I'm getting a warning from ReSharper about a call to a virtual member from my objects constructor.
Why would this be something not to do?
When an object written in C# is constructed, what happens is that the initializers run in order from the most derived class to the base class, and then constructors run in order from the base class to the most derived class (see Eric Lippert's blog for details as to why this is).
Also in .NET objects do not change type as they are constructed, but start out as the most derived type, with the method table being for the most derived type. This means that virtual method calls always run on the most derived type.
When you combine these two facts you are left with the problem that if you make a virtual method call in a constructor, and it is not the most derived type in its inheritance hierarchy, that it will be called on a class whose constructor has not been run, and therefore may not be in a suitable state to have that method called.
This problem is, of course, mitigated if you mark your class as sealed to ensure that it is the most derived type in the inheritance hierarchy - in which case it is perfectly safe to call the virtual method.
In order to answer your question, consider this question: what will the below code print out when the Child object is instantiated?
class Parent
{
public Parent()
{
DoSomething();
}
protected virtual void DoSomething()
{
}
}
class Child : Parent
{
private string foo;
public Child()
{
foo = "HELLO";
}
protected override void DoSomething()
{
Console.WriteLine(foo.ToLower()); //NullReferenceException!?!
}
}
The answer is that in fact a NullReferenceException will be thrown, because foo is null. An object's base constructor is called before its own constructor. By having a virtual call in an object's constructor you are introducing the possibility that inheriting objects will execute code before they have been fully initialized.
The rules of C# are very different from that of Java and C++.
When you are in the constructor for some object in C#, that object exists in a fully initialized (just not "constructed") form, as its fully derived type.
namespace Demo
{
class A
{
public A()
{
System.Console.WriteLine("This is a {0},", this.GetType());
}
}
class B : A
{
}
// . . .
B b = new B(); // Output: "This is a Demo.B"
}
This means that if you call a virtual function from the constructor of A, it will resolve to any override in B, if one is provided.
Even if you intentionally set up A and B like this, fully understanding the behavior of the system, you could be in for a shock later. Say you called virtual functions in B's constructor, "knowing" they would be handled by B or A as appropriate. Then time passes, and someone else decides they need to define C, and override some of the virtual functions there. All of a sudden B's constructor ends up calling code in C, which could lead to quite surprising behavior.
It is probably a good idea to avoid virtual functions in constructors anyway, since the rules are so different between C#, C++, and Java. Your programmers may not know what to expect!
Reasons of the warning are already described, but how would you fix the warning? You have to seal either class or virtual member.
class B
{
protected virtual void Foo() { }
}
class A : B
{
public A()
{
Foo(); // warning here
}
}
You can seal class A:
sealed class A : B
{
public A()
{
Foo(); // no warning
}
}
Or you can seal method Foo:
class A : B
{
public A()
{
Foo(); // no warning
}
protected sealed override void Foo()
{
base.Foo();
}
}
In C#, a base class' constructor runs before the derived class' constructor, so any instance fields that a derived class might use in the possibly-overridden virtual member are not initialized yet.
Do note that this is just a warning to make you pay attention and make sure it's all-right. There are actual use-cases for this scenario, you just have to document the behavior of the virtual member that it can not use any instance fields declared in a derived class below where the constructor calling it is.
There are well-written answers above for why you wouldn't want to do that. Here's a counter-example where perhaps you would want to do that (translated into C# from Practical Object-Oriented Design in Ruby by Sandi Metz, p. 126).
Note that GetDependency() isn't touching any instance variables. It would be static if static methods could be virtual.
(To be fair, there are probably smarter ways of doing this via dependency injection containers or object initializers...)
public class MyClass
{
private IDependency _myDependency;
public MyClass(IDependency someValue = null)
{
_myDependency = someValue ?? GetDependency();
}
// If this were static, it could not be overridden
// as static methods cannot be virtual in C#.
protected virtual IDependency GetDependency()
{
return new SomeDependency();
}
}
public class MySubClass : MyClass
{
protected override IDependency GetDependency()
{
return new SomeOtherDependency();
}
}
public interface IDependency { }
public class SomeDependency : IDependency { }
public class SomeOtherDependency : IDependency { }
Yes, it's generally bad to call virtual method in the constructor.
At this point, the objet may not be fully constructed yet, and the invariants expected by methods may not hold yet.
Because until the constructor has completed executing, the object is not fully instantiated. Any members referenced by the virtual function may not be initialised. In C++, when you are in a constructor, this only refers to the static type of the constructor you are in, and not the actual dynamic type of the object that is being created. This means that the virtual function call might not even go where you expect it to.
Your constructor may (later, in an extension of your software) be called from the constructor of a subclass that overrides the virtual method. Now not the subclass's implementation of the function, but the implementation of the base class will be called. So it doesn't really make sense to call a virtual function here.
However, if your design satisfies the Liskov Substitution principle, no harm will be done. Probably that's why it's tolerated - a warning, not an error.
One important aspect of this question which other answers have not yet addressed is that it is safe for a base-class to call virtual members from within its constructor if that is what the derived classes are expecting it to do. In such cases, the designer of the derived class is responsible for ensuring that any methods which are run before construction is complete will behave as sensibly as they can under the circumstances. For example, in C++/CLI, constructors are wrapped in code which will call Dispose on the partially-constructed object if construction fails. Calling Dispose in such cases is often necessary to prevent resource leaks, but Dispose methods must be prepared for the possibility that the object upon which they are run may not have been fully constructed.
One important missing bit is, what is the correct way to resolve this issue?
As Greg explained, the root problem here is that a base class constructor would invoke the virtual member before the derived class has been constructed.
The following code, taken from MSDN's constructor design guidelines, demonstrates this issue.
public class BadBaseClass
{
protected string state;
public BadBaseClass()
{
this.state = "BadBaseClass";
this.DisplayState();
}
public virtual void DisplayState()
{
}
}
public class DerivedFromBad : BadBaseClass
{
public DerivedFromBad()
{
this.state = "DerivedFromBad";
}
public override void DisplayState()
{
Console.WriteLine(this.state);
}
}
When a new instance of DerivedFromBad is created, the base class constructor calls to DisplayState and shows BadBaseClass because the field has not yet been update by the derived constructor.
public class Tester
{
public static void Main()
{
var bad = new DerivedFromBad();
}
}
An improved implementation removes the virtual method from the base class constructor, and uses an Initialize method. Creating a new instance of DerivedFromBetter displays the expected "DerivedFromBetter"
public class BetterBaseClass
{
protected string state;
public BetterBaseClass()
{
this.state = "BetterBaseClass";
this.Initialize();
}
public void Initialize()
{
this.DisplayState();
}
public virtual void DisplayState()
{
}
}
public class DerivedFromBetter : BetterBaseClass
{
public DerivedFromBetter()
{
this.state = "DerivedFromBetter";
}
public override void DisplayState()
{
Console.WriteLine(this.state);
}
}
The warning is a reminder that virtual members are likely to be overridden on derived class. In that case whatever the parent class did to a virtual member will be undone or changed by overriding child class. Look at the small example blow for clarity
The parent class below attempts to set value to a virtual member on its constructor. And this will trigger Re-sharper warning, let see on code:
public class Parent
{
public virtual object Obj{get;set;}
public Parent()
{
// Re-sharper warning: this is open to change from
// inheriting class overriding virtual member
this.Obj = new Object();
}
}
The child class here overrides the parent property. If this property was not marked virtual the compiler would warn that the property hides property on the parent class and suggest that you add 'new' keyword if it is intentional.
public class Child: Parent
{
public Child():base()
{
this.Obj = "Something";
}
public override object Obj{get;set;}
}
Finally the impact on use, the output of the example below abandons the initial value set by parent class constructor.
And this is what Re-sharper attempts to to warn you, values set on the Parent class constructor are open to be overwritten by the child class constructor which is called right after the parent class constructor.
public class Program
{
public static void Main()
{
var child = new Child();
// anything that is done on parent virtual member is destroyed
Console.WriteLine(child.Obj);
// Output: "Something"
}
}
Beware of blindly following Resharper's advice and making the class sealed!
If it's a model in EF Code First it will remove the virtual keyword and that would disable lazy loading of it's relationships.
public **virtual** User User{ get; set; }
There's a difference between C++ and C# in this specific case.
In C++ the object is not initialized and therefore it is unsafe to call a virutal function inside a constructor.
In C# when a class object is created all its members are zero initialized. It is possible to call a virtual function in the constructor but if you'll might access members that are still zero. If you don't need to access members it is quite safe to call a virtual function in C#.
Just to add my thoughts. If you always initialize the private field when define it, this problem should be avoid. At least below code works like a charm:
class Parent
{
public Parent()
{
DoSomething();
}
protected virtual void DoSomething()
{
}
}
class Child : Parent
{
private string foo = "HELLO";
public Child() { /*Originally foo initialized here. Removed.*/ }
protected override void DoSomething()
{
Console.WriteLine(foo.ToLower());
}
}
I think that ignoring the warning might be legitimate if you want to give the child class the ability to set or override a property that the parent constructor will use right away:
internal class Parent
{
public Parent()
{
Console.WriteLine("Parent ctor");
Console.WriteLine(Something);
}
protected virtual string Something { get; } = "Parent";
}
internal class Child : Parent
{
public Child()
{
Console.WriteLine("Child ctor");
Console.WriteLine(Something);
}
protected override string Something { get; } = "Child";
}
The risk here would be for the child class to set the property from its constructor in which case the change in the value would occur after the base class constructor has been called.
My use case is that I want the child class to provide a specific value or a utility class such as a converter and I don't want to have to call an initialization method on the base.
The output of the above when instantiating the child class is:
Parent ctor
Child
Child ctor
Child
I would just add an Initialize() method to the base class and then call that from derived constructors. That method will call any virtual/abstract methods/properties AFTER all of the constructors have been executed :)
Another interesting thing I found is that the ReSharper error can be 'satisfied' by doing something like below which is dumb to me. However, as mentioned by many earlier, it still is not a good idea to call virtual properties/methods in constructor.
public class ConfigManager
{
public virtual int MyPropOne { get; private set; }
public virtual string MyPropTwo { get; private set; }
public ConfigManager()
{
Setup();
}
private void Setup()
{
MyPropOne = 1;
MyPropTwo = "test";
}
}
I have a problem with my code.
I would expect that since I'm constructing the Implementation object; every time I call Method() I'd use actual Implementation.Method() and not it's abstract's Base.Method(). It does not seem reasonable that I have to downcast to actual implementer or specify interface explicitly (So interfaces are not transitive in C#? I will call the "first proper instance of interface implementer" and not my class?)
I have a structure similar to this (simplified for clarity):
https://dotnetfiddle.net/oYVlQO
using System;
public interface IBase
{
string Method();
}
public abstract class Base : IBase
{
public string Method() { return "Sample"; }
}
public class Implementation : Base // if I add ", IBase" to this it works as expected, but why?
{
new public string Method() { return "Overriden"; }
}
public class Program
{
// and it's used like so...
public static void Main()
{
IBase b = new Implementation();
//Implementation b = new Implementation(); // It works as expected, always using Implementation.Method();
Console.WriteLine(b.Method()); // Produces "Sample", so Base.Method(). Why not implementation?
Console.WriteLine(((Implementation) b).Method()); // Produces "Overriden", so Implementation.Method(); Since when I have to downcast to use overriden method!?
}
}
}
I'm really scratching my head over this; Especially that the same code in Java works "as I would expect" https://repl.it/repls/VitalSpiritedWebpage
I've tried to find it in the c# specs to no avail, maybe I do not have the proper keywords...
In cause of the question, which is:
Why is it that way?
My answer:
Because you don’t override the method but hide it.
The interface is implemented by the baseclass, so the Method is called on the base-class.
To answer the question, which isn’t asked:
How would it work?
Answer:
using System;
public interface IBase
{
string Method();
}
public abstract class Base : IBase
{
public virtual string Method() { return "Sample"; }
}
public class Implementation : Base
{
public override string Method() { return "Overriden"; }
}
You may want to take a look at the part of the C# spec that deals with interface re-implementation.
When you access a member through the interface, it begins its lookup at the most derived type that explicitly implements that interface. In your example, the most derived type is Base and so it calls the method that's present there.
When you added IBase to the list of interfaces explicitly implemented by Implementation it worked, because that's the new starting point for lookup and it finds your new method.
You can either solve your problem by making the base member virtual and then overriding it in derived classes, or you can re-implement the interface by including that in the list for your Implementation class.
So the problem in my sample code is two-fold.
I assumed that in C# methods are "virtual" by default (as is in Java). Since I'm usually coding in Java, I've made an incorrect assumption.
See Is it possible to override a non-virtual method?
If I'd use virtual, I could override the method and achieve exactly the output I expected, as described in doc:
https://learn.microsoft.com/en-us/dotnet/csharp/language-reference/keywords/virtual namely "When a virtual method is invoked, the run-time type of the object is checked for an overriding member. The overriding member in the most derived class is called, which might be the original member, if no derived class has overridden the member."
My code, hovewer, is using method hiding, so unless I inform the compiler about my intention of using my implementation, it'll default to non-hidden method (as resolved by abstract class being the actual, original implementer)
I have a BaseClass, which implements a method to populate itself form a different data structure. SubClasses will add their properties to the base ones. I want all sub-classes and sub-sub...classes to implement their own version of the method and call their parent class to do the same. So this is not just overriding, I want to force the implementation, but each implementation of the method has to be called, not just overridden.
Sort of like:
class BaseClass
{
int id;
virtual void fromDictionary(Dictionary data)
{
id = data["id"];
}
}
class Derived1 : BaseClass
{
string name;
override void fromDictionary(Dictionary data)
{
name = data["name"];
base.fromDictionary(data);
}
}
class Derived2 : Derived1
{
float size;
override void fromDictionary(Dictionary data)
{
size = data["size"];
base.fromDictionary(data);
}
}
Then doing this:
Derived2 object = new Derived2();
object.fromDictionary(dictionary);
Populates all of the object's properties.
I can make the first one virtual and override in the derived ones, but that doesn't force the implementation. And making an abstract base above this base or using an interface wouldn't force ALL levels of inheritance to implement.
The ideal case would be forcing all derived classes to implement their version AND call their parent's version (or even better, have some sort of extension/overriding that automatically calls ALL implementations of the method from the instantiated object's class upwards).
How close to that can I get ?
Seeing why you need this kind of overriding I strongly believe that you should try to move this logic to a constructor because as it looks now:
Derived2 object = new Derived2();
object.fromDictionary(dictionary);
Your object will only be valid if it has a dictionary. So instead of constructing it from a dictionary using a method, you should provide a constructor which receives a dictionary as a parameter:
Derived2 object = new Derived2(dictionary);
Now you have a valid object from the beggining. There are more reasons why you should do this instead of using a method which composes your object, one as you observed is that each subclass will need to call the base method, and having this kind of constructor (assuming that you will not provide a parameterless one) will force the inheritors to call the base one.
Another advantage in using this kind of approach is that you will have a valid object form the beginning instead of making it possible for users of that classes to make invalid objects by forgetting to call the fromDictionary() method.
Thanks for the suggestions everyone, the closest I could get is:
public abstract class DAO
{
public long id { get; set; }
public void fromDictionary(Dictionary<string, object> obj)
{
//Does own part in the method
id = (long)obj["id"];
//Calls most derived implementation
fromDictionaryOperation(obj);
}
//Forces child to implement its part
protected abstract void fromDictionaryOperation(Dictionary<string, object> obj);
}
//Is forced to implement its part, and the base implementation will be executed always
public class Area : DAO
{
public string name { get; set; }
protected override void fromDictionaryOperation(Dictionary<string, object> obj)
{
name = (string)obj["name"];
}
}
//Is NOT forced to implement method, and MUST call base.fromDictionary() for all this to work properly, but is NOT FORCED TO.
public class CircularArea : Area
{
public float radius { get; set; }
protected override void fromDictionaryOperation(Dictionary<string, object> obj)
{
radius = (float)obj["radius"];
base.fromDictionary(obj);
}
}
So all 2nd generation classes will be fine, but subsequent sub-classes wont be forced to implement its part or call the parent implementation. Which means that if in an implementation of a sub-sub-class, the base.fromDictionary() method is not called, then all parent classes implementation, except the first/base class, will be skipped without any compiling warning or error.
To force implementation at all levels, I guess I could put the abstract method in an Interface and make all classes implement the interface, which can't be forced itself, but is as close as I can think of.
If anyone knows a way to completely force ALL of them to implement the same method, that would be top notch.
Given the following code:
public class Base {
public virtual void Method() { }
}
public class Derived : Base {
public override void Method() { }
}
...
var baseMethodInfo = typeof(Base).GetMember("Method")[0];
var derivedMethodInfo = typeof(Derived).GetMember("Method")[0];
Is it possible to determine if the derivedMethodInfo represents a method declaration which overrides another in a base class?
In another question it was observed that had Method been declared abstract (and not implemented) in the base class, derivedMethodInfo.DeclaringType would have turned up as Base, which makes sense after reading #EricLippert's comments. I noticed that in the present example, since the derived class re-declares the method, that derivedMethodInfo.DeclaringType == derivedMethodInfo.ReflectedType, viz. Derived.
There doesn't seem to be any connection between baseMethodInfo and derivedMethodInfo, other than their names are the same and their respective declaring types appear in the same inheritance chain. Is there any better way to make the connection?
The reason I ask is that there appears to be no way to distinguish, through reflection, between the earlier example and the following one:
public class Base {
public virtual void Method() { }
}
public class Derived : Base {
public new void Method() { }
}
In this case as well, the Derived class both declares and reflects a member called Method.
A method shadowing a virtual method will have the VtableLayoutMask flag set in Attributes.
Note that an ordinary virtual method (with no similar name from a base type) will also have this flag set.
This flag appears to indicate that the method introduces a new entry in the VTable.
There's a more specific class MethodInfo which derives from MemberInfo. Note that not all kinds of members can be virtual (fields cannot, for example).
If you say
var derivedMethodInfo = typeof(Derived).GetMethod("Method");
then you can check if
derivedMethodInfo.GetBaseDefinition() == derivedMethodInfo
or not. See documentation for GetBaseDefinition() where they also have a code example.
I'm trying to do the following:
public abstract BaseClass {
public virtual void ReceiveEvent(Event evt)
{
ProcessEvent(evt as dynamic);
}
private void ProcessEvent(object evt)
{
LogManager.Log(#"Received an event that is not being processed!
Dispatch fallback");
}
}
public DerivedClass: BaseClass {
private void ProcessEvent(SpecificEvent evt)
{
LogManager.Log("Processing Event");
}
}
SpecificEvents hit the fallback method instead of the one in the derived class. I use dynamic dispatch within the same class all the time and find it really useful/clean. Will it not work with derived classes as illustrated in the example above?
EDIT:
There seems to be some confusion in the answers. Basically i use the following design all the time:
public class SomeClass{
public void DoSomethingDispatcher(SomeObject obj)
{
ProcessObject(obj as dynamic);
}
private void DoSomething(SomeObjectType1 obj)
{
}
private void DoSomething(SomeObjectType2 obj)
{
}
private void DoSomething(SomeObjectType3 obj)
{
}
private void DoSomething(object obj) //fallback
{
}
}
Works great for when you don't know the exact type beforehand and you don't want to use a big switch statement. Just wondering if this can be implemented with inheritance where the base class holds the fallback method and the derived class holds all the more specific methods.
It's not working for you because even if evt is passed dynamic, ProcessEvent is not declared as virtual. This means that when the call to ProcessEvent is compiled, it is linked to the only implementation of the method that is found in the base class, and the ones in the derived classes will never be executed. Furthermore, you can't simply declare your ProcessEvent as virtual, since the signature will be different in the derived classes.
In order for your code to work as expected you could just override ReceiveEvent in the derived classes leaving it exactly the same:
public override void ReceiveEvent(Event evt)
{
ProcessEvent(evt as dynamic);
}
If you want to manage the unhandled events in the base class, just change the modifier of Process event in the base class to protected (otherwise it can't be executed when called by the overridden version of ReceiveEvents).
If the method is not virtual/abstract in the base class, and the method is not marked as override in the derived class, it will never work.
Also, I dont understand the usage of dynamic here.
What is the type of your "evt" when it hit ProcessEvent ?
You may take a look to Using Type dynamic :
The type is a static type, but an object of type dynamic bypasses
static type checking. In most cases, it functions like it has type
object.
So, evt is not a SpecificEvent.
To get the expected behaviour you should override the virtual method:
public DerivedClass: BaseClass
{
private override void ReceiveEvent(Event evt)
{
// Process your event here.
}
}
With this code, ReceiveEvent in the base class won't be called, thus the fallback ProcessEvent won't be called.
There is no reason to use dynamic.