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What is the difference between a deep copy and a shallow copy?
Breadth vs Depth; think in terms of a tree of references with your object as the root node.
Shallow:
The variables A and B refer to different areas of memory, when B is assigned to A the two variables refer to the same area of memory. Later modifications to the contents of either are instantly reflected in the contents of other, as they share contents.
Deep:
The variables A and B refer to different areas of memory, when B is assigned to A the values in the memory area which A points to are copied into the memory area to which B points. Later modifications to the contents of either remain unique to A or B; the contents are not shared.
Shallow copies duplicate as little as possible. A shallow copy of a collection is a copy of the collection structure, not the elements. With a shallow copy, two collections now share the individual elements.
Deep copies duplicate everything. A deep copy of a collection is two collections with all of the elements in the original collection duplicated.
Try to consider following image
For example Object.MemberwiseClone creates a shallow copy link
and using ICloneable interface you can get deep copy as described here
In short, it depends on what points to what. In a shallow copy, object B points to object A's location in memory. In deep copy, all things in object A's memory location get copied to object B's memory location.
This wiki article has a great diagram.
http://en.wikipedia.org/wiki/Object_copy
Especially For iOS Developers:
If B is a shallow copy of A, then for primitive data it's like B = [A assign]; and for objects it's like B = [A retain];
B and A point to the same memory location
If B is a deep copy of A, then it is like B = [A copy];
B and A point to different memory locations
B memory address is same as A's
B has same contents as A's
Shallow copy: Copies the member values from one object into another.
Deep Copy: Copies the member values from one object into another.
Any pointer objects are duplicated and Deep Copied.
Example:
class String
{
int size;
char* data;
};
String s1("Ace"); // s1.size = 3 s1.data=0x0000F000
String s2 = shallowCopy(s1);
// s2.size =3 s2.data = 0X0000F000
String s3 = deepCopy(s1);
// s3.size =3 s3.data = 0x0000F00F
// (With Ace copied to this location.)
Just for the sake of easy understanding you could follow this article:
https://www.cs.utexas.edu/~scottm/cs307/handouts/deepCopying.htm
Shallow Copy:
Deep Copy:
I haven't seen a short, easy to understand answer here--so I'll give it a try.
With a shallow copy, any object pointed to by the source is also pointed to by the destination (so that no referenced objects are copied).
With a deep copy, any object pointed to by the source is copied and the copy is pointed to by the destination (so there will now be 2 of each referenced object). This recurses down the object tree.
char * Source = "Hello, world.";
char * ShallowCopy = Source;
char * DeepCopy = new char(strlen(Source)+1);
strcpy(DeepCopy,Source);
'ShallowCopy' points to the same location in memory as 'Source' does.
'DeepCopy' points to a different location in memory, but the contents are the same.
{Imagine two objects: A and B of same type _t(with respect to C++) and you are thinking about shallow/deep copying A to B}
Shallow Copy:
Simply makes a copy of the reference to A into B. Think about it as a copy of A's Address.
So, the addresses of A and B will be the same i.e. they will be pointing to the same memory location i.e. data contents.
Deep copy:
Simply makes a copy of all the members of A, allocates memory in a different location for B and then assigns the copied members to B to achieve deep copy. In this way, if A becomes non-existant B is still valid in the memory. The correct term to use would be cloning, where you know that they both are totally the same, but yet different (i.e. stored as two different entities in the memory space). You can also provide your clone wrapper where you can decide via inclusion/exclusion list which properties to select during deep copy. This is quite a common practice when you create APIs.
You can choose to do a Shallow Copy ONLY_IF you understand the stakes involved. When you have enormous number of pointers to deal with in C++ or C, doing a shallow copy of an object is REALLY a bad idea.
EXAMPLE_OF_DEEP COPY_ An example is, when you are trying to do image processing and object recognition you need to mask "Irrelevant and Repetitive Motion" out of your processing areas. If you are using image pointers, then you might have the specification to save those mask images. NOW... if you do a shallow copy of the image, when the pointer references are KILLED from the stack, you lost the reference and its copy i.e. there will be a runtime error of access violation at some point. In this case, what you need is a deep copy of your image by CLONING it. In this way you can retrieve the masks in case you need them in the future.
EXAMPLE_OF_SHALLOW_COPY I am not extremely knowledgeable compared to the users in StackOverflow so feel free to delete this part and put a good example if you can clarify. But I really think it is not a good idea to do shallow copy if you know that your program is gonna run for an infinite period of time i.e. continuous "push-pop" operation over the stack with function calls. If you are demonstrating something to an amateur or novice person (e.g. C/C++ tutorial stuff) then it is probably okay. But if you are running an application such as surveillance and detection system, or Sonar Tracking System, you are not supposed to keep shallow copying your objects around because it will kill your program sooner or later.
What is Shallow Copy?
Shallow copy is a bit-wise copy of an object. A new object is created that has an exact copy of the values in the original object. If any of the fields of the object are references to other objects, only the reference addresses are copied i.e., only the memory address is copied.
In this figure, the MainObject1 has fields field1 of type int, and ContainObject1 of type ContainObject. When you do a shallow copy of MainObject1, MainObject2 is created with field2 containing the copied value of field1 and still pointing to ContainObject1 itself. Note that since field1 is of primitive type, its value is copied to field2 but since ContainedObject1 is an object, MainObject2 still points to ContainObject1. So any changes made to ContainObject1 in MainObject1 will be reflected in MainObject2.
Now if this is shallow copy, lets see what's deep copy?
What is Deep Copy?
A deep copy copies all fields, and makes copies of dynamically allocated memory pointed to by the fields. A deep copy occurs when an object is copied along with the objects to which it refers.
In this figure, the MainObject1 have fields field1 of type int, and ContainObject1 of type ContainObject. When you do a deep copy of MainObject1, MainObject2 is created with field2 containing the copied value of field1 and ContainObject2 containing the copied value of ContainObject1. Note any changes made to ContainObject1 in MainObject1 will not reflect in MainObject2.
good article
Deep Copy
A deep copy copies all fields, and makes copies of dynamically allocated memory pointed to by the fields. A deep copy occurs when an object is copied along with the objects to which it refers.
Shallow Copy
Shallow copy is a bit-wise copy of an object. A new object is created that has an exact copy of the values in the original object. If any of the fields of the object are references to other objects, just the reference addresses are copied i.e., only the memory address is copied.
In object oriented programming, a type includes a collection of member fields. These fields may be stored either by value or by reference (i.e., a pointer to a value).
In a shallow copy, a new instance of the type is created and the values are copied into the new instance. The reference pointers are also copied just like the values. Therefore, the references are pointing to the original objects. Any changes to the members that are stored by reference appear in both the original and the copy, since no copy was made of the referenced object.
In a deep copy, the fields that are stored by value are copied as before, but the pointers to objects stored by reference are not copied. Instead, a deep copy is made of the referenced object, and a pointer to the new object is stored. Any changes that are made to those referenced objects will not affect other copies of the object.
I would like to give example rather than the formal definition.
var originalObject = {
a : 1,
b : 2,
c : 3,
};
This code shows a shallow copy:
var copyObject1 = originalObject;
console.log(copyObject1.a); // it will print 1
console.log(originalObject.a); // it will also print 1
copyObject1.a = 4;
console.log(copyObject1.a); //now it will print 4
console.log(originalObject.a); // now it will also print 4
var copyObject2 = Object.assign({}, originalObject);
console.log(copyObject2.a); // it will print 1
console.log(originalObject.a); // it will also print 1
copyObject2.a = 4;
console.log(copyObject2.a); // now it will print 4
console.log(originalObject.a); // now it will print 1
This code shows a deep copy:
var copyObject2 = Object.assign({}, originalObject);
console.log(copyObject2.a); // it will print 1
console.log(originalObject.a); // it will also print 1
copyObject2.a = 4;
console.log(copyObject2.a); // now it will print 4
console.log(originalObject.a); // !! now it will print 1 !!
'ShallowCopy' points to the same location in memory as 'Source' does. 'DeepCopy' points to a different location in memory, but the contents are the same.
Shallow Cloning:
Definition: "A shallow copy of an object copies the ‘main’ object, but doesn’t copy the inner objects."
When a custom object (eg. Employee) has just primitive, String type variables then you use Shallow Cloning.
Employee e = new Employee(2, "john cena");
Employee e2=e.clone();
You return super.clone(); in the overridden clone() method and your job is over.
Deep Cloning:
Definition: "Unlike the shallow copy, a deep copy is a fully independent copy of an object."
Means when an Employee object holds another custom object:
Employee e = new Employee(2, "john cena", new Address(12, "West Newbury", "Massachusetts");
Then you have to write the code to clone the 'Address' object as well in the overridden clone() method. Otherwise the Address object won't clone and it causes a bug when you change value of Address in cloned Employee object, which reflects the original one too.
var source = { firstName="Jane", lastname="Jones" };
var shallow = ShallowCopyOf(source);
var deep = DeepCopyOf(source);
source.lastName = "Smith";
WriteLine(source.lastName); // prints Smith
WriteLine(shallow.lastName); // prints Smith
WriteLine(deep.lastName); // prints Jones
Shallow Copy- Reference variable inside original and shallow-copied objects have reference to common object.
Deep Copy- Reference variable inside original and deep-copied objects have reference to different object.
clone always does shallow copy.
public class Language implements Cloneable{
String name;
public Language(String name){
this.name=name;
}
public String getName() {
return name;
}
#Override
protected Object clone() throws CloneNotSupportedException {
return super.clone();
}
}
main class is following-
public static void main(String args[]) throws ClassNotFoundException, CloneNotSupportedException{
ArrayList<Language> list=new ArrayList<Language>();
list.add(new Language("C"));
list.add(new Language("JAVA"));
ArrayList<Language> shallow=(ArrayList<Language>) list.clone();
//We used here clone since this always shallow copied.
System.out.println(list==shallow);
for(int i=0;i<list.size();i++)
System.out.println(list.get(i)==shallow.get(i));//true
ArrayList<Language> deep=new ArrayList<Language>();
for(Language language:list){
deep.add((Language) language.clone());
}
System.out.println(list==deep);
for(int i=0;i<list.size();i++)
System.out.println(list.get(i)==deep.get(i));//false
}
OutPut of above will be-
false true true
false false false
Any change made in origional object will reflect in shallow object not in deep object.
list.get(0).name="ViSuaLBaSiC";
System.out.println(shallow.get(0).getName()+" "+deep.get(0).getName());
OutPut- ViSuaLBaSiC C
Imagine there are two arrays called arr1 and arr2.
arr1 = arr2; //shallow copy
arr1 = arr2.clone(); //deep copy
In Simple Terms, a Shallow Copy is similar to Call By Reference and a Deep Copy is similar to Call By Value
In Call By Reference, Both formal and actual parameters of a function refers to same memory location and the value.
In Call By Value, Both formal and actual parameters of a functions refers to different memory location but having the same value.
A shallow copy constructs a new compound object and insert its references into it to the original object.
Unlike shallow copy, deepcopy constructs new compound object and also inserts copies of the original objects of original compound object.
Lets take an example.
import copy
x =[1,[2]]
y=copy.copy(x)
z= copy.deepcopy(x)
print(y is z)
Above code prints FALSE.
Let see how.
Original compound object x=[1,[2]] (called as compound because it has object inside object (Inception))
as you can see in the image, there is a list inside list.
Then we create a shallow copy of it using y = copy.copy(x). What python does here is, it will create a new compound object but objects inside them are pointing to the orignal objects.
In the image it has created a new copy for outer list. but the inner list remains same as the original one.
Now we create deepcopy of it using z = copy.deepcopy(x). what python does here is, it will create new object for outer list as well as inner list. as shown in the image below (red highlighted).
At the end code prints False, as y and z are not same objects.
HTH.
struct sample
{
char * ptr;
}
void shallowcpy(sample & dest, sample & src)
{
dest.ptr=src.ptr;
}
void deepcpy(sample & dest, sample & src)
{
dest.ptr=malloc(strlen(src.ptr)+1);
memcpy(dest.ptr,src.ptr);
}
To add more to other answers,
a Shallow Copy of an object performs copy by value for value types
based properties, and copy by reference for reference types based properties.
a Deep Copy of an object performs copy by value for value types based
properties, as well as copy by value for reference types based
properties deep in the hierarchy (of reference types)
Shallow copy will not create new reference but deep copy will create the new reference.
Here is the program to explain the deep and shallow copy.
public class DeepAndShollowCopy {
int id;
String name;
List<String> testlist = new ArrayList<>();
/*
// To performing Shallow Copy
// Note: Here we are not creating any references.
public DeepAndShollowCopy(int id, String name, List<String>testlist)
{
System.out.println("Shallow Copy for Object initialization");
this.id = id;
this.name = name;
this.testlist = testlist;
}
*/
// To performing Deep Copy
// Note: Here we are creating one references( Al arraylist object ).
public DeepAndShollowCopy(int id, String name, List<String> testlist) {
System.out.println("Deep Copy for Object initialization");
this.id = id;
this.name = name;
String item;
List<String> Al = new ArrayList<>();
Iterator<String> itr = testlist.iterator();
while (itr.hasNext()) {
item = itr.next();
Al.add(item);
}
this.testlist = Al;
}
public static void main(String[] args) {
List<String> list = new ArrayList<>();
list.add("Java");
list.add("Oracle");
list.add("C++");
DeepAndShollowCopy copy=new DeepAndShollowCopy(10,"Testing", list);
System.out.println(copy.toString());
}
#Override
public String toString() {
return "DeepAndShollowCopy [id=" + id + ", name=" + name + ", testlist=" + testlist + "]";
}
}
Taken from [blog]: http://sickprogrammersarea.blogspot.in/2014/03/technical-interview-questions-on-c_6.html
Deep copy involves using the contents of one object to create another instance of the same class. In a deep copy, the two objects may contain ht same information but the target object will have its own buffers and resources. the destruction of either object will not affect the remaining object. The overloaded assignment operator would create a deep copy of objects.
Shallow copy involves copying the contents of one object into another instance of the same class thus creating a mirror image. Owing to straight copying of references and pointers, the two objects will share the same externally contained contents of the other object to be unpredictable.
Explanation:
Using a copy constructor we simply copy the data values member by member. This method of copying is called shallow copy. If the object is a simple class, comprised of built in types and no pointers this would be acceptable. This function would use the values and the objects and its behavior would not be altered with a shallow copy, only the addresses of pointers that are members are copied and not the value the address is pointing to. The data values of the object would then be inadvertently altered by the function. When the function goes out of scope, the copy of the object with all its data is popped off the stack.
If the object has any pointers a deep copy needs to be executed. With the deep copy of an object, memory is allocated for the object in free store and the elements pointed to are copied. A deep copy is used for objects that are returned from a function.
I came to understand from the following lines.
Shallow copy copies an object value type(int, float, bool) fields in to target object and object's reference types(string, class etc) are copied as references in target object. In this target reference types will be pointing to the memory location of source object.
Deep copy copies an object's value and reference types into a complete new copy of the target objects. This means both the value types and reference types will be allocated a new memory locations.
Shallow copying is creating a new object and then copying the non-static fields of the current object to the new object. If a field is a value type --> a bit-by-bit copy of the field is performed; for a reference type --> the reference is copied but the referred object is not; therefore the original object and its clone refer to the same object.
Deep copy is creating a new object and then copying the nonstatic fields of the current object to the new object. If a field is a value type --> a bit-by-bit copy of the field is performed. If a field is a reference type --> a new copy of the referred object is performed. The classes to be cloned must be flagged as [Serializable].
Copying ararys :
Array is a class, which means it is reference type so array1 = array2 results
in two variables that reference the same array.
But look at this example:
static void Main()
{
int[] arr1 = new int[] { 1, 2, 3, 4, 5 };
int[] arr2 = new int[] { 6, 7, 8, 9, 0 };
Console.WriteLine(arr1[2] + " " + arr2[2]);
arr2 = arr1;
Console.WriteLine(arr1[2] + " " + arr2[2]);
arr2 = (int[])arr1.Clone();
arr1[2] = 12;
Console.WriteLine(arr1[2] + " " + arr2[2]);
}
shallow clone means that only the memory represented by the cloned array is copied.
If the array contains value type objects, the values are copied;
if the array contains reference type, only the references are copied - so as a result there are two arrays whose members reference the same objects.
To create a deep copy—where reference type are duplicated, you must loop through the array and clone each element manually.
The copy constructor is used to initialize the new object with the previously created object of the same class. By default compiler wrote a shallow copy. Shallow copy works fine when dynamic memory allocation is not involved because when dynamic memory allocation is involved then both objects will points towards the same memory location in a heap, Therefore to remove this problem we wrote deep copy so both objects have their own copy of attributes in a memory.
In order to read the details with complete examples and explanations you could see the article C++ constructors.
To add just a little more for confusion between shallow copy and simply assign a new variable name to list.
"Say we have:
x = [
[1,2,3],
[4,5,6],
]
This statement creates 3 lists: 2 inner lists and one outer list. A reference to the outer list is then made available under the name x. If we do
y = x
no data gets copied. We still have the same 3 lists in memory somewhere. All this did is make the outer list available under the name y, in addition to its previous name x. If we do
y = list(x)
or
y = x[:]
This creates a new list with the same contents as x. List x contained a reference to the 2 inner lists, so the new list will also contain a reference to those same 2 inner lists. Only one list is copied—the outer list.
Now there are 4 lists in memory, the two inner lists, the outer list, and the copy of the outer list. The original outer list is available under the name x, and the new outer list is made available under the name y.
The inner lists have not been copied! You can access and edit the inner lists from either x or y at this point!
If you have a two dimensional (or higher) list, or any kind of nested data structure, and you want to make a full copy of everything, then you want to use the deepcopy() function in the copy module. Your solution also works for 2-D lists, as iterates over the items in the outer list and makes a copy of each of them, then builds a new outer list for all the inner copies."
source: https://www.reddit.com/r/learnpython/comments/1afldr/why_is_copying_a_list_so_damn_difficult_in_python/
Related
Shallow copy is nothing but when trying to copy and change the contents of copied array {array containing objects and not primitive data types} the contents of the original array also got changed.
In the following case, why the dst[0]= {name : "tom"} is a deep copy and not a shallow copy (the contents of the original array "src" is not changed). On the contrary, the second operation which is dst[0].name= "tom" is creating a shallow copy as expected and changing the contents of src. Can someone please clarify this?
let src = [{
name: 'brat'
}, {
age: 40
}]
let dst = [...src]
// dst[0]={name:"tom"}
dst[0].name = "tom";
console.log(`src ${JSON.stringify(src)}`)
console.log(`dst ${JSON.stringify(dst)}`)
That is because in the below line, you are not changing the copied property of src, instead you are re-assigning both 'key' and 'value' and replacing the old property altogether with a new property.
dst[0]={name:"tom"}
is the same as doing this:
dst[0]={newName:"tom"}
It's considered as dst's new property instead of the one copied from src, and that is the reason the original object does not change, because this 'name' property is not the same one that was copied from the original object.
When you use the {name : "tom" } to re-assign the value to dst, then dst[0]'s memory is released from the heap and a new object replaces dst[0] altogether, it is not the same object that was shallow copied from src[0].
dst[0]={name:"tom"} and dst[0].name = "tom"; are not copies, they're just assignments. One changes the dst array (sets a new object as the first element), the other changes the first object (sets a new string for the .name property).
The shallow copy happens in the line let dst = [...src], and if you're not changing that it's always a shallow copy where dst contains the same mutable objects as src.
In this code... mapObj.fetchTimeObjs should NOT change right?!?!
Somehow mapObj.fetchTimeObjs gets changed when this function is run:
function clockErasePast(){
var now = new Date().getTime();
var tmpFetchTimeObjs = [];
for(var i =0; i<mapObj.fetchTimeObjs.length; i++){
tmpFetchTimeObjs.push(mapObj.fetchTimeObjs[i]);
if(mapObj.fetchTimeObjs[i].start < now){tmpFetchTimeObjs[i].start = now;}
}
return tmpFetchTimeObjs;
}
tmpFetchTimeObjs[i]
will contain only reference to the mapObj.fetchTimeObjs[i].
If you will change tmpFetchTimeObjs[i], the mapObj.fetchTimeObjs[i] will be changed, because you will have only one object which has two references. And if it will changed from one reference, it will be changed for the second reference too.
Let's consider an object which has two references. Here I change the object from one reference, and get the update for the second reference, because they refer to the same object.
var objA = { name: 'Bob', age: 25};
var objB = objA;
objB.age = 30;
console.log(objA.age);
To get independent object you need to create them. You can use Object.assign() function, which will copy any enumerable properties from into the destination(first parameter) object and returns it.
You can create with
var obj = Object.assign({}, mapObj.fetchTimeObjs[i]);
tmpFetchTimeObjs.push(obj);
The objects you push to the new array, are the same as the original array has, so if you mutate them, that mutation is visible to both arrays. This is what is called doing a shallow copy.
You could make a copy at one deeper level, where you would also create new objects and copy the original object's properties (like start) into those new objects. This you can easily do with Object.assign:
tmpFetchTimeObjs.push(Object.assign({}, mapObj.fetchTimeObjs[i]));
If these objects themselves have nested objects, then the problem you had would occur at deeper levels. If this is an issue, then you should look to deep clone solutions, like provided in this Q&A.
You need to clone the TimeObjs, since variables only keep the reference to the TimeObjs. We don't know the structure of your TimeObjs, so if the TimeObjs doesn't contain other objects, Object.assign() will work, otherwise maybe you need a deep clone method, such as jQuery.extend().
I have an "empty" array that will be populated with data later in the code. But before it reaches that stage there's a section where the default content is copied to a temporary array, so the original can be changed and later receive the relevant data that was stored in the copy.
The problem is when I use slice and delete the section in the original array, the temporary one is affected as well.
var array1 = [["text", [[[1,2],[3,4],[5,6]]], 0]];
var array2 = array1[0].slice(0);
//alert(array2[1][0]) // Output: 1,2,3,4,5,6
array1[0][1][0] = new Array();
//alert(array2[1][0]) // Output:
http://jsfiddle.net/Mbv6j/4/
I can use a workaround to copy each section of the array separately rather than all at once, but I still would like to understand why this is happening.
This is expected behaviour. Have a look at the documentation.
You only get a shallow copy of the original array:
The slice() method returns a shallow copy of a portion of an array into a new array object.
For the arrays, object references are stored, so just the references get copied. For the String you will not observe this behaviour.
For object references (and not the actual object), slice copies object references into the new array. Both the original and new array refer to the same object. If a referenced object changes, the changes are visible to both the new and original arrays.
For strings and numbers (not String and Number objects), slice copies strings and numbers into the new array. Changes to the string or number in one array does not affect the other array.
Taken from here:
For object references (and not the actual object), slice copies object
references into the new array. Both the original and new array refer
to the same object. If a referenced object changes, the changes are
visible to both the new and original arrays.
My guess is that since your array contains arrays of arrays, they are probably being represented as object references; thus, slice is copying the references, not the objects. It only does a shallow copy, not a deep copy. If the items in your array weren't objects, you wouldn't encounter this problem.
this is my very first post! I have a quick question in regarding inheritance in javascript.
Of course the 'extend2' method is used to inherit child objects from the parent object using a for-in loop.
var extend2 = function (child, parent) {
var c = child.prototype;
var p = parent.prototype;
for (var i in p) {
c[i] = p[i];
}
}
I'm currently reading "Object Oriented Javascript" by Stoyan Stefanov. It's an awesome book.
Can anyone give me a good detailed explanation of how the child's prototype object is not entirely overwritten or replaced but it's just augmented?
How is it that when the objects inherit, they copy (primitive data types) instead of being looked up as a reference by using the extend2 function?
This would really help thanks!
Primitive data types in javascript are passed via value, rather than reference. Thus when you copy a value it is actually copying it, not referring to it.
Traditionally this is because a primitive was literally encoded in the memory in such a way (so the primitive int 7 would be encoded in memory as 0x7. When dealing with objects, however, they are encoded as a pointer to the memory location where the actualy object is. Thus, when you copy the value it is merely a copy of the reference pointer, not the object that that pointer referrs to.
As for the fact that the child's prototype is not replaced, that is because a prototype in java is merely another object. So a prototype may look like:
{
someField: 5
}
Which would indicate that instances of the object would initialize with a field called someField whose value is 5. with the above code, each entry in the object is copied to the child prototype, but nothing is deleted. Thus if the child prototype looks like:
{
someField: 10
someOtherField: 3
}
Then performing the above extend2 command will overwrite someField, but not someOtherField, so the resulting prototype would be:
{
someField: 5
someOtherField: 3
}
Today I came across this problem in javascript and do not know why it happens.
var a = {
prop: {
bool: true
}
};
console.log(a.prop.bool); // logs true
var b = a;
b.prop.bool = false;
console.log(a.prop.bool); // logs false ¿?
The expression { prop: ... } expression is evaluated once to create one object.
a and b both are references to that single object.
See What's the difference between passing by reference vs. passing by value? and http://en.wikipedia.org/wiki/Reference_(computer_science)
References are widely used in programming, especially to efficiently pass large or mutable data as arguments to procedures, or to share such data among various uses.
EDIT
clone from underscore does a shallow copy.
Create a shallow-copied clone of the object. Any nested objects or arrays will be copied by reference, not duplicated.
To create a deep copy, the easiest way is probably to serialize and deserialize. This will do weird things if a has reference cycles though.
var b = JSON.parse(JSON.stringify(a));
You've created a reference to the same object. When you do this, any changes of variable b will affect the object stored in variable a.
You will need to do a 'clone' of the object to change it so you have two objects instead of one with two references.
when you assign b to a the assignment is by reference meaning b references the same location in memory as a so when b is updated and you observe a it looks like a is also updated.