linux&c++ R&D

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Understand decorator design pattern

    The Decorator Pattern is used for adding additional functionality to a particular object as opposed to a class of objects. It is easy to add functionality to an entire class of objects by subclassing an object, but it is impossible to extend a single object this way. With the Decorator Pattern, you can add functionality to a single object and leave others like it unmodified.
   A Decorator, also known as a Wrapper, is an object that has an interface identical to an object that it contains. Any calls that the decorator gets, it relays to the object that it contains, and adds its own functionality along the way, either before or after the call. This gives you a lot of flexibility, since you can change what the decorator does at runtime, as opposed to having the change be static and determined at compile time by subclassing. Since a Decorator complies with the interface that the object that it contains, the Decorator is indistinguishable from the object that it contains.  That is, a Decorator is a concrete instance of the abstract class, and thus is indistinguishable from any other concrete instance, including other decorators.   This can be used to great advantage, as you can recursively nest decorators without any other objects being able to tell the difference, allowing a near infinite amount of customization.



An example:

#include <iostream>
class IntCompute{
public:
    virtual 
int Compute(int,int=0;
}
;
class IntComputeAdd:public IntCompute{
public:
int Compute(int a,int b){
    
return a+b;
}

}
;
class DecorateMinusAdd:public IntCompute{
private:
    IntComputeAdd intadd;
public:
    
int Compute(int a,int b)
    
{
      
return -(intadd.Compute(a,b));

    }

}
;
class DecorateSumSquare:public IntCompute{
private:
    IntComputeAdd intadd;
public:
    
int Compute(int a,int b)
    
{
        
int result=intadd.Compute(a,b);
        
return result*result;

    }

}
;
int main()
{
 
DecorateMinusAdd minadd;
DecorateSumSquare sumsquare;
std::cout
<<minadd.Compute(10,20)<<std::endl;
std::cout
<<sumsquare.Compute(10,20)<<std::endl;
 
return 0;
}

posted on 2007-05-11 21:38 丑石 阅读(450) 评论(0)  编辑 收藏 引用 所属分类: software engineering


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