How to implement HAS-A relationship?

[Andy James]

If class A shares HAS-A relationship with class B, is definition of instance variables the only way that these classes can share HAS-A relationship:

class A{}
class B {
    A var = new A();     // class B HAS-A class A
}

or can class B define variables of class A as local variable to implement HAS-A relationship?

class A{}
class B {
    void aMethod() {
        A var = new A();     // Is this HAS-A relationship?
    }
}

Cheers,
Andy

[Kevin Florish]

You can implement a HAS-A relationship as long as you can create an instance of the class and this depends on what access you have to that class.

So for Class B to have a Class A, yes it would need to be part of its object state and not in a method but you still need to have access to incorporate it.

This post on encapsualtion may help with your understanding of access modifers encapsulation discussion

[Andy James]

Thanks, Kevin.

So can I infer that if class B is accessible in class A, the following is the only way to implement relationship "B HAS-A A"?

class A{}   
class B {   
    A var = new A();     // class B HAS-A class A   
}

Class B can't define variable of class A as method parameter, local variable or static variable to implement relation "B HAS-A A". RIght?

Cheers,
Andy

[Kevin Florish]

Yes I think my answer wasn't very clear.

So to put it another way for Class B to have a HAS-A relationship with class A then it needs to have an instance of class A as part of its object state, in other words as one of its instance variables.

Local variables only generally exist as long as the method they reside in exists on the stack, unless the local variable is an object and that object is referenced elsewhere after the method ends and is popped off the stack.

Static variables are class variables and so pertain to the class as a whole, not an instance of the class.

Method parameters are just a signature we use to pass arguments across to a method.

hth Kevin.

[Seetharaman Venkatasamy]

Andy James wrote:can class B define variables of class A as local variable to implement HAS-A relationship?

class A{}
class B {
    void aMethod() {
        A var = new A();     // Is this HAS-A relationship?
    }
}

Cay Horstmann calls this relationship as *uses* relationship(Dependency) , and
he claimed that Aggregation is a stronger form of this uses relationships, since aggregation has another object as instance variable.
and also he mentioned uses relationship denoted as -----> in UML diagram . (for more details read BIG JAVA(3rd edition) - chapter:Object-Oriented Design, page-540)

hence, shall we call it as *weak* HAS-A? :-)

[Andy James]

Thanks for all the replies. It is getting interesting. A quick question:

Does the HAS-A relationship also applies to the base classes? For example, if class A extends Base and B HAS-A A, does it also imply that class B HAS-A Base?

class Base {}
class A extends Base {}      
class B {      
    A var = new A();     // Because A extends Base and B HAS-A, does B HAS-A Base hold true?
}

cheers,
Andy

[Seetharaman Venkatasamy]

Andy James wrote:
Does the HAS-A relationship also applies to the base classes? For example, if class A extends Base and B HAS-A A, does it also imply that class B HAS-A Base?

class Base {}
class A extends Base {}      
class B {      
    A var = new A();     // Because A extends Base and B HAS-A, does B HAS-A Base hold true?
}

Yup, because A IS-A Base .

[Kevin Florish]

Just to expand on Seetharaman's answer a little bit.

Class B IS-A Base through inheritance not composition, so B inherits Base as it does Object through an implied extends through Base.

So B IS-A Base as opposed to HAS-A Base, whereas B HAS-A A.

But as I write this I realize that anything used from Base for A will appear in the instance variable of B, so I suppose it is also composition, I have confused myself