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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.
Related
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.
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.
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.
I was looking at this question, and aside from a rather odd way to enumerate something, the op was having trouble because the enumerator is a struct. I understand that returning or passing a struct around uses a copy because it is a value type:
public MyStruct GetThingButActuallyJustCopyOfIt()
{
return this.myStructField;
}
or
public void PretendToDoSomething(MyStruct thingy)
{
thingy.desc = "this doesn't work as expected";
}
So my question is if MyStruct implements IMyInterface (such as IEnumerable), will these types of methods work as expected?
public struct MyStruct : IMyInterface { ... }
//will caller be able to modify the property of the returned IMyInterface?
public IMyInterface ActuallyStruct() { return (IMyInterface)this.myStruct; }
//will the interface you pass in get its value changed?
public void SetInterfaceProp(IMyInterface thingy)
{
thingy.desc = "the implementing type is a struct";
}
Yes, that code will work, but it needs explanation, because there is a whole world of code that will not work, and you're likely to trip into that unless you know this.
Before I forget: Mutable structs are evil. OK, with that out of the way, let's move on.
Let's take a simple example, you can use LINQPad to verify this code:
void Main()
{
var s = new MyStruct();
Test(s);
Debug.WriteLine(s.Description);
}
public void Test(IMyInterface i)
{
i.Description = "Test";
}
public interface IMyInterface
{
string Description { get; set; }
}
public struct MyStruct : IMyInterface
{
public string Description { get; set; }
}
When executing this, what will be printed?
null
OK, so why?
Well, the problem is this line:
Test(s);
This will in fact box that struct and pass the boxed copy to the method. You're successfully modifying that boxed copy, but not the original s variable, which was never assigned anything, and is thus still null.
OK, so if we change just one line in the first piece of code:
IMyInterface s = new MyStruct();
Does this change the outcome?
Yes, because now you're boxing that struct here, and always use the boxed copy. In this context it behaves like an object, you're modifying the boxed copy and writing out the contents of the boxed copy.
The problem thus crops up whenever you box or unbox that struct, then you get copies that live separate lives.
Conclusion: Mutable structs are evil.
I see two answers about using ref here now, and this is barking up the wrong tree. Using ref means you've solved the problem before you added ref.
Here's an example.
If we change the Test method above to take a ref parameter:
public void Test(ref IMyInterface i)
Would this change anything?
No, because this code is now invalid:
var s = new MyStruct();
Test(ref s);
You'll get this:
The best overloaded method match for 'UserQuery.Test(ref UserQuery.IMyInterface)' has some invalid arguments
Argument 1: cannot convert from 'ref UserQuery.MyStruct' to 'ref UserQuery.IMyInterface'
And so you change the code to this:
IMyInterface s = new MyStruct();
Test(ref s);
But now you're back to my example, just having added ref, which I showed is not necessary for the change to propagate back.
So using ref is orthogonal, it solves different problems, but not this one.
OK, more comments regarding ref.
Yes, of course passing a struct around using ref will indeed make the changes flow throughout the program.
That is not what this question was about. The question posted some code, asked if it would work, and it would. In this particular variant of code it would work. But it's so easy to trip up. And pay particular note that the question was regarding structs and interfaces. If you leave interfaces out of it, and pass the struct around using ref, then what do you have? A different question.
Adding ref does not change this question, nor the answer.
Within the CLR, every value-type definition actually defines two kinds of things: a structure type, and a heap object type. A widening conversion exists from the structure type to the boxed object type, and a narrowing conversion exists from Object to the structure type. The structure type will behave with value semantics, and the heap object type will behave with mutable reference semantics. Note that the heap object types associated with all non-trivial structure types [i.e. those with any non-default states] are always mutable, and nothing in the structure definition can cause them to be otherwise.
Note that value types may be constrained, cast, or coerced to interface types, and cast or coerced to reference types. Consider:
void DoSomethingWithDisposable<T,U>(ref T p1,
List<int>.Enumerator p2) where T:IDisposable
{
IDisposable v1a = p1; // Coerced
Object v1b = p1; // Coerced
IDisposable v2a = (IDisposable)p2; // Cast
Object v2b = (Object)p2; // Cast
p1.Dispose(); // Constrained call
}
void blah( List<string>.Enumerator p1, List<int>.Enumerator p2) // These are value types
{
DoSomethingWithDisposable(p1,p2); // Constrains p1 to IDisposable
}
Constraining a generic type to an interface type does not affect its behavior as a value type. Casting or coercing an a value type to an interface or reference type, however, will create a new instance of the heap object type and return a reference to that. That reference will then behave with reference-type semantics.
The behavior of value types with generic constraints can at times be very useful, and such usefulness can apply even when using mutating interfaces, but unfortunately there's no way to tell the compiler that a value type must remain as a value type, and that the compiler should warn if it would find itself converting it to something else. Consider the following three methods:
bool AdvanceIntEnumerator1(IEnumerator<int> it)
{ return it.MoveNext(); }
bool AdvanceIntEnumerator2(ref T it) where T:IEnumerator<int>
{ return it.MoveNext(); }
bool AdvanceIntEnumeratorTwice<T>(ref T it) where T:IEnumerator<int>
{ return it.MoveNext() && AdvanceIntEnumerator1(it); }
If one passes to the first piece of code a variable of type List<int>.Enumerator, the system will copy its state to a new heap object, call MoveNext on that object, and abandon it. If one passes instead a variable of type IEnumerator<int> which holds a reference to a heap object of type List<int>.Enumerator, it will call MoveNext on that instance, which the calling code will still retain.
If one passes to the second piece of code a variable of type List<int>.Enumerator, the system will call MoveNext on that variable, thus changing its state. If one passes a variable of type IEnumerable<T>, the system will call MoveNext on the object referred to by that variable; the variable won't be modified (it will still point to the same instance), but the instance to which it points will be.
Passing to the third piece of code a variable of type List<int>.Enumerator will cause MoveNext to be called on that variable, thus changing its state. If that returns true, the system will copy the already-modified variable to a new heap object and call MoveNext on that. The object will then be abandoned, so the variable will only be advanced once, but the return value will indicate whether a second MoveNext would have succeeded. Passing the third piece of code a variable of type IEnumerator<T> which holds a reference to a List<T>.Enumerator, however, will cause that instance to be advanced twice.
No, interface is a contract, to make it work properly you need to use ref keyword.
public void SetInterfaceProp(ref IMyInterface thingy)
{
thingy.desc = "the implementing type is a struct";
}
What matters here is a real type that stays inside that interface wrap.
To be more clear:
even if code with method SetInterfaceProp defined like
public void SetInterfaceProp(IMyInterface thingy)
{
thingy.desc = "the implementing type is a struct";
}
will work:
IMyInterface inter= default(MyStruct);
SetInterfaceProp(inter);
this one will not :
MyStruct inter = default(MyStruct);
SetInterfaceProp(inter);
You can not gurantee that the caller of your method will always use IMyInterface, so to guarantee expected behavior, in this case, you can define ref keyword, that will guarantee that in both cases method would run as expected.
Why is the last line not allowed?
IEnumerable<double> doubleenumerable = new List<double> { 1, 2 };
IEnumerable<string> stringenumerable = new List<string> { "a", "b" };
IEnumerable<object> objects1 = stringenumerable; // OK
IEnumerable<object> objects2 = doubleenumerable; // Not allowed
Is this because double is a value type that doesn't derive from object, hence the covariance doesn't work?
Does that mean that there is no way to make this work:
public interface IMyInterface<out T>
{
string Method();
}
public class MyClass<U> : IMyInterface<U>
{
public string Method()
{
return "test";
}
}
public class Test
{
public static object test2()
{
IMyInterface<double> a = new MyClass<double>();
IMyInterface<object> b = a; // Invalid cast!
return b.Method();
}
}
And that I need to write my very own IMyInterface<T>.Cast<U>() to do that?
Why is the last line not allowed?
Because double is a value type and object is a reference type; covariance only works when both types are reference types.
Is this because double is a value type that doesn't derive from object, hence the covariance doesn't work?
No. Double does derive from object. All value types derive from object.
Now the question you should have asked:
Why does covariance not work to convert IEnumerable<double> to IEnumerable<object>?
Because who does the boxing? A conversion from double to object must box the double. Suppose you have a call to IEnumerator<object>.Current that is "really" a call to an implementation of IEnumerator<double>.Current. The caller expects an object to be returned. The callee returns a double. Where is the code that does the boxing instruction that turns the double returned by IEnumerator<double>.Current into a boxed double?
It is nowhere, that's where, and that's why this conversion is illegal. The call to Current is going to put an eight-byte double on the evaluation stack, and the consumer is going to expect a four-byte reference to a boxed double on the evaluation stack, and so the consumer is going to crash and die horribly with an misaligned stack and a reference to invalid memory.
If you want the code that boxes to execute then it has to be written at some point, and you're the person who gets to write it. The easiest way is to use the Cast<T> extension method:
IEnumerable<object> objects2 = doubleenumerable.Cast<object>();
Now you call a helper method that contains the boxing instruction that converts the double from an eight-byte double to a reference.
UPDATE: A commenter notes that I have begged the question -- that is, I have answered a question by presupposing the existence of a mechanism which solves a problem every bit as hard as a solution to the original question requires. How does the implementation of Cast<T> manage to solve the problem of knowing whether to box or not?
It works like this sketch. Note that the parameter types are not generic:
public static IEnumerable<T> Cast<T>(this IEnumerable sequence)
{
if (sequence == null) throw ...
if (sequence is IEnumerable<T>)
return sequence as IEnumerable<T>;
return ReallyCast<T>(sequence);
}
private static IEnumerable<T> ReallyCast<T>(IEnumerable sequence)
{
foreach(object item in sequence)
yield return (T)item;
}
The responsibility for determining whether the cast from object to T is an unboxing conversion or a reference conversion is deferred to the runtime. The jitter knows whether T is a reference type or a value type. 99% of the time it will of course be a reference type.
To understand what is allowed and not allowed, and why things behave as they do, it is helpful to understand what's going on under the hood. For every value type, there exists a corresponding type of class object, which--like all objects--will inherit from System.Object. Each class object includes with its data a 32-bit word (x86) or 64-bit longword (x64) which identifies its type. Value-type storage locations, however, do not hold such class objects or references to them, nor do they have a word of type data stored with them. Instead, each primitive-value-type location simply holds the bits necessary to represent a value, and each struct-value-type storage location simply holds the contents of all the public and private fields of that type.
When one copies a variable of type Double to one of type Object, one creates a new instance of the class-object type associated with Double and copies all the bytes from the original to that new class object. Although the boxed-Double class type has the same name as the Double value type, this does not lead to ambiguity because they can generally not be used in the same contexts. Storage locations of value types hold raw bits or combinations of fields, without stored type information; copying one such storage location to another copies all bytes, and consequently copies all public and private fields. By contrast, heap objects of types derived from value types are heap objects, and behave like heap objects. Although C# regards the contents of value-type storage locations as though they are derivatives of Object, under the hood the contents of such storage locations are simply collections of bytes, effectively outside the type system. Since they can only be accessed by code which knows what the bytes represent, there is no need to store such information with the storage location itself. Although the necessity for boxing when calling GetType on a struct is often described in terms of GetType being a non-shadowed, non-virtual function, the real necessity stems from the fact that the contents of a value-type storage location (as distinct from the location itself) don't have type information.
Variance of this type is only supported for reference types. See http://blogs.msdn.com/b/csharpfaq/archive/2010/02/16/covariance-and-contravariance-faq.aspx