英文:
Using multiplicity without navigability in UML is correct diagram?
问题
I can see many UML example like above diagram.
I can understand that
- Class A has navigability to B with multiplicity of 10.
But I can't understand that - Class B has no navigability to A but has multiplicity of 5..7.
Is it possible has multiplicity without navigability in UML?
I can implement case 1 like below.
class A {
public:
B b[10];
}
But how can I implement case 2? No navigability but has multiplicity?
I am totally confused of case 2. What am I missing?
英文:
I can see many UML example like above diagram.
I can understand that
- Class A has navigability to B with multiplicity of 10.
But I can't understand that - Class B has no navigability to A but has multiplicity of 5..7.
Is it possible has multiplicity wihtout navigability in UML?
I can implement case 1 like below.
class A {
public:
B b[10];
};
But how can I implement case 2? No navigability but has multiplicity?
I am totally confused of case 2. What am I missing?
答案1
得分: 2
是的，可以在没有导航性的情况下拥有多重性。
导航性只是对在运行时轻松查找另一端的实例做出的承诺。没有导航性的承诺（甚至在另一端用X明确表示没有导航性）并不会改变关联的存在，因此也不会改变多重性。
第二种情况的实现，例如，可以使用在B中指向A的指针向量来实现导航回去（导航性会存在，但只是没有指示出来）。另一种实现方式可能是在A中使用指向B的指针向量，以便B可以被多个A共享，而其他机制确保它被3到7个A共享（没有导航性，但多重性被强制执行）。
英文:
Yes, it is possible to have a multiplicity without navigability.
Navigability only makes a promise about the ease at runtime to find instances at the other end. The fact the there is no promise on navigability (or even an explicit absence of navigability with a X at the opposite end) does not change the existence of an association and hence the multiplicity.
The implementation of 2) could for example be done with a vector of pointers to A in B to navigate back (the navigability would exist, but is simply not indicated). Another implementation coukd be to use in A a vector of pointers to B, so that B can be shared, and some other mechanism make sure it is shared by 3 to 7 A's (no navigability, but multiplicity enforced)
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