
         },

         //Contact 2
         new Contact {
            Name = "contactName2",
            Age = 41,
            Email = "emailaddress2@email.com"
         },

         //Contact 3
         new Contact("contactName3",
            42,
            "emailaddress3@email.com"),
      };

      foreach (var contact in contacts_array)
         Console.WriteLine(contact);

      Console.ReadLine();
   }
}

public class Contact
{
    public static int totalContacts = 0;
    public int Id { get; private set; }
    public string Name { get; set; }
    public int? Age { get; set; }
    public string Email { get; set; }

    public Contact()
    {  
        new Contact("ANONYMOUS", null, "ANONYMOUS@unknown.com");  
    }

    public Contact(string name, int? age, string email)
    {
        Id = Contact.totalContacts++;
        Name = name;
        Age = age;
        Email = email;
    }

    public override string ToString()
    {
        return string.Format("[Contact: Id={0}, Name={1}, Age={2}, Email={3}]",
                             Id, Name, Age, Email);
    }
}

输出:

[Contact: Id=0, Name=, Age=, Email=]  
[Contact: Id=0, Name=contactName1, Age=40, Email=emailaddress1@email.com]  
[Contact: Id=0, Name=contactName2, Age=41, Email=emailaddress2@email.com]  
[Contact: Id=3, Name=contactName3, Age=42, Email=emailaddress3@email.com] 

问题:

为什么第二个和第三个联系人中的 Contact.ID == 0 而不是分别为 1 和 2,尽管始终调用参数化构造函数并始终递增 ID 属性?

4

1 回答 1

6

您的默认构造函数并没有按照您的想法执行:

public Contact()
{
    new Contact("ANONYMOUS", null, "ANONYMOUS@unknown.com");
}

这将构造一个新的Contact然后丢弃它,当前实例将获得所有默认值。这是您所追求的语法:

public Contact()
  : this("ANONYMOUS", null, "ANONYMOUS@unknown.com")
{
}
于 2010-09-24T06:19:13.690 回答