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In an interview I was asked why are objects in java created dynamically?

I can't understand this question, can any body explain this?

Sajan Chandran
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user2783700
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    That makes two of us – Bart Sep 16 '13 at 11:53
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    Make it three. Has the interviewer at least told you what (s)he means by "dynamically" in this context? – mikołak Sep 16 '13 at 11:54
  • flied over me. add me – Adi Sep 16 '13 at 11:57
  • Is the interviewer a techy or just some random business guy who has no clue about any Computing terminology, because he just sounds like an idiota. – blackpanther Sep 16 '13 at 11:57
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    "Dynamically" is such a meaningless word. In many cases, you can simply remove it from a sentence and the sentence still has the same meaning. In this case as well: The question is about as meaningful, whether or not you include the "dynamically". – Joachim Sauer Sep 16 '13 at 11:57
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    The most sensible interpretation of this is "why are all objects created on the heap and none on the stack" (unlike in C++ or C# structs). But the question is open to interpretation, so it would be best to ask for a clarification. – nd. Sep 16 '13 at 11:59
  • If the interviewer doesn't know the answer to this question, why is he asking you? Rather than James Gosling for example? If he does know, what makes him think you know? Or should know? And why does he care? – user207421 Sep 16 '13 at 12:00
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    @JoachimSauer Do you mean that you can dynamically include or exclude the word "dynamically" in a sentence and still maintain the same meaning? – Theodoros Chatzigiannakis Sep 16 '13 at 12:02
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    @TheodorosChatzigiannakis: that's what I dynamically meant. – Joachim Sauer Sep 16 '13 at 12:03
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    i think interviewer doesn't know the answer to this question.. – gifpif Sep 16 '13 at 12:17
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    It isn't even true. String objects for string literals are created statically, in effect. – user207421 Sep 17 '13 at 23:26

4 Answers4

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The person could be referring to the fact that Java does not know implicit object creation on the stack like C++ does.

std::string string;

Creates an object on the stack in C++ while

String string;

just creates a reference in Java, but no object is created (and no constructor called).

If you are interested about that topic I suggest to read more about Java's memory model.

Jan Henke
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    They *could* mean that, but personally I'd not rate it highly if the candidate just guessed and didn't ask for clarification at this point. – Joachim Sauer Sep 16 '13 at 12:11
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It seems to me that the question asks why objects in Java are created only with dynamically allocated memory (using the new keyword) as opposed to being created with statically allocated memory as well (like in C++, for example). If (and it's a big if) this is what the question meant, there are some things you can answer to that.

Before answering, one must note that the premise of the question is not entirely right (you could even say that it's wrong). Java objects aren't created strictly dynamically. If escape analysis can prove that a reference doesn't escape a given scope, it may be compiled so that it uses static allocation.

Given the above, one possible answer is abstraction. The stack and the heap (which are traditionally associated with static and dynamic memory, respectively) are in fact implementation details (even if we are so used to them). Java attempts to hide that and thus it doesn't give you terms like static or dynamic memory to work with - it doesn't even give you memory, it gives you objects.

Another answer (again, given the note) is the real world usage of objects. A lot of the time in real world scenarios, objects do need to escape their initial scope, making dynamic allocation the only valid candidate.

Theodoros Chatzigiannakis
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A very open-ended question, some context is required to give a good answer.

It's not specific to Java, but to all programming languages in general. Every object can be though of as dynamic, as memory allocated for an object, is allocated in run-time, from the resources available on the executing platform. Even constants, static classes and similar concepts use memory only when a file was actually executed.

Static objects may exist in some hardware hardcoded values, such as some parts of BIOS ot TPL module. This values are read-only - there is no known way to change the values of the object.

However, this is only a single answer that depends on what I think dynamic means. It is subjective and different people may come up with other valid answers. It's not a good question for an interview, unless some additional context was given.

oleksii
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If you take the question literall i would say:

So you dont have to load everything at once (on the init fase) and keep it in the memory till the end of the app/program. which gives you a more efficient memory usage.

;-)

W vd L
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