C# compiler oddity with delegate constructors

2019-02-04 22:41发布

Based on the following question, I found some odd behaviour of the c# compiler.

The following is valid C#:

static void K() {}

static void Main()
{
  var k = new Action(new Action(new Action(K))));
}

What I do find strange is the compiler 'deconstructing' the passed delegate.

The ILSpy output is as follows:

new Action(new Action(new Action(null, ldftn(K)), ldftn(Invoke)).Invoke);

As one can see, it automatically decides to use the Invoke method of the delegate. But why?

As it is, the code is unclear. Do we have a triply-wrapped delegate (actual) or is the inner delegate just 'copied' to the outer ones (my initial thought).

Surely if the intent was like the compiler emitted the code, one should have written:

var k = new Action(new Action(new Action(K).Invoke).Invoke);

Similar to the decompiled code.

Can anyone justify the reason for this 'surprising' transformation?

Update:

I can only think of one possible use-case for this; delegate type conversion. Eg:

delegate void Baz();
delegate void Bar();
...
var k = new Baz(new Bar( new Action (K)));

Perhaps the compiler should emit a warning if the same delegate types are used.

3条回答
欢心
2楼-- · 2019-02-04 22:56

The spec (section 7.6.10.5) says:

  • The new delegate instance is initialized with the same invocation list as the delegate instance given by E.

Now suppose the compiler translated it to something similar to your suggestion of:

new Action( a.Target, a.Method)

That would only ever create a delegate with an invocation list of a single method call. For a multi-cast delegate, it would violate the spec.

Sample code:

using System;

class Program
{
    static void Main(string[] args)
    {
        Action first = () => Console.WriteLine("First");
        Action second = () => Console.WriteLine("Second");

        Action both = first + second;
        Action wrapped1 =
            (Action) Delegate.CreateDelegate(typeof(Action),
                                             both.Target, both.Method);
        Action wrapped2 = new Action(both);

        Console.WriteLine("Calling wrapped1:");
        wrapped1();

        Console.WriteLine("Calling wrapped2:");
        wrapped2();
    }
}

Output:

Calling wrapped1:
Second
Calling wrapped2:
First
Second

As you can see, the real behaviour of the compiler matches the spec - your suggested behaviour doesn't.

This is partly due to the somewhat odd "sometimes single-cast, sometimes multi-cast" nature of Delegate, of course...

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\"骚年 ilove
3楼-- · 2019-02-04 22:57

When you try to treat a delegate as a method, the compiler actually uses the delegate's Invoke() method. So, for example, the two lines below compile to the exact same IL (both call Invoke()):

k();
k.Invoke();

I assume the oddity you're seeing is a consequence of this. The delegate constructor expects a method (or rather, a method group), but it gets a delegate instead. So it treats it as a method and uses the Invoke() method.

As for the meaning, it is delegate that calls delegate that calls the actual method. You can verify this yourself by accessing the delegate's Method and Target properties. In the case of the outer-most delegate, Method is Action.Invoke and Target the inner delegate.

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Anthone
4楼-- · 2019-02-04 23:00
  • delegate is a class
  • Action delegate has a constructor like so

    public extern Action(object @object, IntPtr method);

  • Since K is a static method there is no need to pass object to inner most action instance as first argument and hence it passes null

  • Since second argument is pointer to function therefore it passes pointer of K method using ldftn function
  • for the remaining Action instances the object is passed is inner Action and the second parameter is the Invoke method since when you call a delegate you're actually calling the Invoke method

Summary

var action = new Action(K) => Action action = new Action(null, ldftn(K))
new Action(action) => new Action(action, ldftn(Action.Invoke))

I hope this explains what is happening?

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