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Wrong answer? HFE Page 432, question 4

 
Hai Lin
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Hi, guys,
Feel big doubt for the HFE page 432, question 4

Given the container-managed unidirectional relationship:
Foo (0-1) --> Bar (0-1)
And the object relations:
f1-->b1
f2-->b2

What will be true after the following code runs? (Choose all the that apply.)
f2.setBar(f1.getBar());

A. f1.getBar() == null
B. b2.getFoo() == null
C. b1.getFoo() == null
D. none of the above

the answer given by the book is A, B. Actually, why B is correct? Since the Foo and Bar are unidirectional relationship, that mean, foo can getBar(), but bar cannot get Foo, there should NOT have method like b2.getFoo() existing.

Thanks a lot for your advice.

Hai
[ August 12, 2004: Message edited by: Hai Lin ]
 
alzamabar
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Originally posted by Hai Lin:
Hi, guys,
Feel big doubt for the HFE page 432, question 4

Given the container-managed unidirectional relationship:
Foo (0-1) --> Bar (0-1)
And the object relations:
f1-->b1
f2-->b2

What will be true after the following code runs? (Choose all the that apply.)
f2.setBar(f1.getBar());

A. f1.getBar() == null
B. b2.getFoo() == null
C. b1.getFoo() == null
D. none of the above

the answer given by the book is A, B. Actually, why B is correct? Since the Foo and Bar are unidirectional relationship, that mean, foo can getBar(), but bar cannot get Foo, there should NOT have method like b2.getFoo() existing.

Thanks a lot for your advice.
Hai

[ August 12, 2004: Message edited by: Hai Lin ]


Actually, although your observation is *logically* true, there is nothing in the specs that say that you can't go from Bar to Foo. I'm referring to chapter '10.3.7.2 One-to-one unidirectional relationships', page 138.

It would be interesting to hear from the author (or co-authors) the explanation to this question.
 
alzamabar
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Originally posted by Marco Tedone:


Actually, although your observation is *logically* true, there is nothing in the specs that say that you can't go from Bar to Foo. I'm referring to chapter '10.3.7.2 One-to-one unidirectional relationships', page 138, but, that said, I think that answer B was not correct too.

 
Dan T
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Originally posted by Hai Lin:
Hi, guys,
Feel big doubt for the HFE page 432, question 4

Given the container-managed unidirectional relationship:
Foo (0-1) --> Bar (0-1)
And the object relations:
f1-->b1
f2-->b2

What will be true after the following code runs? (Choose all the that apply.)
f2.setBar(f1.getBar());

A. f1.getBar() == null
B. b2.getFoo() == null
C. b1.getFoo() == null
D. none of the above

the answer given by the book is A, B. Actually, why B is correct? Since the Foo and Bar are unidirectional relationship, that mean, foo can getBar(), but bar cannot get Foo, there should NOT have method like b2.getFoo() existing.

Thanks a lot for your advice.

Hai

[ August 12, 2004: Message edited by: Hai Lin ]


This is because every Foo can only have 1 Bar, and every Bar can only have 1 Foo. Thus, if u set B1 to F2, F2 will not have relatioship with B2 because F2 can only refer to 1 Bar, which is B1. Now, B2 refers to nothing.
 
Lionel Orellana
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Have a look at
Unconfirmed error reports and comments from readers

for HFEJB.
 
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