Why output is 5?

Question

I am trying to add 2 more functions to my delegate but it seems it doesn't process the functions I added. Following program outputs 5. I expect 10. Why is this the case?

using System;

namespace abc
{
public delegate int Del(int k);

class Class1
{
    public int Add2(int value)
    {
        return value = value + 2;
    }

    public int Add3(int value)
    {
        return value = value + 3;
    }
}

class Program
{
    static void Main(string[] args)
    {
        Class1 c = new Class1();
        Del mydel = c.Add2;
        mydel += c.Add3;
        mydel += c.Add2;

        Console.WriteLine(mydel(3));
        Console.ReadLine();
    }
}
}

`mydel(3)` doesn't actually change the value of the argument, so you get the result of the last function which is `Add2(3)` which is 3+2 so 5 — Rafalon, Apr 24 '17 at 14:13

score 5 · Accepted Answer · edited May 23 '17 at 12:25

5

What happens when you call mydel(3) is this:

Add2(3) is called. This returns 3 + 2 = 5, the return value of which is discarded.
Add3(3) is called. This returns 3 + 3 = 6, the return value of which is discarded.
Add2(3) is called. This returns 3 + 2 = 5, the return value of which is printed.

When you chain delegates, the return value of one is NOT passed as a parameter to a chained one.

Note that the delegates are called in LIFO order.

If you want to see how to actually chain functions together using delegates, see this thread.

Alternatively, as per the linked answer (and as mentioned by Toskr in the comments below), you can use GetInvocationList() and DynamicInvoke() to chain the methods together, but this is a very unusual thing to do:

static void Main()
{
    Class1 c = new Class1();
    Del mydel = c.Add2;
    mydel += c.Add3;
    mydel += c.Add2;

    int result = 3;

    foreach (var func in mydel.GetInvocationList())
    {
        result = (int)func.DynamicInvoke(result);
    }

    Console.WriteLine(result);
}

edited May 23 '17 at 12:25

Community

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answered Apr 24 '17 at 14:13

Matthew Watson

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beat me to it, i was half way through roughly this :) – BugFinder Apr 24 '17 at 14:13
1

also you forgot the first call to `Add2(3)`, and last is also `Add2(3)` not `Add3(2)`. – Rafalon Apr 24 '17 at 14:14
Is it possible to save the outputs and use it on next call? – Lyrk Apr 24 '17 at 14:16
@Rafalon Thanks; fixed. :) – Matthew Watson Apr 24 '17 at 14:18
@Lyrk Of course. But why? You should really have a good reason for doing hacks like this. Multicast delegates are designed for events, and events shouldn't have return values at all. – Luaan Apr 24 '17 at 14:18
@Lyrk See the link at the end of my answer. – Matthew Watson Apr 24 '17 at 14:20
Just because I am wondering. I am studying C# and there are lots of things flying in my mind. – Lyrk Apr 24 '17 at 14:20
@MatthewWatson in order : `Add2(3) -> 3+2 = 5` discarded | `Add3(3) -> 3+3 = 6` discarded | `Add2(3) -> 3+2 = 5` printed – Rafalon Apr 24 '17 at 14:28
Disappointed :( There is a way for him to do this without using the more convoluted way, but the question is closed. He can simply get the invocation list and concat them to a single int for a response value. – Toskr Apr 24 '17 at 14:30

Why output is 5?

1 Answers1