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Object Passing Reference vs Value [closed]

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To give context to the code imagine a collection of objects inside a cube. The objects are placed randomly and can affect each other. Several series of test events are planned then executed against the cube of objects. Only the best result is kept. This is not the real problem but a simplified version to focus the question.
Sample code
class Loc{
double UpDown
double LeftRight
double FrontBack
}
class Affects{
string affectKey
List<string> impacts //scripts that execute against properties
}
class Item{
Loc startLoc
Loc endLoc
List<string> affectedBy
string resultText // summary of analysis of changes
}
class ItemColl{
List<Item> myItems
}
class main{
ItemColl items
List<string> actions
void ProcessAffects(ItemColl tgt, List<string> acts){
// take actions against the tgt set and return
}
int IsBetter(ItemColl orig, List<Items> altered){
// compares the collection to determine "better one"
// positive better, negative worse, zero for no change
}
void DoThings(){
// original code
ItemColl temp = items
ProcessAffects(temp,actions)
IsBetter(temp,actions)
// the result was always zero - admittedly a duh error
}
}
When I added an alternate constructor that copied the object passed in and did the same to all subordinate objects, as in
class ItemColl{
public ItemColl(){}
public ItemColl (ItemColl clone){
// do a deep copy
}
// partial code from main DoThings
// replaced ItemColl temp = items
// with
ItemColl temp = new ItemColl(items)
it solved the problem that lead me to first question. (Thanks to the people who answered that question kindly.) What I am stuck on is whether or not there are other options to consider? I am hoping this restatement has a better focus and if I am not taking advantage of some newer efficiencies I would like to know.
I removed the old question entirely and re-phrased post face-palm.

Before you get into parameters, you need some background:
Background
There are two kinds of objects in .NET-land, Reference types and Value types. The main difference between the two is how assignment works.
Value Types
When you assign a value type instance to a variable, the value is copied to the variable. The basic numeric types (int, float, double, etc) are all value types. As a result, in this code:
decimal dec1 = 5.44m;
decimal dec2 = dec1;
dec1 = 3.1415m;
both decimal variables (dec and dec2) are wide enough to hold a decimal valued number. In each case, the value is copied. At the end, dec1 == 3.145m and dec2 == 5.44m.
Nearly all value types are declared as a struct (yes, if you get access to the .NET sources, int is a struct). Like all .NET types, they act when boxed as if they are derived from the object base class (their derivation is through System.ValueType. Both object (aka System.Object) and System.ValueType are reference types, even though the unboxed types that derive from System.ValueType are value types (a little magic happens here).
All value types are sealed/final - you can't sub-class them. You also can't create a default constructor for them - they come with a default constructor that initializes them to their default value. You can create additional constructors (which don't hide the built-in default constructor).
All enums are value types as well. They inherit from System.Enum but are value types and behave mostly like other value types.
In general, value types should be designed to be immutable; not all are.
Reference Types
Variables of reference types hold references, not values. That said, it sometimes help to think of them holding a value - it's just that that value is a reference to an object on the managed heap.
When you assign to a variable of reference type, you are assigning the reference. For example:
public class MyType {
public int TheValue { get; set; }
// more properties, fields, methods...
}
MyType mt1 = new MyType() {TheValue = 5};
MyType mt2 = mt1;
mt1.TheValue = 42;
Here, the mt1 and mt2 variables both contain references to the same object. When that object is mutated in the final line of code, you end up with two variables both referring to an object whose TheValue property is 42.
All types declared as a class are reference types. In general, other than the numeric types, enums and bools, most (but not all) of the types that you normally encounter will be reference types.
Anything declared to be a delegate or an event are also reference types under the covers. Someone mentioned interface. There is no such thing as an object typed purely as an interface. Both structs and classes may be declared to implement an interface - it doesn't change their value/reference type nature, but a struct stored as an interface will be boxed.
Difference in Constructor Behavior
One other difference between Reference and Value Types is what the new keyword means when constructing a new object. Consider this class and this struct:
public class CPoint {
public float X { get; set; }
public float Y { get; set; }
public CPoint (float x, float y) {
X = x;
Y = y;
}
}
public struct SPoint {
public float X { get; set; }
public float Y { get; set; }
public CPoint (float x, float y) {
X = x;
Y = y;
}
}
They are basically the same, except that CPoint is a class (a reference type) and SPoint is a struct (a value type).
When you create an instance of SPoint using the two float constructor (remember, it gets a default constructor auto-magically), like this:
var sp = new SPoint (42.0, 3.14);
What happens is that the constructor runs and creates a value. That value is then copied into the sp variable (which is of type SPoint and large enough to hold a two-float SPoint).
If I do this:
var cp = new CPoint (42.0, 3.14);
Something very different happens. First, memory is allocated on the managed heap large enough to hold a CPoint (i.e., enough to hold two floats plus the overhead of the object being a reference type). Then the two-float constructor runs (and that constructor is the only constructor - there is no default constructor (the additional, programmer-written constructor hides the compiler generated default constructor)). The constructor initializes that newCPoint in the memory allocated on the managed heap. Finally, a reference to that newly create object is created and copied to the variable cp.
Parameter Passing
Sorry the preamble took so long.
Unless otherwise specified, all parameters to functions/methods are passed by value. But, don't forget that the value of a variable of reference type is a reference.
So, if I have a function declared as (MyType is the class declared above):
public void MyFunction(decimal decValue, MyType myObject) {
// some code goes here
}
and some code that looks like:
decimal dec1 = 5.44m;
MyType mt1 = new MyType() {TheValue = 5};
MyFunction (dec1, mt1);
What happens is that the value of dec1 is copied to the function parameter (decValue) and available for use within MyFunction. If someone changes the value of the decValue within the function, no side effects outside the function occurs.
Similarly, but differently, the value of mt1 is copied to the method parameter myObject. However, that value is reference to a MyType object residing on the managed heap. If, within the method, some code mutates that object (say: myObject.TheValue=666;), then the object to which both the mt1 and myObject variables refer is mutated, and that results in a side effect viewable outside of the function. That said, everything is still being passed by value.
Passing Parameters by Reference
You can pass parameters by reference in two ways, using either the out or ref keywords. An out parameter does not need to be initialized before the function call (while a ref parameter must be). Within the function, an out parameter must be initialized before the function returns - ref parameters may be initialized, but they do not need to be. The idea is that ref parameters expect to pass in and out of the function (by reference). But out parameters are designed simply as a way to pass something out of the function (by reference).
If I declare a function like:
public void MyByRefFunction(out decimal decValue, ref MyType myObject) {
decValue = 25.624; //decValue must be intialized - it's an out parameter
myObject = new MyType (){TheValue = myObject.TheValue + 2};
}
and then I call it this way
decimal dec1; //note that it's not initalized
MyType mt1 = new MyType() {TheValue = 5};
MyType mt2 = mt1;
MyByRefFunction (out dec1, ref mt1);
After that call, dec1 will contain the value 25.624; that value was passed out of the function by reference.
Passing reference type variables by reference is more interesting. After the function call, mt1 will no longer refer to the object created with TheValue equal to 5, it will refer to the newly created object with TheValue equal to 5 + 2 (the object created within the function). Now, mt1 and mt2 will refer to different object with different TheValue property values.
With reference types, when you pass a variable normally, the object you pass it may mutate (and that mutation is visible after the function returns). If you pass a reference by reference, the reference itself may mutate, and the value of the reference may be different after the function returns.

All custom objects (derived from tobject) are "Reference type".
Nope. See the docs pages for Reference Types and Value Types
The following keywords are used to declare reference types:
class
interface
delegate
C# also provides the following built-in reference types:
dynamic
object
string
A value type can be one of the two following kinds:
a structure type ...
an enumeration type ...
So any time you make a class, it's always a Reference type.
EVERY type inherits from Object - Value Types and Reference Types.
Even if you pass it to a function with a reference parameter, as with the RefChange function both items are changed and both have exactly the same values in the integer list.
The ref keyword just forces your parameter to be passed by reference. Using ref with a Reference Type allows you to reassign the original passed in reference. See What is the use of “ref” for reference-type variables in C#?
.
Do not confuse the concept of passing by reference with the concept of reference types. The two concepts are not the same. A method parameter can be modified by ref regardless of whether it is a value type or a reference type. There is no boxing of a value type when it is passed by reference.
Source
Of course, the ref keyword is important when you pass in a Value Type, such as a struct.
If you want to pass a copy of an object, create an overloaded constructor to which you pass the original object and inside the constructor manage the duplication of the values that matter.
That's called a Copy Constructor, and is a long-established pattern, if you want to use it. In fact, there is a new c# 9.0 feature all about it: records.

well i cant comment since my reputation is too low, but value types are usually in built types such as int, float ...
everything else is reference type. reference type is always a shallow copy regardless of ref keyword.
ref keyword mainly served for value-type or act as a safeguard.
if u want to deep copy, Icloneable is very useful.

Misunderstanding of C# class vs struct

I was trying to implement an automatic differentiation C# lib based on a Python example but I have an issue with the lambda / local function (kind of vague, not sure why it doesn't work).
I even implemented the thing in another language: Kotlin succesfully ... So I probably have a misunderstanding of some programming part either something related to narrow / deep copy or lambda / local function.
public static Value operator +(Value a, Value b)
{
var ret = new Value(a.Data + b.Data, new[]{a, b});
void B()
{
a.Gradient += ret.Gradient;
b.Gradient += ret.Gradient;
}
ret._backward = B;
return ret;
}
I give this little piece of code but it's better to look at the whole: full code with tests in 3 languages
The Gradient doesn't change strangely, I don't know how to give a better explanation of the problem.

Try changing Value from being struct i.e. value type to class i.e. reference type:
public class Value
{
...
}
As docs state:
A variable of a value type contains an instance of the type. This differs from a variable of a reference type, which contains a reference to an instance of the type. By default, on assignment, passing an argument to a method, and returning a method result, variable values are copied. In the case of value-type variables, the corresponding type instances are copied
so changes to fields of local a and b inside operator + function will not affect the values outside.

C# pass by value vs. pass by reference

Consider the following code (I have purposefully written MyPoint to be a reference type for this example)
public class MyPoint
{
public int x;
public int y;
}
It is universally acknowledged (in C# at least) that when you pass by reference, the method contains a reference to the object being manipulated, whereas when you pass by value, the method copies the value being manipulated, thus the value in global scope is not affected.
Example:
void Replace<T>(T a, T b)
{
a = b;
}
int a = 1;
int b = 2;
Replace<int>(a, b);
// a and b remain unaffected in global scope since a and b are value types.
Here is my problem; MyPoint is a reference type, thus I would expect the same operation on Point to replace a with b in global scope.
Example:
MyPoint a = new MyPoint { x = 1, y = 2 };
MyPoint b = new MyPoint { x = 3, y = 4 };
Replace<MyPoint>(a, b);
// a and b remain unaffected in global scope since a and b...ummm!?
I expected a and b to point to the same reference in memory...can someone please clarify where I have gone wrong?

Re: OP's Assertion
It is universally acknowledged (in C# at least) that when you pass by reference, the method contains a reference to the object being manipulated, whereas when you pass by value, the method copies the value being manipulated ...
TL;DR
There's more to it than that. Unless you pass variables with the ref or out keywords, C# passes variables to methods by value, irrespective of whether the variable is a value type or a reference type.
If passed by reference, then the called function may change the variable's address at the call-site (i.e. change the original calling function's variable's assignment).
If a variable is passed by value:
if the called function re-assigns the variable, this change is local to the called function only, and will not affect the original variable in the calling function
however, if changes are made to the variable's fields or properties by the called function, it will depend on whether the variable is a value type or a reference type in order to determine whether the calling function will observe the changes made to this variable.
Since this is all rather complicated, I would recommend avoiding passing by reference if possible (instead, if you need to return multiple values from a function, use a composite class, struct, or Tuples as a return type instead of using the ref or out keywords on parameters)
Also, when passing reference types around, a lot of bugs can be avoided by not changing (mutating) fields and properties of an object passed into a method (for example, use C#'s immutable properties to prevent changes to properties, and strive to assign properties only once, during construction).
In Detail
The problem is that there are two distinct concepts:
Value Types (e.g. int) vs Reference Types (e.g. string, or custom classes)
Passing by Value (default behaviour) vs Passing by Reference(ref, out)
Unless you explicitly pass (any) variable by reference, by using the out or ref keywords, parameters are passed by value in C#, irrespective of whether the variable is a value type or reference type.
When passing value types (such as int, float or structs like DateTime) by value (i.e. without out or ref), the called function gets a copy of the entire value type (via the stack).
Any change to the value type, and any changes to any properties / fields of the copy will be lost when the called function is exited.
However, when passing reference types (e.g. custom classes like your MyPoint class) by value, it is the reference to the same, shared object instance which is copied and passed on the stack.
This means that:
If the passed object has mutable (settable) fields and properties, any changes to those fields or properties of the shared object are permanent (i.e. any changes to x or y are seen by anyone observing the object)
However, during method calls, the reference itself is still copied (passed by value), so if the parameter variable is reassigned, this change is made only to the local copy of the reference, so the change will not be seen by the caller. This is why your code doesn't work as expected
What happens here:
void Replace<T>(T a, T b) // Both a and b are passed by value
{
a = b; // reassignment is localized to method `Replace`
}
for reference types T, means that the local variable (stack) reference to the object a is reassigned to the local stack reference b. This reassign is local to this function only - as soon as scope leaves this function, the re-assignment is lost.
If you really want to replace the caller's references, you'll need to change the signature like so:
void Replace<T>(ref T a, T b) // a is passed by reference
{
a = b; // a is reassigned, and is also visible to the calling function
}
This changes the call to call by reference - in effect we are passing the address of the caller's variable to the function, which then allows the called method to alter the calling method's variable.
However, nowadays:
Passing by reference is generally regarded as a bad idea - instead, we should either pass return data in the return value, and if there is more than one variable to be returned, then use a Tuple or a custom class or struct which contains all such return variables.
Changing ('mutating') a shared value (and even reference) variable in a called method is frowned upon, especially by the Functional Programming community, as this can lead to tricky bugs, especially when using multiple threads. Instead, give preference to immutable variables, or if mutation is required, then consider changing a (potentially deep) copy of the variable. You might find topics around 'pure functions' and 'const correctness' interesting further reading.
Edit
These two diagrams may help with the explanation.
Pass by value (reference types):
In your first instance (Replace<T>(T a,T b)), a and b are passed by value. For reference types, this means the references are copied onto the stack and passed to the called function.
Your initial code (I've called this main) allocates two MyPoint objects on the managed heap (I've called these point1 and point2), and then assigns two local variable references a and b, to reference the points, respectively (the light blue arrows):
MyPoint a = new MyPoint { x = 1, y = 2 }; // point1
MyPoint b = new MyPoint { x = 3, y = 4 }; // point2
The call to Replace<Point>(a, b) then pushes a copy of the two references onto the stack (the red arrows). Method Replace sees these as the two parameters also named a and b, which still point to point1 and point2, respectively (the orange arrows).
The assignment, a = b; then changes the Replace methods' a local variable such that a now points to the same object as referenced by b (i.e. point2). However, note that this change is only to Replace's local (stack) variables, and this change will only affect subsequent code in Replace (the dark blue line). It does NOT affect the calling function's variable references in any way, NOR does this change the point1 and point2 objects on the heap at all.
Pass by reference:
If however we we change the call to Replace<T>(ref T a, T b) and then change main to pass a by reference, i.e. Replace(ref a, b):
As before, two point objects allocated on the heap.
Now, when Replace(ref a, b) is called, while mains reference b (pointing to point2) is still copied during the call, a is now passed by reference, meaning that the "address" to main's a variable is passed to Replace.
Now when the assignment a = b is made ...
It is the the calling function, main's a variable reference which is now updated to reference point2. The change made by the re-assignment to a is now seen by both main and Replace. There are now no references to point1
Changes to (heap allocated) object instances are seen by all code referencing the object
In both scenarios above, no changes were actually made to the heap objects, point1 and point2, it was only local variable references which were passed and re-assigned.
However, if any changes were actually made to the heap objects point1 and point2, then all variable references to these objects would see these changes.
So, for example:
void main()
{
MyPoint a = new MyPoint { x = 1, y = 2 }; // point1
MyPoint b = new MyPoint { x = 3, y = 4 }; // point2
// Passed by value, but the properties x and y are being changed
DoSomething(a, b);
// a and b have been changed!
Assert.AreEqual(53, a.x);
Assert.AreEqual(21, b.y);
}
public void DoSomething(MyPoint a, MyPoint b)
{
a.x = 53;
b.y = 21;
}
Now, when execution returns to main, all references to point1 and point2, including main's variables a and b, which will now 'see' the changes when they next read the values for x and y of the points. You will also note that the variables a and b were still passed by value to DoSomething.
Changes to value types affect the local copy only
Value types (primitives like System.Int32, System.Double) and structs (like System.DateTime, or your own structs) are allocated on the stack, not the heap, and are copied verbatim onto the stack when passed into a call. This leads to a major difference in behaviour, since changes made by the called function to a value type field or property will only be observed locally by the called function, because it only will be mutating the local copy of the value type.
e.g. Consider the following code with an instance of the mutable struct, System.Drawing.Rectangle
public void SomeFunc(System.Drawing.Rectangle aRectangle)
{
// Only the local SomeFunc copy of aRectangle is changed:
aRectangle.X = 99;
// Passes - the changes last for the scope of the copied variable
Assert.AreEqual(99, aRectangle.X);
} // The copy aRectangle will be lost when the stack is popped.
// Which when called:
var myRectangle = new System.Drawing.Rectangle(10, 10, 20, 20);
// A copy of `myRectangle` is passed on the stack
SomeFunc(myRectangle);
// Test passes - the caller's struct has NOT been modified
Assert.AreEqual(10, myRectangle.X);
The above can be quite confusing and highlights why it is good practice to create your own custom structs as immutable.
The ref keyword works similarly to allow value type variables to be passed by reference, viz that the 'address' of the caller's value type variable is passed onto the stack, and assignment of the caller's assigned variable is now directly possible.

C# is actually pass by value. You get the illusion it's pass by reference, because when you pass a reference type you get a copy of the reference (the reference was passed by value). However, since your replace method is replacing that reference copy with another reference, it's effectively doing nothing (The copied reference goes out of scope immediately). You can actually pass by reference by adding the ref keyword:
void Replace<T>(ref T a, T b)
{
a = b;
}
This will get you your desired result, but in practice is a little strange.

In C# all the params that you pass to a method are passed by value.
Now before you shout keep on reading:
A value-type's value is the data that is copied while a reference type's value is actually a reference.
So when you pass an objects reference to a method and change that object then the changes will reflect outside the method as well since you are manipulating the same memory the object was allocated.
public void Func(Point p){p.x = 4;}
Point p = new Point {x=3,y=4};
Func(p);
// p.x = 4, p.y = 4
Now Lets look at this method:
public void Func2(Point p){
p = new Point{x=5,y=5};
}
Func2(p);
// p.x = 4, p.y = 4
So no changed occurred here and why? Your method simply created a new Point and changed p's reference(Which was passed by value) and therefore the change was local. You didn't manipulate the point, you changed the reference and you did locally.
And there comes the ref keyword that saves the day:
public void Func3(ref Point p){
p = new Point{x=5,y=5};
}
Func3(ref p);
// p.x = 5, p.y = 5
The same occurred in your example. You assigned a point with a new reference, but you did it locally.

C# is passing reference types objects not by reference, but rather it's passing the reference by value. Meaning you can mess around with their insides, but you can't change the assignment itself.
Read this great piece by Jon Skeet for deeper understanding.

Have a look on behavior by a simple program in C#:
class Program
{
static int intData = 0;
static string stringData = string.Empty;
public static void CallByValueForValueType(int data)
{
data = data + 5;
}
public static void CallByValueForRefrenceType(string data)
{
data = data + "Changes";
}
public static void CallByRefrenceForValueType(ref int data)
{
data = data + 5;
}
public static void CallByRefrenceForRefrenceType(ref string data)
{
data = data +"Changes";
}
static void Main(string[] args)
{
intData = 0;
CallByValueForValueType(intData);
Console.WriteLine($"CallByValueForValueType : {intData}");
stringData = string.Empty;
CallByValueForRefrenceType(stringData);
Console.WriteLine($"CallByValueForRefrenceType : {stringData}");
intData = 0;
CallByRefrenceForValueType(ref intData);
Console.WriteLine($"CallByRefrenceForValueType : {intData}");
stringData = string.Empty;
CallByRefrenceForRefrenceType(ref stringData);
Console.WriteLine($"CallByRefrenceForRefrenceType : {stringData}");
Console.ReadLine();
}
}
Output:

You're not understanding what passing by reference means. Your Replace method is creating a copy of the Point object--passing by value (which is actually the better way of doing it).
To pass by reference, so that a and b both reference the same point in memory, you need add "ref" to the signature.

You dont get it right.
It is similar like Java - everything is passed by value! But you do have to know, what the value is.
In primitive data types, the value is the number itself. In other cases it is reference.
BUT, if you copy reference to another variable, it holds same reference, but does not reference the variable (thus it is not pass by reference known in C++).

By default c# passes ALL arguements by value... that is why a and b remain unaffected in global scope in your examples. Here's a reference for those down voters.

To add more detail...in .NET, C# methods, using the default "pass by value" assigned to all parameters, reference types act differently in two scenarios. In the case of all reference types using classes (System.Object types), a copy of the "pointer" (to a memory block) to the original class or object is passed in and assigned to the method's parameter or variable name. This pointer is a value, too, and copied on the stack in memory where all value types are store. The value of the object isn't stored just a copy of its pointer, that points back to the original cl;ass object. I believe this is a 4-byte value. That's what is physically passed and stored in methods for all reference types. So, you now have a new method parameter or variable with a pointer assigned to it still pointing back to the original class object outside the method. You can now do two things to the new variable with the copied pointer value:
You can change the ORIGINAL object outside the method by changing its properties iniside your method. If "MyObject" is your variable with the copied pointer, you would do MyObject.myproperty = 6;, which changed the "myproperty" inside the original object outside the method. You did this as you passed in a pointer to the original object and assigned it to a new variable in your method. Note that this DOES change the referenced object outside the method.
Or, setting your variable with copied pointer to a new object and new pointer like so: MyObject = new SomeObject(); Here, we destroyed the old copied pointer assigned the variable above and assigned it to a new pointer to a new object! Now we have lost connection to the outside object and only changing a new object.

Why reversing a string array inside method does not persist

So I've got a simple string array. Passed it to a function that reverses its input and then displayed the content of the array. I was expecting the contents of the array to reverse since arrays are passed by reference, but the string array did not change.
string[] words = { "Metal", "Gear", "is", "Awesome!" };
mutateArray(ref words);
foreach (string word in words)
Console.Write(word + " ");
This is my mutateArray function:
public static void mutateArray(ref string[] arr)
{
arr = arr.Reverse().ToArray();
}
I know that the mutateArray method changes to the array will persist once I state that the parameter must be passed in with the keyword ref.
Aren't all arrays passed in by reference by default?
Why do the changes persist when the keyword ref is involved?
What's the difference between passing a reference type (classes, interfaces, array, delegates) by value vs passing them by reference (with the keyword ref)?

The reason this doesn't work how you expect is that Reverse() does not actually reverse the contents of the array in place but rather makes a new list with the reversed contents of the original. That's why it works once you pass the array by reference: then, you're actually replacing the entire original array in the calling method with a new one created in mutateArray.
If you had a method that did the reversing in-place, you could pass in the original array (not using ref), and after the method call, the array would be in reverse order.

All parameters are passed by value by default in C#. For reference types like array, this means the reference is passed by value.
ref causes the variable to be passed by reference into the function. This effectively means the arr parameter in mutateArray is an alias for words in the caller. This is why an assignment to arr results in a change in words after mutateArray has exited.
Passing a reference type by value into a function means a copy of the reference is made. Without the ref modifier, arr in mutateArray is a different variable containing a reference to the same object as words in the caller. Assigning to arr in this case has no effect on words in the caller. Note that you can mutate the array through the shared reference, but arr and words are separate storage locations.

You're confusing a Reference Type with passing by Reference. The ref keyword can be applied to both Value and Reference types. Even Reference types aren't passed by reference by default. Instead the reference is passed by value to the method.
Based on the documentation from MSDN, they should. That's the whole purpose of using the ref keyword with a Reference Type.
The difference is that when you pass by Reference Type by Reference, you are able to change the reference of the original variable rather than just the instance inside your method. Check the previously linked documentation for more details.

words is a reference to an array. Just consider it to contain the memory address of that array.
When you give it to MutateArray as a parameter (without the ref keyword) its VALUE will be copied into arr. So arr is a different variable as words, but they contain the same value (= memory address). This means that they refer to the same object (the string array).
You can change the contents of the object, but words (and arr) will still be referring to it.
If you assign arr to a different object, then its value changes, so it will refer to a different object than words.
However, if you use the ref keyword, then arr and words are the SAME variable. That means if you change arr's value (= assign it to a new object), you are also changing words's value, so words will refer to the same, new, object.
Maybe all of this is not technically 100% correct, but it's the way I like to think about it in order to understand how it works.
The ref keyword is why the following Swap method works in C#; without it, it would just change the inner variables of the Swap method (and do nothing basicallly)
public void Swap<T>(ref T a, ref T b) {
T temp = a;
a = b;
b = temp;
}

Your question is totally understandable and for a moment I was confused as well.
So here is the explanation.
Short explanation
When you pass a reference type to a method and make a change to its property, the object outside the method CAN SEE the property change because the object itself remains the same.
When you pass a reference type to a method and make a change to the instance itself, the object outside the method CANNOT SEE the change because inside the method you basically started pointing to another object. So the object changed inside the method and remained there as a stranger.
Long explanation
Suppose you have a reference type instance where you get the value from database.
using (var context = new MyAdventureWorksEntities2())
{
Product p = context.Products.Where(item => item.ProductID == 1000).First();
Console.WriteLine(p.Name); // p.Name = "INITIAL NAME"
UpdateName(p);
Console.WriteLine(p.Name);
}
And here is your UpdateName method:
public static void UpdateName(Product p)
{
p.Name = "UPDATED NAME";
}
This code emits the following result:
INITIAL NAME
UPDATED NAME
HOWEVER, if you change the method to the following:
public static void UpdateName(Product p)
{
using (var context = new MyAdventureWorksEntities2())
{
p = context.Products.Where(item => item.ProductID == 1003).First();
// p.Name = "ANOTHER PRODUCT NAME"
}
}
your result will be:
INITIAL NAME
INITIAL NAME
Note that I didn't touch the ref keyword at all.
And perhaps after those examples the short description will be much more comprehensible.

C# is passing objects by value?

For some reason, (I am new to C# and know java and c++) C# keeps copying objects when I want to pass by value. I have an arraylist of a Vector2 class, and whenever I want to increment a value, I have to do this:
Vector2 d = (Vector2) myObjects[i];
d.Y++;
myObjects [i] = d;
I want to be able to do this:
Vector2 d = (Vector2) myObjects[i];
d.Y++;
and be done. I searched the web and surprisingly no answers.
BTW the vector is a struct.

In C#, instances of classes are passed as references, whereas instances of structs are passed by copy (by default).
The answer was just where it was supposed to be: http://msdn.microsoft.com/en-us/library/vstudio/ms173109.aspx
A class is a reference type. When an object of the class is created, the variable to which the object is assigned holds only a reference to that memory. When the object reference is assigned to a new variable, the new variable refers to the original object. Changes made through one variable are reflected in the other variable because they both refer to the same data.
A struct is a value type. When a struct is created, the variable to which the struct is assigned holds the struct's actual data. When the struct is assigned to a new variable, it is copied. The new variable and the original variable therefore contain two separate copies of the same data. Changes made to one copy do not affect the other copy.

You are experiencing one of the effects of value-types. Because it copies itself by value, rather than by reference when assigned to new variables or passed as an argument.
You can pass a struct or other value type by ref, using the ref keyword in your method signature, unfortunately you can't use it for treating a variable in the same stack frame as a reference (i.e. you can't just say ref int test = yourArray[0], but must make something like:
public void SomeMethod(ref Vector2 input) {
// now you are modifying the original vector2
}
public void YourOriginalMethod()
{
SomeMethod(yourArray[20]);
}
In response to the comment below, from http://msdn.microsoft.com/en-us/library/14akc2c7.aspx:
Do not confuse the concept of passing by reference with the concept of
reference types. The two concepts are not the same. A method parameter
can be modified by ref regardless of whether it is a value type or a
reference type. There is no boxing of a value type when it is passed
by reference.