ArrayList vs Vector passing by reference

Question

In Java, if you have the following code, list1 and list2 end up pointing to the same object, so a change in list2 will result in a change in list1.

ArrayList<Integer> list1 = new ArrayList<Integer>();
list1.add(1);
list1.add(2);
list1.add(3);
ArrayList<Integer> list2 = new ArrayList<Integer>();
list2 = list1;
list2.set(0, 10);

for(int i =0; i<list1.size(); i++){
    System.out.println(list1.get(i));
}
for(int i =0; i<list2.size(); i++){
    System.out.println(list2.get(i));
}

And the output ends up being:

However, in C++, the behavior is different?

vector<int> list1;
list1.push_back(1);
list1.push_back(2);
list1.push_back(3);
vector<int> list2;
list2 = list1;
list2.at(0) = 10;
for(int i =0; i<list1.size(); i++){
    cout << list1[i] << endl;
}

for(int i=0; i<list2.size(); i++){
    cout << list2[i] << endl;
}

The output ends up being:

Can someone explain?

In your Java code you assign *references*. in your C++ code you assign *objects*. Make it `vector&` or `vector*` and you'll get Java in C++! ;) — Ivan Aksamentov - Drop, Jan 24 '16 at 04:09
More details: [The Definitive C++ Book Guide and List](http://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list) — Ivan Aksamentov - Drop, Jan 24 '16 at 04:11

Ted Hopp · Accepted Answer · 2016-01-24T05:55:31.847

3

To elaborate on my comment:

In Java, the = operator on (assignment-compatible) reference types simply copies the reference value. (Think pointer if you're coming from the C/C++ world.) The key thing to keep in mind is that in Java, no variable is ever an object; it is either a primitive value or a reference to an object.

In your C++ example, the = operator is working on actual objects, not on references. (With other data types, the story might be different.) From the docs for std::vector::operator=:

Assigns new contents to the container, replacing its current contents, and modifying its size accordingly.

So in C++, list2 remains a separate vector object after the assignment, just with a copy of the contents of list1. In Java, the second ArrayList object you assigned to list2 becomes garbage once you reassign list2 to be a reference to the same object as list1.

edited Jan 24 '16 at 05:55

answered Jan 24 '16 at 04:09

Ted Hopp

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Just one question. When I write ArrayList list2 = list1, why does this behavior now mimic that of C++? – Danny Liu Jan 24 '16 at 04:19
@DannyLiu - It does? That should definitely not be happening. Please check your code again. If it is indeed behaving like the C++ code, please edit your question and post the revised version and its output. – Ted Hopp Jan 24 '16 at 04:31
Oh shoot it doesn't. My bad – Danny Liu Jan 24 '16 at 04:37
@DannyLiu - Good. Balance has been restored to the Force. :) – Ted Hopp Jan 24 '16 at 04:41

score 1 · Answer 2 · answered Jan 24 '16 at 04:17

1

To explain simply:

Java uses reference semantics. In this case you have two names (list1, list2) for the same object after the line list2 = list1.

C++ uses value semantics. The values of list1 are copied to list2 and you have a new unrelated object (list2) that may be mutated without affecting the copied object(list1).

answered Jan 24 '16 at 04:17

evansjohnson

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Of course, you can do the same thing in Java with `list2 = new ArrayList(list1);`. – Ted Hopp Jan 25 '16 at 03:19

score 1 · Answer 3 · answered Jan 24 '16 at 05:51

1

The simplest way to have the same behavior in C++ is to declare list2 as a reference - however, this requires that the assignment occur in list2's declaration and not after:

vector<int> &list2 = list1;

answered Jan 24 '16 at 05:51

Dave Doknjas

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ArrayList vs Vector passing by reference

3 Answers3