This question already has answers here:
Making a generic property
(3 answers)
Closed 9 years ago.
I have a class:
public class class1
{
public string Property1 {get;set;}
public int Property2 {get;set;}
}
Which will be instantiated:
var c = new class1();
c.Property1 = "blah";
c.Property2 = 666;
So bear with me (I am new to generics), I need another class with a property of a generic type so that Property1 or Property2 can be used to set Property3:
public class Class2
{
public GenericType Property3 {get;set;}
}
I want to be able to:
var c2 = new class2();
c2.Property3 = c1.Property2 // Any property of any type.
#bytenik I think the originator is asking that class3 be defined to contain a generic property. That way when he / she has a property from class1 or class2 which in this case is a string / int that class3's property could handle either case.
public class Class3<T>
{
public T Property3 {get;set;}
}
I think the intent is the poster wants to do this:
Class3.Property3 = Class2.Property2
I think the poster will need to cast it to type T for this to work though.
Look at the link that was posted for an example: Making a generic property
Here is what you can do:
namespace GenericSO
{
public class Class1
{
public int property1 { get;set;}
}
public class Class2<T>
{
public T property2 { get; set; }
}
class Program
{
static void Main(string[] args)
{
Class1 c1 = new Class1();
c1.property1 = 20;
Class2<int> c2 = new Class2<int>();
c2.property2 = c1.property1;
}
}
}
Notice how your template property2 gets the value of property1.
You have to tell it what kind of generic.
public class class1<T>
{
public T Property3 {get;set;}
}
Regarding to edited version of the question:
If you need a property, which can be set with any type, the most reasonable solution here is to simply use property of type Object. For C# compiler there is no way to find out instance of which exactly type you've previously pushed into property setter.
I think you may have misunderstood generics. Another word that could be used is "template" but that is avoided because it is used for more advanced things in C++.
The following will create a generic class of a currently undefined type T.
public class Class2<T>
{
public T Property3 { get; set; }
}
To use this you need to specify the missing type:
var x = new Class2<int>();
This will create an object that has a property Property3 that is of type int.
... or ...
var y = new Class2<string>();
This will create an object that has a property Property3 that is of type string.
From your question I believe you actually want a type where you can assign any type to it at runtime, but this is not what generics provide.
Related
This question already has answers here:
What is difference between Init-Only and ReadOnly in C# 9?
(4 answers)
Closed 1 year ago.
I can define a class like below:
public class MyClass
{
public int Id { get; }
public MyClass(int id) => Id = id;
}
And I have to define the Id from the constructor and it will be read-only.
But if I want to use Init only setters in the C# 9.0, what does it and how can I use it?
public class MyClass
{
public int Id { get; init; }
}
Init only setters provide consistent syntax to initialize members of an object. Property initializers make it clear which value is setting which property. The downside is that those properties must be settable.
With that, you don't need to provide the value at the beginning and the constructor and you can do it afterward:
var myClass = new MyClass
{
Id = 10
}
and it will be sealed and you cannot change it anymore.
myClass.Id = 43; // not allowed
read more info
In a nutshell:
var obj = new MyClass
{
Id = 42 // totally fine
};
obj.Id = 43; // not OK, we're not initializing
Trivial in this case and not much different to using a constructor parameter, but useful in some more complex scenarios where you don't want 200 constructor parameters, but you do want it to be outwardly immutable once constructed.
This question already has answers here:
Generic list of generic objects
(3 answers)
Closed 4 years ago.
I have an object that I want to put into a List or Collection. Is there a way to do that without having the T specified?
I want to do something like this: List<CommonProperty<T>>
Here's the object for reference:
internal class CommonProperty<T>
{
public string Name { get; set; }
public PropType Type { get; set; }
public List<T> PossibleValues { get; set; }
private T _value;
public T Value
{
get { return _value; }
set
{
if (!_value.Equals(value))
{
_value = value;
}
}
}
}
No, you can't use open generic types like that.
You could have a List<CommonProperty<T>> within a context where T is already a type parameter:
public class Foo<T>
{
static void Bar()
{
// This is fine, but is not what you're looking for - it uses
// the type parameter T as the type argument
List<CommonProperty<T>> list = new List<CommonProperty<T>>();
}
}
Typically the solution here is to have a non-generic base class or interface which the generic class or interface derives from:
// Consider making it abstract
internal class CommonProperty
{
public string Name { get; set; }
public PropType Type { get; set; }
}
internal class CommonProperty<T> : CommonProperty
{
public List<T> PossibleValues { get; set; }
private T _value;
public T Value
{
get => _value;
set
{
// TODO: Is this really necessary?
if (!_value.Equals(value))
{
_value = value;
}
}
}
}
You can then create a List<CommonProperty>... although be aware that it's entirely possible that you could end up with an element which wasn't a CommonProperty<T> at all that way.
From the List<CommonProperty> you'd be able to retrieve the names and types of all the properties - but the values wouldn't be available without casting to the specific type. You could have an abstract object Value { get; set; } property in the base class, which was then overridden in the derived class, potentially - but it's not clear whether that's necessary or helpful in your use case.
I think the closest you can get is to define an interface to match an un-typed (specific) CommonProperty, using Object instead of the T. Then have your CommonProperty<T> implement that interface. Now you can use the interface with your list.
But this isn't great. You'll lose a lot of nice type checking, and have to do more casting. If this is the primary way you'll use these objects, there's not much point to having a generic class at all anymore.
It is not possible to put a mixture of generics instantiated with different type arguments into the same collection. Such collection would not be useful anyway, because the caller would be expected to supply T for each item at compile time.
Suppose you could do what you want, i.e.
// imagine that you could do this
List<CommonProperty<T>> mixedList = GetAllProperties();
You would be forced to supply T once you start using items from that list, i.e.
foreach (CommonProperty<T> prop in mixedList) {
... // ^^^
... // Here you would need to specify T, but you don't know which one
}
In other words, such list would be unusable.
A list of properties of a specific type, on the other hand, would be useful:
List<CommonProperty<string>> stringPropList = GetPropertiesOfType<string>();
foreach (CommonProperty<string> prop in stringPropList ) {
...
}
Therefore, one solution to your problem would be building GetPropertiesOfType<T> method returning List<CommonProperty<T>> bound to type T from the caller.
An alternative approach would be to make CommonProperty non-generic, and let the caller check the type at runtime:
internal class CommonProperty {
public string Name { get; set; }
public PropType Type { get; set; }
public List<T> GetPossibleValues<T> { get; set; }
private object _value;
public T GetValue<T>() {
return (T)_value;
}
public void SetValue<T>(T val) {
_value = val;
}
}
I have previously created methods that take a property as input using the following syntax:
public void MyMethod<T>(Expression<Func<T>> property) { }
public int MyProperty { get; set; }
public void SomeOtherMethod() {
MyMethod(() => MyProperty);
}
That way I can, for example, cast the property to a MemberExpression and get the fully qualified name of the property.
However, is it possible to do the same in a class constructor, or have a property that stores another property, like how the property variable does in MyMethod()?
I have tried the following, but that didn't work because the type parameter is not specified in either cases, but I wouldn't even know which type to use... especially because the type is magically inferred in the topmost example.
In constructor
public class MyClass<T> {
public MyClass(Expression<Func<T>> property) { }
}
public int MyProperty { get; set; }
var myClass = new MyClass(() => MyProperty);
As property
public class MyClass<T> {
public Expression<Func<T>> SomeProperty { get; set; }
}
public int MyProperty { get; set; }
var myClass = new MyClass() { SomeProperty = () => MyProperty };
So, how can I have a property expression in a constructor or property of its own without having the use the type definition?
Is it possible to [infer generic parameters] in a class constructor?
In C# generic inference is not supported on constructors, so you have to set it explicitly:
var myClass = new MyClass<int>(() => MyProperty);
The main reason it's not supported is because there are often many types that could be used as the generic constraint due to implicit type conversions, so the compiler would have to choose the "best" generic parameters based on the information that it has. The costs associated with designing, programming, testing, documenting, and integrating this logic outweigh the benefit, or at least don't provide a relative benefit that exceeds the other features that MS could spend time developing.
In this case you know the best generic parameter type because the MyProperty property returns an int. Technically you could use any type that is implicitly convertible from int (object, double, etc.) but int is the best fit.
Could I maybe use object as a catch-all?
Sure, but then your code isn't generic. You can just do
public class MyClass {
public MyClass(Expression<Func<object>> property) { }
}
You need to supply the type parameter of T
var myClass = new MyClass<int>(() => MyProperty);
var myClass = new MyClass<int>() { SomeProperty = () => MyProperty };
Otherwise it won't know what type of expression it should get in the constructor.
This question already has an answer here:
emit class with a property of the same type as declaring emitted type
(1 answer)
Closed 8 years ago.
I want to generate Types via reflection at runtime that reference each other.
With static code I would do this
public class Order
{
public int Id { get; set; }
public Customer Customer { get; set; }
}
public class Customer
{
public int Id { get; set; }
public Order Order { get; set; }
}
I can sucessfully generate my Order and my OrderDetails Type with the value type properties.
The code looks like this
var aName = new System.Reflection.AssemblyName("DynamicAssembly");
var ab = AppDomain.CurrentDomain.DefineDynamicAssembly(
aName, System.Reflection.Emit.AssemblyBuilderAccess.Run);
var mb = ab.DefineDynamicModule(aName.Name);
var tb = mb.DefineType("Order",
System.Reflection.TypeAttributes.Public, typeof(Object));
var pbId = tb.DefineProperty("Id", PropertyAttributes.None, typeof(int), null);
Now I am stuck at this line:
var pbCustomer = tb.DefineProperty("Customer", PropertyAttributes.None, ???, null);
I am required to pass the type of the property to the DefineProperty method but the type does not exist at this point.
Now I could just create a type builder for customer at this point and use tb.CreateType() to get the type but that would not help since Customer needs a reference to Order, too.
Your last paragraph is roughly right, but TypeBuilder derives from Type, so you don't need to call CreateType. That is, create type builders for each of the recursive types, then define the properties passing the respective builders themselves as the return types.
Is there any way to make a property that is defined in a class compulsory?
I want the compiler to complain if the object is declared and used without the compulsory property.
Best approach I would think would be to not specify a default constructor.
public class MyClass
{
public MyClass(string myProp)
{
MyProp = myProp;
}
public string MyProp { get; set; }
}
A way might be to implement a constructor that forces the creation of an object to have an initial value for a property:
class Demo
{
public int P {get; set;}
Demo(int p)
{
P = p;
}
}
Now the client can only create the object by passing a value for p:
var d = new Demo(42);
How about adding a constructor to the class that would take a value for the property?
public class MyClass
{
public int Mandatory {get; set;}
public int Optional {get; set;}
public MyClass(int mandatory)
{
Mandatory = mandatory;
}
public MyClass(int mandatory, int optional)
{
Mandatory = mandatory;
Optional = optional;
}
}
This way, the class can be instantiated using either the one or two-parameter constructor, so the user will have to specify a value for the Mandatory property.
MyClass x = new MyClass(); // does not compile
MyClass x = new MyClass(1); //Mandatory = 1; Optional = 0 (default value)
MyClass x = new MyClass(1,2);//Mandatory = 1; Optional = 2
No, you can't force somebody to provide information to a property.
However, you can go the normal route, and make the constructor require the information. That way, the object can't be constructed without providing the information you need.
You can just add a constructor to your class that accepts one parameter - the value for your property. Like this, the property is always initialized after an instance of the class is created.
make default constructor private and define constructor that will get needed value for your property
public YourClass(string value)
{
YourCompulsaryProperty = value;
}
public string YourCompulsaryProperty{get; set;}