C# reference type - string and object

Question

I am studying reference and value types in C#. According to documentation string and object are reference types. When we execute following code:

 static void Main(string[] args)
    {
        int a = 30;
        object o = a;
        AlterObject(o);

        Console.WriteLine(o);
        Console.Read();
    }

    static void AlterSObject(object testO)
    {
        int b = 130;
        testO = b;
    }

It prints output as 30. Please explain if object is reference type, why value of object is not changed in function.

Because the function doesn't change anything on the object referenced by the `testO` variable. It changes the variable to point to a new reference. Had you done something like `testO.SomeProperty = "some value";` then that property would have changed on the referenced object. — David, Nov 07 '18 at 14:19
I can´t see where you use a string. Ths 3 of the 4 duplicates don´t match the question, IMHO. — MakePeaceGreatAgain, Nov 07 '18 at 14:36

MakePeaceGreatAgain · Answer 1 · 2018-11-07T14:33:35.027

2

When you pass the value 30 as object, you create a new instance which points to 30 (this is called boxing), which you pass to the method.

This reference itself is passed - as every argument in .NET - by value. Thus you can´t change the reference, but only its fields or properties. You could re-reference any other object within your method, but because the parameter is still passed by value this won´t be reflected in the calling code.

But even by using the ref-keyword a won´t change, because it has nothing to do with the boxed object:

static void AlterSObject(ref object testO)
{
    int b = 130;
    testO = b;
}

Now o (which is the boxed object and has nothing to do with a) would reflect whatever you assigned to testO within AlterSObject. However a will not reflect it. See my fiddle for test.

edited Nov 07 '18 at 14:33

answered Nov 07 '18 at 14:23

MakePeaceGreatAgain

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If I create 2 objects like following - object o = 30; object o2 = o; o = 60; Why o and o2 does not share a memory location ? – AkshayP Nov 07 '18 at 14:48
1

@AkshayP Because you don´t *modify* the underlying object (`30`) , but just reference **another one** (`60`). So `o2` still points to the exact same instance (`30`). – MakePeaceGreatAgain Nov 07 '18 at 14:51

score 0 · Answer 2 · answered Nov 07 '18 at 14:46

You never declared AlterObject()'s parameters as references, so its passing by value. C# will use a pass-by-value approach unless you designate the given parameter as a reference using the ref keyword. You will also need to either overload(as I did in the following sample code) or rewrite your method to accept reference values.

public class test{
static void Main(string[] args){
    int a = 30;
    object o = a;

    AlterObject(ref o);

    System.Console.WriteLine(o);
    System.Console.Read();
}

static void AlterObject(object testO){
    int b = 130;
    testO = b;
}

static void AlterObject(ref object testO){
    int b = 130;
    testO = b;
}
}

C# reference type - string and object

2 Answers2