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I'm trying to do some practice for an exam and need some help (did I do them correctly? If not, why?) on these problems about java primitives and java objects. These are all true or false.

  1. The following variable declaration is a reference to an object which is dynamically allocated and stored in the heap: int x = 7;

    False, because it is pass by value since int is primitive

  2. The following variable declaration is a reference to an object which is dynamically allocated and stored in the heap: Integer x = 7;

    True, because it is referencing an object stored on the heap

  3. If you pass the variable ‘x’ as declared in (1) to a method, that variable is passed by reference, and its value may be modified by the called function.

    False,because Java only does pass by value

  4. If you pass the variable ‘x’ as declared in (2) to a method, a copy of that variable is created in the heap, and passed to the function, so that the function’s object reference is pointing to a distinct place in memory.

    True, because variable is in the stack but it is pointing to a place in the heap

Thank you all for the help.

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  • I'm not convinced by 4) -- IIRC, a copy of the reference is placed on the stack, though the object referred to is still in the heap. Commented Dec 11, 2013 at 22:07
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    4) seems false to me; a copy of the reference is passed but that copy points to the same object which in on the heap Commented Dec 11, 2013 at 22:07
  • Do some tests, printing the values, before and after internal changes and you will get your answer. Commented Dec 15, 2013 at 13:05

4 Answers 4

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(4) is false for two reasons:

  • "a copy of that variable is created in the heap" is false. Only objects are on the heap. The local variable in a function is not on the heap. The value of that variable is simply passed, i.e. copied into the stack frame of the called function.
  • "the function’s object reference is pointing to a distinct place in memory." is false. The function's reference will point to the same place in memory. That's the point of passing -- the value of the variable inside the function is the same as the value that was passed. The value of a reference is where it points.
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4) If you pass the variable ‘x’ as declared in (2) to a method, a copy of that variable is created in the heap, and passed to the function, so that the function’s object reference is pointing to a distinct place in memory.

Okay, this can be a little sketchy. If dealing with objects, you are not creating a copy of the object and passing the copy to the method. You are creating a copy of the reference to the object, and passing that over by value.

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Actually there is a reason 2 is wrong - but its probably not the reason the person who set it expects.

 Integer x = 7;

Will get turned into:

 Integer x = Integer.valueOf(7);

Which will re-use a cached value for all integers in the range -128 to +127 inclusive. (It may reuse them for other values too).

So you will get a reference to an object which is present in the JVM implementation dependant cached storage for common Integer values.

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so you believe that this cache is not on the heap?
It probably is on the heap but it will definitely not be "dynamically allocated" (after the first time at least depending on whether it's lazy allocation or not of the cache) on the heap.
So you said the object can be lazily allocated. So how is it allocated when it is allocated, if not dynamically? Literally, the statement said that the "object ... is dynamically allocated". It doesn't say when the object is dynamically allocated.
It's not very dynamic if it only ever happens once...you might as well say my house is dynamically placed - since it was dynamic at the point they decided to build it there even though it never moved since. Dynamic allocation would imply a new one was created every time you did Integer x=7 - which is not the case.
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False because Integer is an object. So you will be passing just the reference to the object address in the function. It is standard in Java.

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