How to add another action to an action in C#?

Question

Why this code will print only "A" and "B", but not "C" ?

Action act = null;
act += () => MessageLog.Message("A");
act += () => MessageLog.Message("B");
Action<Action> add = a => a += () => MessageLog.Message("C");
add(act);
act.Invoke();

Howrah

Answer 1

Delegates are immutable. The += operator creates a new delegate with an invocation list consisting of the invocation list from the left side, followed by the invocation list from the right hand side of the operator. It then stores a reference to that delegate in the variable on the left hand side. So

act += () => MessageLog.Message("A");

is equivalent to

act = act + (Action) (() => MessageLog.Message("A"));

which is in turn equivalent to:

act = (Action) Delegate.Combine(act, (Action) (() => MessageLog.Message("A"));

So, now we can come to your add delegate. This lambda expression:

a => a += () => MessageLog.Message("C");

... modifies the parameter a to refer to the newly combined delegate... and then ignores it.

Instead, you need to return the combined delegate (so change add to be a Func<Action, Action>) and then use the return value when you invoke add:

Action act = null;
act += () => MessageLog.Message("A");
act += () => MessageLog.Message("B");
Func<Action, Action> add = a => a + () => MessageLog.Message("C");
act = add(act);
act.Invoke();

Ogive

So, in comparison, this is just like passing an "int" to a function and inside, adding some value to that "int"? Anyway all that helped me and as always, I get stuck on the simplest of things. Thanks! – Howrah

@krokots: Yes, exactly. – Ogive

Answer 2

As you probably know, A += B is just shorthand for A = A + B, so your add delegate is really just:

a => a = a + () => MessageLog.Message("C")

The + creates the combined delegate a + () => MessageLog.Message("C"), and = assigns it back to a. Notice how you are just reassigning the parameter a, and now it should be clear that this doesn't change the variable that is passed in (act) at all.

Other than returning the combined delegate, you can also pass act by reference:

delegate void ByRefAction<T>(ref T parameter);

...

    Action act = null;
    act += () => Console.WriteLine("A");
    act += () => Console.WriteLine("B");
    ByRefAction<Action> add = (ref Action a) => a += () => Console.WriteLine("C");
    add(ref act);
    act.Invoke();

Deficiency