将值分配给属性时出现空错误

Question

尝试创建一个自定义对象，该对象将允许我在记录中拾取错误。

public class gridIntegerField 
{ 
    private int value; 
    private bool isValid; 
    private string message; 



    public int Value 
    { 
     get { return this.value; } 
     set { this.value = value; } 
    } 
    public bool IsValid 
    { 
     get { return isValid; } 
     set { isValid = value; } 
    } 
    public string Message 
    { 
     get { return message; } 
     set { message = value; } 
    } 


} 

public class gridRecord  
{ 
    private gridIntegerField printRun; 

    public gridIntegerField PrintRun 
    { 
     get { return printRun; } 
     set { printRun = value; } 
    } 

} 

当创建对象，试图把我得到如下因素误差值...

型“System.NullReferenceException”未处理的异常发生在基于XML Console.exe

代码为创建对象...

gridRecord spr = new gridRecord(); 
spr.PrintRun.Value = 200; 
spr.PrintRun.IsValid = true; 
spr.PrintRun.Message = "No Errors"; 

Console.WriteLine(spr.PrintRun.Value.ToString()); 
Console.WriteLine(spr.PrintRun.IsValid.ToString()); 
Console.WriteLine(spr.PrintRun.Message.ToString()); 
Console.ReadKey(); 

错误发生在这行代码

spr.PrintRun.Value = 200; 

Answer 1

因为在gridRecord实例化后没有实例化printRun字段。

你可以在构造方法中做到这一点。

public class gridRecord  
{ 
    private gridIntegerField printRun; 

    public gridIntegerField PrintRun 
    { 
     get { return printRun; } 
     set { printRun = value; } 
    } 

    //Add a constructor method 
    public gridRecord() 
    { 
     //and instantiate the printRun. 
     printRun = new gridIntegerField(); 
    }  
}