While the accepted answer addresses this (as do some others), I wanted to give an answer dedicated to what it seems like you're actually asking, which is about the semantics of variable assignment.
Variables in C# are simply pieces of memory that are set aside to hold a single value. It's important to note that there's no such thing as a "value variable" and a "reference variable", because variables only hold values.
The distinction between "value" and "reference" comes with the type. A Value Type (VT) means that the entire piece of data is stored within the variable.
If I have an integer variable named abc
that holds the value 100
, then that means that I have a four-byte block of memory within my application that stores the literal value 100
inside it. This is because int
is a value type, and thus all of the data is stored within the variable.
On the other hand, if I have a string
variable named foo
that holds the value "Adam"
, then there are two actual memory locations involved. The first is the piece of memory that stores the actual characters "Adam"
, as well as other information about my string (its length, etc.). A reference to this location is then stored within my variable. References are very similar to pointers in C/C++; while they are not the same, the analogy is sufficient for this explanation.
So, to sum it up, the value for a reference type is a reference to another location in memory, where the value for a value type is the data itself.
When you assign something to a variable, all you're changing is that variable's value. If I have this:
string str1 = "foo";
string str2 = str1;
Then I have two string variables that hold the same value (in this case, they each hold a reference to the same string, "foo"
.) If then do this:
str1 = "bar";
Then I have changed the value of str1
to a reference to the string "bar"
. This doesn't change str2
at all, since its value is still a reference to the string "foo"
.
=
does not change the object, but rather changes the reference to point at a different object. Your code has two references:me
andcopy
, and since you're only assigning a different object tome
,copy
still points at the old object. – Aleece