C# 如何在C中创建一个惰性属性,将类的字符串属性作为参数传递给它?
是否有方法将Name属性作为参数传递给延迟BOM表初始化C# 如何在C中创建一个惰性属性,将类的字符串属性作为参数传递给它?,c#,lazy-initialization,C#,Lazy Initialization,是否有方法将Name属性作为参数传递给延迟BOM表初始化 public class Item { private Lazy<BOM> _BOM = new Lazy<BOM>(); // How to pass the Name as parameter ??? public Item(string name) { this.Name = name; } public string Name {
public class Item
{
private Lazy<BOM> _BOM = new Lazy<BOM>(); // How to pass the Name as parameter ???
public Item(string name)
{
this.Name = name;
}
public string Name { get; private set; }
public BOM BOM { get { return _BOM.Value; } }
}
public class BOM
{
public BOM (string name)
{
}
}
您可以使用Lazy来实现这一点。这也意味着,正如Zohar的评论所建议的那样,实例化必须移动到构造函数中,因为非静态字段不能从字段初始值设定项引用
public class Item
{
private Lazy<BOM> _BOM;
public Item(string name)
{
this.Name = name;
_BOM = new Lazy<BOM>(() => new BOM(Name));
}
public string Name { get; private set; }
public BOM BOM { get { return _BOM.Value; } }
}
public class BOM
{
public BOM(string name)
{
}
}
您可以使用Lazy来实现这一点。这也意味着,正如Zohar的评论所建议的那样,实例化必须移动到构造函数中,因为非静态字段不能从字段初始值设定项引用
public class Item
{
private Lazy<BOM> _BOM;
public Item(string name)
{
this.Name = name;
_BOM = new Lazy<BOM>(() => new BOM(Name));
}
public string Name { get; private set; }
public BOM BOM { get { return _BOM.Value; } }
}
public class BOM
{
public BOM(string name)
{
}
}
不要实例化Lazy at声明,而是在Item的构造函数中实例化它:
不要实例化Lazy at声明,而是在Item的构造函数中实例化它:
该死快了半分钟!:-@ZoharPeled我喝的咖啡太多了我还不够:-该死!快了半分钟!:-@ZoharPeled我喝了太多咖啡我喝得不够:-