LINQ - Does the Where expression return new instance or reference to object instance.

Question

This is probably a basic question for some, but it affects how I design a piece of my program.

I have a single collection of type A:

IEnumerable<A> myCollection;

I am filtering my collection on 2 different criteria:

IEnumerable<A> subCollection1 = myCollection.Where(x => x.Count > 10);
etc.

Now, I know that the .Where expression will return a new instance of IEnumerable, but does the new collection contain the same reference to an instance of type A that 'myCollection' references, or are new copies of type A created? If new instances of type 'A' are created, is there a way to say that 'subCollection1' references the same instances of A as 'myCollection' references?

Edit: To Add further clarification.

I am looking for a way so that when I make a change to an instance of 'A' in 'subCollection1', that it is also modified for 'myCollection'.

Answer 1

The instances are the same if they are classes, but copies if they are structs/valuetypes.

int, byte, double are value types and any structs (like System.Drawing.Point and self defined structs). But strings, all of your own classes, basically "the rest" is reference types.

Note: LINQ uses the same rules as all other assignments.

For objects:

Person p1 = new Person();
p1.Name = "Mr Jones";
Person p2 = p1;
p2.Name = "Mr Anderssen";
// Now p1.Name is also "Mr Anderssen"

For structs

Point p1 = new Point();
p1.x = 5;
Point p2 = p1;
p2.x = 10;
// p1.x is still 5

The same rules applies when using LINQ.

Answer 2

They are same objects. Where only filters, Select produces (can produce) new instances.

Answer 3

Making a new object that is a reference type is non-trivial. LINQ would have no idea how to do it. LINQ always returns the same instances when dealing with reference types.

Answer 4

I just wanted to add to some of the other answers -- in general, when I'm not sure of something but require a particular behavior, I'll add a unit test for it. You could easily put this into a test and then check for equality, which will tell you if you're looking at a reference of the object in the original container. Some may argue that this is stupid because you "should just know" what happens, but for me I know I will either be 1) unsure because I'm not an awesome programmer, and 2) there are always nights where I have to burn the midnight oil, and it's good to have the reassurance that something behaves as you need it to.