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# [C#] Object passed to method as parameter -> questions

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4 replies to this topic

### #1elodman  Members

Posted 21 September 2011 - 08:55 AM

Greetings.

Can you explain why
1) method nochange() doesn't change the object argument, although the object is of reference-type passed by value?
2) method change2() can change the field of the object passed without REF?

More comments in the code below.

Thanx. Bye.


using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace ref_obj
{
class clsTest
{
public int a;

public clsTest(int i)
{
a = i;
}

//this will not change the argument
public void noChange(clsTest obj)
{
clsTest newObj = new clsTest(0);
obj = newObj; // this has no effect outside of noChange()
}

//this will change what the argument refers to
public void change1(ref clsTest obj)
{
clsTest newObj = new clsTest(0);
obj = newObj;//this affects the calling argument
}

// changing Object argument field without REF
public void change2(clsTest obj)
{
obj.a = 33;// this changes a field of the called argument
}
}
}

namespace ref_obj
{
class Program
{
static void Main(string[] args)
{
clsTest ob = new clsTest(100);

Console.WriteLine("ob.a before call: " + ob.a);

ob.noChange(ob);
Console.WriteLine("ob.a after nochange call: " + ob.a);

ob.change1(ref ob);
Console.WriteLine("ob.a after change1 call: " + ob.a);

ob.change2(ob);
Console.WriteLine("ob.a after change2 call: " + ob.a);

}
}
}


OUTPUT:

ob.a before call: 100
ob.a after call to NoChange(): 100
ob.a after call to Change1(): 0
ob.a after call to Change2(): 33
That was the time, the Golden Age, when C-64 and Amiga ruled!

### #2speciesUnknown  Members

Posted 21 September 2011 - 09:09 AM

Greetings.

Can you explain why
1) method nochange() doesn't change the object argument, although the object is of reference-type passed by value?
2) method change2() can change the field of the object passed without REF?

More comments in the code below.

Thanx. Bye.


using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace ref_obj
{
class clsTest
{
public int a;

public clsTest(int i)
{
a = i;
}

//this will not change the argument
public void noChange(clsTest obj)
{
clsTest newObj = new clsTest(0);
obj = newObj; // this has no effect outside of noChange()
}

//this will change what the argument refers to
public void change1(ref clsTest obj)
{
clsTest newObj = new clsTest(0);
obj = newObj;//this affects the calling argument
}

// changing Object argument field without REF
public void change2(clsTest obj)
{
obj.a = 33;// this changes a field of the called argument
}
}
}

namespace ref_obj
{
class Program
{
static void Main(string[] args)
{
clsTest ob = new clsTest(100);

Console.WriteLine("ob.a before call: " + ob.a);

ob.noChange(ob);
Console.WriteLine("ob.a after nochange call: " + ob.a);

ob.change1(ref ob);
Console.WriteLine("ob.a after change1 call: " + ob.a);

ob.change2(ob);
Console.WriteLine("ob.a after change2 call: " + ob.a);

}
}
}


OUTPUT:

ob.a before call: 100
ob.a after call to NoChange(): 100
ob.a after call to Change1(): 0
ob.a after call to Change2(): 33

It is important to get your head around this stuff early on, so good job on noticing it and asking.

1)
The reason for this behaviour is that reference types are accessed through their reference, but a reference itself is actually a value type. in your function NoChange, a copy of the reference (which is usually just an integer ID of the object, depending on the language) is created. What you are doing is then modifying the local copy of the reference. The original reference cannot be changed.

This is where the ref keyword comes in. You are now passing a reference to a reference. Now, you can change the original.

2)
In this case, you are not modifying the reference, which was passed by value, but you are modifying something elsewhere in memory, using your copy of the reference to access it. Remember, only the reference itself is passed by value, the new reference still points to the same referee.
Don't thank me, thank the moon's gravitation pull! Post in My Journal and help me to not procrastinate!

### #3elodman  Members

Posted 22 September 2011 - 07:04 AM

2)
In this case, you are not modifying the reference, which was passed by value, but you are modifying something elsewhere in memory, using your copy of the reference to access it. Remember, only the reference itself is passed by value, the new reference still points to the same referee.

Thank you very much!

Although I don't really understand why point 2 happens as is - so feel free to tell me more on it - , but I've found some other sources for the subject for the curious ones:
Jon Skeet's info on C# parameter passing (he has a famous book : C# in depth)

and:

Originally Posted by MSDNA variable of a reference type does not contain its data directly; it contains a reference to its data. When you pass a reference-type parameter by value, it is possible to change the data pointed to by the reference, such as the value of a class member. However, you cannot change the value of the reference itself; that is, you cannot use the same reference to allocate memory for a new class and have it persist outside the block. To do that, pass the parameter using the ref (or out) keyword. For simplicity, the following examples use ref.

That was the time, the Golden Age, when C-64 and Amiga ruled!

### #4rip-off  Moderators

Posted 22 September 2011 - 07:39 AM

Here is a crappy ASCII art version, showing the various in scope object references and their values. The drawing indicates the current state of the program after the previous line.

Object identity is shown by a "id" value. For the ref parameter, you can see I've called it "Main.ob" to indicate that this is the actual object. The reference variable is just a temporary alias for that name.
public void noChange(clsTest obj)
{
// [this]----[id: 1, a: 100]
//           |
// [obj ]----+

clsTest newObj = new clsTest(0);
// [this]------[id: 1, a: 100]
//             |
// [obj ] -----+
//
// [newObj]----[id: 2, a: 0]

obj = newObj;
// [this]------[id: 1, a: 100]
//
// [obj ]------+
//             |
// [newObj]----[id: 2, a: 100]
}

public void change1(ref clsTest obj)
{
// [this   ]----[id: 1, a: 100]
//              |
// [Main.ob]----+

clsTest newObj = new clsTest(0);
// [this]---------[id: 1, a: 100]
//                |
// [Main.ob]------+
//
// [newObj]-------[id: 3, a: 0]

obj = newObj;
// [this]---------[id: 1, a: 100]
//
// [Main.ob]------+
//                |
// [newObj]-------[id: 3, a: 0]
}

public void change2(clsTest obj)
{
// [this]----[id: 3, a: 0]
//           |
// [obj ]----+

obj.a = 33;
// [this]----[id: 3, a: 33]
//           |
// [obj ]----+
}

static void Main(string[] args)
{
clsTest ob = new clsTest(100);
// [ob]----[id: 1, a: 100]

ob.noChange(ob);
// [ob]----[id: 1, a: 100]

ob.change1(ref ob);
// [ob]----[id: 3, a: 0]

ob.change2(ob);
// [ob]----[id: 3, a: 33]
}

Hopefully this clarifies, and not confuses!

### #5elodman  Members

Posted 22 September 2011 - 09:14 AM

Thank thee Rip-off, hope I will understand one day...

other fine examples and info from msdn:
http://msdn.microsof...y/s6938f28.aspx
That was the time, the Golden Age, when C-64 and Amiga ruled!

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