How can I implicitly convert another struct to my Type ?

Question

Hello,

As it is MyClass x = 120;, is it possible to create such a custom class? If so, how can I do that?

Answer 1

Not sure if this is what you want but you may get there by implementing the implicit operator: http://msdn.microsoft.com/en-us/library/z5z9kes2(VS.71).aspx

Answer 2

Create an implicit operator:

http://msdn.microsoft.com/en-us/library/z5z9kes2.aspx

For example:

public struct MyStruct // I assume this is what you meant, since you mention struct in your title, but use MyClass in your example. 
{
    public MyClass(int i) { val = i; }
    public int val;
    // ...other members

    // User-defined conversion from MyStruct to double
    public static implicit operator int(MyStruct i)
    {
        return i.val;
    }
    //  User-defined conversion from double to Digit
    public static implicit operator MyStruct(int i)
    {
        return new MyStruct(i);
    }
}

"Is this a good idea?" is debatable. Implicit conversions tend to break accepted standards for programmers; generally not a good idea. But if you're doing some large value library, for example, then it might be a good idea.

Answer 3

yes, here's a short example ...

  public struct MyCustomInteger
  {
     private int val;
     private bool isDef;
     public bool HasValue { get { return isDef; } } 
     public int Value { return val; } } 
     private MyCustomInteger() { }
     private MyCustomInteger(int intVal)
     { val = intVal; isDef = true; }
     public static MyCustomInteger Make(int intVal)
     { return new MyCustomInteger(intVal); }
     public static NullInt = new MyCustomInteger();

     public static explicit operator int (MyCustomInteger val)
     { 
         if (!HasValue) throw new ArgumentNullEception();
         return Value;
     }
     public static implicit operator MyCustomInteger (int val)
     {  return new MyCustomInteger(val); }
  }

Answer 4

It's generally considered a bad idea to use implicit operators, as they are, after all, implicit and run behind your back. Debugging code littered with operator overloads is a nightmare. That said, with something like this:

public class Complex
{
    public int Real { get; set; }
    public int Imaginary { get; set; }

    public static implicit operator Complex(int value)
    {
        Complex x = new Complex();
        x.Real = value;
        return x;
    }
}

you could use:

Complex complex = 10;

or you could ever overload the + operator

public static Complex operator +(Complex cmp, int value)
{
  Complex x = new Complex();
  x.Real = cmp.Real + value;
  x.Imaginary = cmp.Imaginary;
  return x;
 }

and use code like

complex +=5;