CODEWITHSUNDEEP
c#eventsdelegatesienumerable
Jader Dias
Jader Dias

Reputation: 90495

Why events behave differently in this case?

The following program

static IEnumerable<Action> Create()
{
    foreach (var i in Enumerable.Range(0, 2))
    {
        yield return () => { Console.Write(i); };
    }
}

static void Main(string[] args)
{
    foreach (var func in Create())
    {
        func();
    }

    Console.ReadLine();
}

ouputs

01

And this program

static event Action SomethingHappened;

static void Register()
{
    foreach (var i in Enumerable.Range(0, 2))
    {
        SomethingHappened += () => { Console.Write(i); };
    }
}

static void Main(string[] args)
{
    Register();
    SomethingHappened();
    Console.ReadLine();
}

outputs

11

Why is that so? How to make the program 2 output 01?

Upvotes: 0

Views: 55

Answers (2)

Brian Rasmussen
Brian Rasmussen

Reputation: 116411

In the second program the variable i is captured by the lambda. To get the correct behavior make a local copy before using it in the lambda. E.g. 

foreach (var i in Enumerable.Range(0, 2))
{
    var local = i;
    SomethingHappened += () => { Console.Write(local); };
}

Upvotes: 1

Jon Skeet
Jon Skeet

Reputation: 1501143

You're capturing the loop variable in your lambda expression. That means when the delegate is finally invoked, it will use the latest value of the variable... which will always be 1. Try this:

foreach (var i in Enumerable.Range(0, 2))
{
    int copy = i;
    SomethingHappened += () => { Console.Write(copy); };
}

... then read Eric Lippert's blog post about it.

Upvotes: 3

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