[review]Design Pattern:Prototype

Prototype

Use the current instance of a type to create a same instance with the Clone but not new.

When we use this pattern 

I have no idea now, need to dig more in the future.

Roles in this pattern 

  • Prototype: define an interface used to clone itself
  • ConcretePrototype: implement the interface of the Prototype and what's more, implement the Clone method to make the self-clone.
  • Client:Use the ConcretePrototype to clone itself.

Demo

namespace Prototype
{
public abstract class Prototype
{
private string id;

public string Id
{
get { return this.id; }
}

public Prototype(string id)
{
this.id = id;
}

public abstract Prototype Clone();
}

public class ConcretePrototype : Prototype
{
public ConcretePrototype(string id) : base(id) { }

public override Prototype Clone()
{
return (Prototype)this.MemberwiseClone();
}
}

public class Client
{
static void Main(string[] args)
{
ConcretePrototype c1 = new ConcretePrototype("me");
ConcretePrototype c2 = (ConcretePrototype)c1.Clone();

Console.WriteLine(c2.Id);//show "me" here
}
}
}

This is a common example to demonstrate this pattern, but in ASP.NET, you don't need to do like this, because there is an interface called as ICloneable available.

The ICloneable is equivalent to the Prototype defined above. So in ASP.NET, you can simplify the example above like below:

namespace Prototype
{

public class ConcretePrototype : ICloneable
{
private string id;

public string Id
{
get { return this.id; }
}
public ConcretePrototype(string id)
{
this.id = id;
}

public object Clone()
{
return (Object)this.MemberwiseClone();
}

}

public class Client
{
static void Main(string[] args)
{
ConcretePrototype c1 = new ConcretePrototype("me");
ConcretePrototype c2 = (ConcretePrototype)c1.Clone();

Console.WriteLine(c2.Id);//show "me" here
}
}
}

The new version is simpler and briefer than the old one with the help of the inherent interface.

there are two concepts we got to know:

  • Shallow copy
  • Deep copy
I don't wanna detail them know here, can google or read the book



转载于:https://www.cnblogs.com/sanjia/archive/2011/11/28/2266837.html

出现这个错误的原因是在导入seaborn包时,无法从typing模块中导入名为'Protocol'的对象。 解决这个问题的方法有以下几种: 1. 检查你的Python版本是否符合seaborn包的要求,如果不符合,尝试更新Python版本。 2. 检查你的环境中是否安装了typing_extensions包,如果没有安装,可以使用以下命令安装:pip install typing_extensions。 3. 如果你使用的是Python 3.8版本以下的版本,你可以尝试使用typing_extensions包来代替typing模块来解决该问题。 4. 检查你的代码是否正确导入了seaborn包,并且没有其他导入错误。 5. 如果以上方法都无法解决问题,可以尝试在你的代码中使用其他的可替代包或者更新seaborn包的版本来解决该问题。 总结: 出现ImportError: cannot import name 'Protocol' from 'typing'错误的原因可能是由于Python版本不兼容、缺少typing_extensions包或者导入错误等原因造成的。可以根据具体情况尝试上述方法来解决该问题。<span class="em">1</span><span class="em">2</span><span class="em">3</span> #### 引用[.reference_title] - *1* *2* *3* [ImportError: cannot import name ‘Literal‘ from ‘typing‘ (D:\Anaconda\envs\tensorflow\lib\typing....](https://blog.youkuaiyun.com/yuhaix/article/details/124528628)[target="_blank" data-report-click={"spm":"1018.2226.3001.9630","extra":{"utm_source":"vip_chatgpt_common_search_pc_result","utm_medium":"distribute.pc_search_result.none-task-cask-2~all~insert_cask~default-1-null.142^v93^chatsearchT3_2"}}] [.reference_item style="max-width: 100%"] [ .reference_list ]
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