Why does the mutable StringBuilder behave like the immutable string when a reference is changed?

Question

The "C# 4.0 IN A NUTSHELL" 4th edition book by the Albaharis states on page 249: ". . . calling object.ReferenceEquals guarantees normal referential equality."

So, I decided to test this out.

First I tried value types like this.

 int aa1 = 5;
 int aa2 = aa1;
MessageBox.Show("object.ReferenceEquals(aa1,aa2) is: " + object.ReferenceEquals(aa1, aa2)); 

And just as I expected, the result was false: object.ReferenceEquals(aa1, aa2) is: False

Life was good. Then I tried a mutable reference type like this.

System.Text.StringBuilder sblr1 = new System.Text.StringBuilder();
sblr1.Append("aaa");
System.Text.StringBuilder sblr2 = sblr1;          
MessageBox.Show("object.ReferenceEquals(sblr1,sblr2) is: " + object.ReferenceEquals(sblr1, sblr2)); 

And just as I expected, the result was true object.ReferenceEquals(sblr1, sblr2) is: True

Life was still good. Then I figured that since it is a mutable reference type, then if I change one variable to null, then both should be null. So I tried the following.

System.Text.StringBuilder sblr1 = new System.Text.StringBuilder();
sblr1.Append("aaa");
System.Text.StringBuilder sblr2 = sblr1;
sblr1 = null;
MessageBox.Show("object.ReferenceEquals(sblr1,sblr2) is: " + object.ReferenceEquals(sblr1, sblr2)); 

And I expected them to both be null. But the result I got was False: object.ReferenceEquals(sblr1, sblr2) is: False

Now life was not so good. I thought that if it overrode the memory location of sblr1, then it would be overriding the memory location of sblr2 also.

Then I thought that maybe they were pointing to two different nulls, so I tried this:

System.Text.StringBuilder sblr1 = new System.Text.StringBuilder();
sblr1.Append("aaa");
System.Text.StringBuilder sblr2 = sblr1;
sblr2 = null;
MessageBox.Show("sblr1 == " + sblr1 + " and sblr2 == " + sblr2);

But here, only one was pointing to a null like this. sblr1 == aaa and sblr2 ==

Only one was null.

It was displaying the behavior I'd expect from an immutable reference type like a string object. With a string object, I can do something like this:

string aa1 = "aaX";
string aa2 = "aaX";
MessageBox.Show("object.ReferenceEquals(aa1,aa2) is: " + object.ReferenceEquals(aa1, aa2)); 

And they will both reference the same thing like this. object.ReferenceEquals(aa1, aa2) is: True because "aaX" only gets written to the assembly once.

But if I do this:

string aa1 = "aaX";
string aa2 = "aaX"; 
aa1 = null; 
MessageBox.Show("After aa1 is null(" + aa1 + "), then aa2 is: " + aa2);

Then they point to different things like this: After aa1 is null (), then aa2 is: aaX

That's because string objects are immutable. The memory location doesn't get overriden. Rather, the variable points to a different location in Heap memory where the new value exists. Changing aa1 to null in the above example means that aa1 will point to a different location on the Heap memory.

Why then is the mutable reference type behaving just the same as the immutable reference type?

---

Edit 4:03PM and 4:08

I've recently tried this:

System.Text.StringBuilder sblr1 = new System.Text.StringBuilder();
sblr1.Append("aaa");

// sblr1 and sblr2 should now both point to the same location on the Heap that has "aaa".
System.Text.StringBuilder sblr2 = sblr1;

System.Text.StringBuilder sblr3 = new System.Text.StringBuilder();
sblr3.Append("bbb");

sblr1 = sblr3;
MessageBox.Show("sblr1 == " + sblr1 + " and sblr2 == " + sblr2 + " and sblr3 == " + sblr3);

Which gave me: sblr1 == bbb and sblr2 == aaa and sblr3 == bbb

That's more like the result I was expecting. I see now, thanks to the comments, that I abscent mindedly expected null to act like a memory location.

Answer 1

I thought that if it overrode the memory location of sblr1, then it would be overriding the memory location of sblr2 also.

This is your misunderstanding.

When you write this:

System.Text.StringBuilder sblr2 = sblr1;

You're assigning the sblr2 variable to be a reference to the same instance of StringBuilder as the one pointed to by sblr1. The two variables now point to the same reference.

You then write:

sblr1 = null;

This changes the sblr1 variable to now be a null reference. You didn't change the instance in memory at all.

This has nothing to do with whether the reference is a mutable type or not. You're changing the variables, not the instance which they are referencing.

As for your string example:

That's because string objects are immutable. The memory location doesn't get overridden

This actually is not true. The fact that you're setting one string variable to null doesn't really have anything to do with the string being immutable. That's a separate concern.

Why then is the mutable reference type behaving just the same as the immutable reference type?

The behavior you're seeing has nothing to do with mutability. It is the standard behavior for all reference types (whether immutable or mutable). Mutability is a different issue.

The main issue with mutability is this:

Suppose you have a class, like so:

class Foo
{
    public int Bar { get; set; }
}

If you write this:

Foo a = new Foo();
a.Bar = 42;
Foo b = a;
b.Bar = 54;
Console.WriteLine(a.Bar); // Will print 54, since you've changed the same mutable object

With immutable types, this can't happen, since you can't change Bar - instead, if you make an immutable class:

class Baz
{
    public Baz(int bar) { this.Bar = bar; }

    public int Bar { get; private set; }
}

You would need to write:

Baz a = new Baz(42);
Baz b = a;

// This isn't legal now:
// b.Bar = 54;
// So you'd write:
b = new Baz(54); // Creates a new reference

Alternatively, you could make the class return a new reference on a "change" operation, ie:

class Baz
{
    public Baz(int bar) { this.Bar = bar; }

    public int Bar { get; private set; }

    public Baz Alter(int newValue) { return new Baz(newValue); } // May copy other data from "this"
}

Then when you'd write:

Baz a = new Baz(42);
Baz b = a.Alter(54); // b is now a new instance

This is what happens with string - all of the methods return a new instance, since a string is immutable, so you can never "change" the existing copy.

Answer 2

This has nothing to do with mutability. The rules involved here are the same for all reference types. A non-ref non-out variable (or member, or slot in a collection) of a reference type is a reference (duh). That means it refers to some object. It does not refer to another reference, or to a location where another reference is (e.g. to a variable). When you assign to a variable (or member, or slot in a collection), you change what reference is in that place; you do not overwrite any part of any object (except, of course, the member you assign to, if it's a member).

In your code, there are two variables srbl1 and srbl2, each of which stores a reference to the same string builder object. Assigning to either changes overwrites one of those reference (e.g. with null, or with a reference to a different object).

Answer 3

Changing a reference is just changing what something refers to. It doesn't change the object itself.

One way to look at it is to imagine an array of integers:

int[] foo = new int[] {0, 1, 2, 3, 4, 5};

You can create two indexes that refer to items in the array:

int ix = 1;
int iy = ix;

And then foo[ix] == foo[iy].

If you then write foo[ix] = 42, then foo[ix] == foo[iy] is still true because you changed the value that the indexes referred to.

But if you change the index so that ix = 3, then ix and iy are referring to different things.

Reference types work exactly the same way.

When you write sblr1 = new StringBuilder(), you've created a new StringBuilder object instance and made sblr1 point to it. If you then write sblr2 = sblr1, you're just making sblr2 point to the same thing. And then sblr1 = null just says, "sblr1 isn't pointing to anything anymore." It doesn't actually affect the item it refers to, just as changing the index to an array doesn't affect the value of the item being indexed.