Is there a workaround for overloading the assignment operator in C#?

Question

Unlike C++, in C# you can\'t overload the assignment operator.

I\'m doing a custom Number class for arithmetic operations with very large numbers and I want it to h

Answer 1

An earlier post suggested this:

public static implicit operator Foo(string normalString) { }

I tried this approach... but to make it work you need this:

public static implicit operator Foo(Foo original) { }

and the compiler won't let you have an implicit conversion function from your exact type, nor from any base type of yourself. That makes sense since it would be a backdoor way of overriding the assignment operator, which C# doesn't want to allow.

Answer 2

Maybe what you're looking for can be solved using C# accessors.

http://msdn.microsoft.com/en-us/library/aa287786(v=vs.71).aspx

Answer 3

Here is a solution that worked for myself :

public class MyTestClass
{
   private int a;
   private string str;

   public MyTestClass()
   {
      a = 0;
      str = null;
   }

   public MyTestClass(int a, string str)
   {
      this.a = a;
      this.str = str;
   }

   public MyTestClass Clone
   {
      get
      {
         return new MyTestClass(this.a, this.str);
      }
   }
}

Somewhere else in the code :

MyTestClass test1 = new MyTestClass(5, "Cat");
MyTestClass test2 = test1.Clone;

Answer 4

you can use the 'implicit' keyword to create an overload for the assignment:

Suppose you have a type like Foo, that you feel is implicitly convertable from a string. You would write the following static method in your Foo class:

public static implicit operator Foo(string normalString)
{
    //write your code here to go from string to Foo and return the new Foo.
}

Having done that, you can then use the following in your code:

Foo x = "whatever";

Answer 5

You won't be able to work around it having the C++ look, since a = b; has other semantics in C++ than in C#. In C#, a = b; makes a point to the same object like b. In C++, a = b changes the content of a. Both has their ups and downs. It's like you do

MyType * a = new MyType();
MyType * b = new MyType(); 
a = b; /* only exchange pointers. will not change any content */

In C++ (it will lose the reference to the first object, and create a memory leak. But let's ignore that here). You cannot overload the assign operator in C++ for that either.

The workaround is easy:

MyType a = new MyType();
MyType b = new MyType();

// instead of a = b
a.Assign(b);

Disclaimer: I'm not a C# developer

You could create a write-only-property like this. then do a.Self = b; above.

public MyType Self {
    set {
        /* copy content of value to this */
        this.Assign(value);
    }
}

Now, this is not good. Since it violates the principle-of-least-surprise (POLS). One wouldn't expect a to change if one does a.Self = b;

Answer 6

It's still not at all clear to me that you really need this. Either:

• Your Number type should be a struct (which is probable - numbers are the most common example of structs). Note that all the types you want your type to act like (int, decimal etc) are structs.

or:

• Your Number type should be immutable, making every mutation operation return a new instance, in which case you don't need the data to be copied on assignment anyway. (In fact, your type should be immutable whether or not it's a struct. Mutable structs are evil, and a number certainly shouldn't be a mutable reference type.)