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So Im going through Illustrated C# 2012, I basically get everything but this section of the chapter, below is the source code: 

class MyClass
{
    public int Val = 20;
}

class Program
{
    static void RefAsParameter(MyClass f1)
    {
        f1.Val = 50;
        Console.WriteLine( "After member assignment: {0}", f1.Val );
        f1 = new MyClass();
        Console.WriteLine( "After new object creation: {0}", f1.Val );
    }

    static void Main()
    {
        MyClass a1 = new MyClass();
        Console.WriteLine( "Before method call: {0}", a1.Val );
        RefAsParameter( a1 );
        Console.WriteLine( "After method call: {0}", a1.Val );
    }
}

This code produces the following output:

Before method call: 20
After member assignment: 50
After new object creation: 20
After method call: 50

So... I understand basically most of it up to the last Console.WriteLine(). why is it 50 "after method call". Since it created a new "MyClass()" shouldn't it still be 20? Clearly im missing something majorly obvious. Is it because f1.Val changed the Myclass Public value "for good" or what?

Slightly confused on this. Thanks. I understand "references" in general this one im just kinda stumped on.

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  • 1
    it passed f1(reference type)and assign 50 then f1 "points" to a new myclass but a1 (which is referencing what f1 was until he got "ressign")gets the new 50. Commented Aug 27, 2013 at 2:07

3 Answers 3

Reset to default
3

The line

f1 = new MyClass();

reassigns the variable f1 to point to a new MyClass. It does not change the a1 variable.

If you change the method to this:

static void RefAsParameter( ref MyClass f1 )

Then it will change the a1 variable, and you will get:

After method call: 20
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4 Comments

Ah I think I understand, a1 is originally 20 when the New MyClass() is first called, then we pass a1 into the RefAsParamFunction, and make a1 = 50 and print that out. We then create a BRAND new MyClass with F1 that is 20 and print that out (so we print 50). a1 is still = 50 since it was passed by value and the copy is now 50. I think I get it...if what I said above makes sense haha.
Sounds like you've got it to me!
But: "a1 is still = 50 since it was passed by value" - yeah: but remember, since we're talking about "reference types", what was actually passed by value was a reference. There was no copy made of the actual object, just a copy of a reference that points to it. When you executed f1.val = 50;, just one value was changed. It's just that you have two references pointing to that value: a1 and f1.
Ah gotcha, so we got a copy of the Reference pointing to the same value. Cool :)
3

Behold.. my awesome paint skillz:

enter image description here

(Click here for the full image)

When f1 is used within the method, it is a new reference that points to the same object in memory. When you re-assign the reference with f1 = new MyClass();, you are effectively saying "okay, this reference now points to a new instance". This leaves a1 untouched.. so when the function unwinds and hits the Console.WriteLine, your a1 reference still references the old area of memory.. which now has a property set at 50.

To have the behaviour you expect, you pass the object in as a ref parameter:

static void RefAsParameter(ref MyClass f1)
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Bless your paint skillz ;-)
1

When an object is referenced, it doesn't mean it is copying itself. It means we are pointing to the same object with two different names. Here a1 and f1 are both referencing the same object that's why if we are changing the value using f1 so the value is changing for a1 as well.

Later you are creating another instance and assigning the reference to f1. So now f1 will have no longer access to the previous object, and now a1 and f1 will not reference the same object.

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