Is there a way to make my new Class mimic the behavior of an int or any Valuetype?
I want something, so these assignments are valid
MyClass myClass = 1;
int i = myClass;
var x = myClass; // And here is important that x is of type int!
The MyClass looks roughly like this
public class MyClass {
public int Value{get;set;}
public int AnotherValue{get;set;}
public T GetSomething<T>() {..}
}
Every assignment of MyClass should return the Variable Value as Type int.
So far i found implicit operator int and implicit operator MyClass (int value). But this is not 'good enough'.
I want that MyClass realy behaves like an int. So var i = myClass lets i be an int.
Is this even possible?
If you´d created a cast from your class to int as this:
public static implicit operator int(MyClass instance) { return instance.Value; }
you could implicetly cast an instance of MyClass to an int:
int i = myClass;
However you can not expect the var-keyword to guess that you actually mean typeof int instead of MyClass, so this does not work:
var x = myClass; // x will never be of type int
Apart from this I would highly discourage from an implicit cast as both types don´t have anything in common. Make it explicit instead:
int i = (int) myClass;
See this excellent answer from Marc Gravell for why using an explicit cast over an implicit one. Basically it´s about determing if data will be lost when converting the one in the other. In your case you´re losing any information about AnotherValue, as the result is just a primitive int. When using an explicit cast on the other hand you claim: the types can be converted, however we may lose information of the original object and won´t care for that.
Is this even possible?
No.
You can convert and cast types to other types implicitly and explicitly, but to have one type supersede another type without any inheritance or interface implementation flies in the face of object-oriented programming principles.
The var keyword is a convenience that tries to guess the type of a value with as much precision as possible. Therefore it can't be forced to represent int when the type is MyClass.
You might consider something like this: var x = (int) myClass;
Related
Please have a look at the following code. Why do I get a compile-error?
I don't get it!
Casting is a way of telling the compiler that I know more about the objects than it does. And in this case, I know for fact, that "x" does actually contain an instance of "SomeClass". But the compiler seems to be unwilling to accept that information.
https://dotnetfiddle.net/0DlmXf
public class StrangeConversion
{
public class SomeClass { }
public interface ISomeInterface { }
public class Implementation : SomeClass, ISomeInterface { }
public void Foo<T>() where T : class
{
T x = (T)Factory();
//Compile-error: Cannot convert type 'T' to 'SomeClass'
SomeClass a = (SomeClass)x;
//This is perfectly fine:
SomeClass b = (SomeClass)(object)x;
if (x is SomeClass c)
{
//This works as well and 'c' contains the reference.
}
}
private object Factory()
{
return new Implementation();
}
}
Edit:
#Charles Mager has the correct answer in the comment: There does not seem to be a valid reason. The language designers just didn't want to allow this cast.
I fixed using the as casting e.g.
SomeClass a = x as SomeClass;
This Answer explains is very well https://stackoverflow.com/a/132467/16690008
Essentially it's because it would throw an exception if T is not type of that class
It's hard to make sense of exactly what you're trying to achieve, but it seems like a generic constraint is what you're after
public void Foo<T>()
where T : SomeClass // constrain T to be inheriting from SomeClass
{
T x = Factory<T>(); // x is guaranted to be SomeClass
}
private T Factory<T>()
where T : SomeClass // same here
{
return new Implementation();
}
You constrain the generic to only reference types by specifying where T : class, but the compiler needs to know with certainty if the cast is possible. This means you are asking the compiler to trust that SomeClass can be cast from any reference type you pass to it, which is something it won't do. The microsoft docs state that for the class generic type constraint:
The type argument must be a reference type. This constraint applies also to any class, interface, delegate, or array type.
Its important to note that SomeClass b = (SomeClass)(object)x; works because of the cast to object which is the root of the object hierarchy. But as you can see from the list of supported reference types, SomeClass a = (SomeClass)x; has to account for things such as delegates, array types, etc., at which point the compiler will throw you the error
Don't do SomeClass b = (SomeClass)(object)x;, it is much cleaner to make proper use of type constraints along with the as & is operators which were designed for this exact purpose of type checking and safe casting
Short answer:
This behaviour is correct according to the spec. The spec is just bad here since this might convert a compile-error into a runtime-error.
Long answer:
I did some more research on the matter. This is an oversight in the language's spec.
C# uses the same syntax for two totally different things:
int i = (int)1.9
This converts the double 1.9 to an integer. The value is actually changed.
object o = "abc";
string s = (string) o;
This looks the same, but does not change the object referenced by "o" at all. It does only convert the type of the reference.
When it comes to generics, this kind of ambiguity is an issue:
function f(T x) {
var x = (string) x;
}
What should the language do if T is "int"?
That's why the spec forces the developer to cast to object first:
function f(T x) {
var x = (string)(object)x;
}
Now, the behaviour is clear: X might still be a value-type. But if it is, it will be converted to a reference-type first.
This ambiguity does not exist in my example, since T is guaranteed to be a reference type:
public void Foo<T>() where T : class
Thus the cast to object is not necessary. It could even be harmful if the "where" specifies an actual type. In that case, the forced cast to object might convert a compile-time-error (impossible cast) to a runtime-error.
Unfortunately, the people who created the spec, did not see this issue and did not include it.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Is it safe for structs to implement interfaces?
Take this code:
interface ISomeInterface
{
public int SomeProperty { get; }
}
struct SomeStruct : ISomeInterface
{
int someValue;
public int SomeProperty { get { return someValue; } }
public SomeStruct(int value)
{
someValue = value;
}
}
and then I do this somewhere:
ISomeInterface someVariable = new SomeStruct(2);
is the SomeStruct boxed in this case?
Jon's point is true, but as a side note there is one slight exception to the rule; generics. If you have where T : ISomeInterface, then this is constrained, and uses a special opcode. This means the interface can be used without boxing. For example:
public static void Foo<T>(T obj) where T : ISomeInterface {
obj.Bar(); // Bar defined on ISomeInterface
}
This does not involve boxing, even for value-type T. However, if (in the same Foo) you do:
ISomeInterface asInterface = obj;
asInterface.Bar();
then that boxes as before. The constrained only applies directly to T.
Yes, it is. Basically whenever you need a reference and you've only got a value type value, the value is boxed.
Here, ISomeInterface is an interface, which is a reference type. Therefore the value of someVariable is always a reference, so the newly created struct value has to be boxed.
I'm adding this to hopefully shed a little more light on the answers offered by Jon and Marc.
Consider this non-generic method:
public static void SetToNull(ref ISomeInterface obj) {
obj = null;
}
Hmm... setting a ref parameter to null. That's only possibly for a reference type, correct? (Well, or for a Nullable<T>; but let's ignore that case to keep things simple.) So the fact that this method compiles tells us that a variable declared to be of some interface type must be treated as a reference type.
The key phrase here is "declared as": consider this attempt to call the above method:
var x = new SomeStruct();
// This line does not compile:
// "Cannot convert from ref SomeStruct to ref ISomeInterface" --
// since x is declared to be of type SomeStruct, it cannot be passed
// to a method that wants a parameter of type ref ISomeInterface.
SetToNull(ref x);
Granted, the reason you can't pass x in the above code to SetToNull is that x would need to be declared as an ISomeInterface for you to be able to pass ref x -- and not because the compiler magically knows that SetToNull includes the line obj = null. But in a way that just reinforces my point: the obj = null line is legal precisely because it would be illegal to pass a variable not declared as an ISomeInterface to the method.
In other words, if a variable is declared as an ISomeInterface, it can be set to null, pure and simple. And that's because interfaces are reference types -- hence, declaring an object as an interface and assigning it to a value type object boxes that value.
Now, on the other hand, consider this hypothetical generic method:
// This method does not compile:
// "Cannot convert null to type parameter 'T' because it could be
// a non-nullable value type. Consider using 'default(T)' instead." --
// since this method could take a variable declared as, e.g., a SomeStruct,
// the compiler cannot assume a null assignment is legal.
public static void SetToNull<T>(ref T obj) where T : ISomeInterface {
obj = null;
}
The MSDN documentation tells us that structs are value, not reference types. They are boxed when converting to/from a variable of type object. But the central question here is: what about a variable of an interface type? Since the interface can also be implemented by a class, then this must be tantamount to converting from a value to a reference type, as Jon Skeet already said, therefore yes boxing would occur. More discussion on an msdn blog.
I'm currently trying to get my head around casting and boxing. As i understand it currently:
Boxing - Value Type to Reference Type (ie int to object)
Unboxing - Reference Type to Value Type (ie object to int)
Type Casting - Seems to me at the moment to be similar to boxing but
allows you to assign what type of object you would like the Reference
Type to be of. (ie int to customObjectType)
The example im working with at the moment to try and get my around it.
Say I have 2 classes, a method in one class calls the constructor of the other.
//1st class
public class FirstClass
{
//code for fields,constructor & other methods
public void CallOtherClass(int firstClassID)
{
int x = firstClassID;
SecondClass test = new SecondClass(x);
}
}
//2nd class
public class SecondClass
{
public SecondClass(FirstClass firstClass)
{
//set some fields
}
}
Ok, so in the above scenario we would have a problem as the method CallOtherClass attempts to set the constructor of SecondClass, however the constructor of SecondClass takes a parameter of type FirstClass and all we can provide is an int.
So as i understand it this would be a good time to use type casting? Something like below.
//1st class
public class FirstClass
{
//code for fields,constructor & other methods
public void CallOtherClass(int firstClassID)
{
int x = firstClassID;
FirstClass a;
a = (FirstClass)x;
SecondClass test = new SecondClass(a);
}
}
//2nd class
public class SecondClass
{
public SecondClass(FirstClass firstClass)
{
//set some fields
}
}
In my head this seems to me to change the type of x to a reference type of FirstClass. Obviously my understanding is way off some where along the lines as it produces an error
"Cannot convert type 'int' to 'Namespace.FirstClass"
Any thoughts?
Type casting isn't boxing or unboxing, but it may cause either.
As an int isn't a FirstClass, that is, int doesn't inherit or extend FirstClass, hence you can't cast it to be of type FirstClass.
Typecasting causes conversion, only if it is possible to do so.
So you can go from an int to a double and vice versa, with possible side effects. But you can't go from an int to FirstClass.
Boxing, wraps a value or reference type in wrapper object. At least that's how I think of it. Not sure how the internals work, but my guess is the assignment operator "=" or casting implicitly returns the wrapped value when unboxing, and the wrapper object when boxing.
I have an enum like:
public enum Test:int
{
A=1,
B=2
}
So here I know my enum is an int type but if I want to do something like following:
int a = Test.A;
this doesn't work.
If I have a class like:
public class MyTest
{
public static int A =1;
}
I can say ,
int a = MyTest.A;
Here I don't need to cast A to int explicitly.
So here I know my enum is an int type
No, it's not. It has an underlying type of int, but it's a separate type. Heck, that's half the point of having enums in the first place - that you can keep the types separate.
When you want to convert between an enum value and its numeric equivalent, you cast - it's not that painful, and it keeps your code cleaner in terms of type safety. Basically it's one of those things where the rarity of it being the right thing to do makes it appropriate to make it explicit.
EDIT: One oddity that you should be aware of is that there is an implicit conversion from the constant value 0 to the enum type:
Test foo = 0;
In fact, in the MS implementation, it can be any kind of constant 0:
Test surprise = 0.0;
That's a bug, but one which it's too late to fix :)
I believe the rest for this implicit conversion was to make it simpler to check whether any bits are set in a flags enum, and other comparisons which would use "the 0 value". Personally I'm not a fan of that decision, but it's worth at least being aware of it.
"The underlying type specifies how much storage is allocated for each enumerator. However, an explicit cast is needed to convert from enum type to an integral type".
With your updated example:
public class MyTest
{
public static int A =1;
}
And usage:
int a = MyTest.A;
That's not how enums look. Enums look more like (comments are places where we differ from a real enum):
public struct MyTest /* Of course, this isn't correct, because we'll inherit from System.ValueType. An enum should inherit from System.Enum */
{
private int _value; /* Should be marked to be treated specially */
private MyTest(int value) /* Doesn't need to exist, since there's some CLR fiddling */
{
_value = value;
}
public static explicit operator int(MyTest value) /* CLR provides conversions automatically */
{
return value._value;
}
public static explicit operator MyTest(int value) /* CLR provides conversions automatically */
{
return new MyTest(value);
}
public static readonly MyTest A = new MyTest(1); /* Should be const, not readonly, but we can't do a const of a custom type in C#. Also, is magically implicitly converted without calling a constructor */
public static readonly MyTest B = new MyTest(2); /* Ditto */
}
Yes, you can easily get to the "underlying" int value, but the values of A and B are still strongly typed as being of type MyTest. This makes sure you don't accidentally use them in places where they're not appropriate.
The enum values are not of int type. int is the base type of the enum. The enums are technically ints but logically (from the perspective of the C# language) not. int (System.Int32) is the base type of all enums by default, if you don't explicitly specify another one.
You enum is of type Test. It is not int just because your enum has integers values.
You can cast your enum to get the int value:
int a = (int) Test.A;
Unlike C++, in C# you can't overload the assignment operator.
I'm doing a custom Number class for arithmetic operations with very large numbers and I want it to have the look-and-feel of the built-in numerical types like int, decimal, etc. I've overloaded the arithmetic operators, but the assignment remains...
Here's an example:
Number a = new Number(55);
Number b = a; //I want to copy the value, not the reference
Is there a workaround for that issue?
you can use the 'implicit' keyword to create an overload for the assignment:
Suppose you have a type like Foo, that you feel is implicitly convertable from a string.
You would write the following static method in your Foo class:
public static implicit operator Foo(string normalString)
{
//write your code here to go from string to Foo and return the new Foo.
}
Having done that, you can then use the following in your code:
Foo x = "whatever";
It's still not at all clear to me that you really need this. Either:
Your Number type should be a struct (which is probable - numbers are the most common example of structs). Note that all the types you want your type to act like (int, decimal etc) are structs.
or:
Your Number type should be immutable, making every mutation operation return a new instance, in which case you don't need the data to be copied on assignment anyway. (In fact, your type should be immutable whether or not it's a struct. Mutable structs are evil, and a number certainly shouldn't be a mutable reference type.)
You won't be able to work around it having the C++ look, since a = b; has other semantics in C++ than in C#. In C#, a = b; makes a point to the same object like b. In C++, a = b changes the content of a. Both has their ups and downs. It's like you do
MyType * a = new MyType();
MyType * b = new MyType();
a = b; /* only exchange pointers. will not change any content */
In C++ (it will lose the reference to the first object, and create a memory leak. But let's ignore that here). You cannot overload the assign operator in C++ for that either.
The workaround is easy:
MyType a = new MyType();
MyType b = new MyType();
// instead of a = b
a.Assign(b);
Disclaimer: I'm not a C# developer
You could create a write-only-property like this. then do a.Self = b; above.
public MyType Self {
set {
/* copy content of value to this */
this.Assign(value);
}
}
Now, this is not good. Since it violates the principle-of-least-surprise (POLS). One wouldn't expect a to change if one does a.Self = b;
Instead of making a copy of the data when passing the reference you could make the class immutable. When the class is immutable having multiple references to it isn't a problem since it can't be changed.
Operations that changes the data would of course return new instances.
An earlier post suggested this:
public static implicit operator Foo(string normalString) { }
I tried this approach... but to make it work you need this:
public static implicit operator Foo(Foo original) { }
and the compiler won't let you have an implicit conversion function from your exact type, nor from any base type of yourself. That makes sense since it would be a backdoor way of overriding the assignment operator, which C# doesn't want to allow.
Here is a solution that worked for myself :
public class MyTestClass
{
private int a;
private string str;
public MyTestClass()
{
a = 0;
str = null;
}
public MyTestClass(int a, string str)
{
this.a = a;
this.str = str;
}
public MyTestClass Clone
{
get
{
return new MyTestClass(this.a, this.str);
}
}
}
Somewhere else in the code :
MyTestClass test1 = new MyTestClass(5, "Cat");
MyTestClass test2 = test1.Clone;
Maybe what you're looking for can be solved using C# accessors.
http://msdn.microsoft.com/en-us/library/aa287786(v=vs.71).aspx