Overloading increment operator

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I was just working on overloading the increment operator. I was hoping to have both pre-increment and post-increment. Unfortunately I see that the same method is called regardless of whether the operator is in pre- or post-position. So I checked the spec and was surprised to find:

"
Note that the operator returns the value produced by adding 1 to the operand, just like the postfix increment and decrement operators (§7.5.9), and the prefix increment and decrement operators (§7.6.5). Unlike in C++, this method need not, and, in fact, must not, modify the value of its operand directly.
"

What the heck good is an increment operator that doesn't increment?! If I want to do x += 1 I have the addition operator for that and it works just fine. I want x++ and ++x just like with the built-in numeric types; x = x++ or x = ++x would be ridiculous, but it appears that's all we're allowed.

Am I missing something? Does anyone have a work-around?

P.S. Or is just that it's 00:34 and I should go get some sleep? It seems to be working as expected somehow? [confused]

modified on Tuesday, February 24, 2009 2:33 AM

Posted 23-Feb-09 20:27pm

PIEBALDconsult

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DaveyM69 · Accepted Answer · 2009-02-23T22:51:00

When calling the increment operator, the object is assigned a new reference, that reference has the new values.

Example. We have a class IntWrapper

public class IntWrapper
{
    public int IntValue
    {
        get;
        set;
    }
}

and we want to use it like this

IntWrapper a = new IntWrapper();
IntWrapper b = a;
a++;
Console.WriteLine("a = {0}, b = {1}", a.IntValue, b.IntValue);

Here are two possible implementations for the increment operator overload in our class, the second is the correct way

public static IntWrapper operator ++(IntWrapper instance)
{
    // WRONG! a and b will be incremented
    instance.IntValue++;
    return instance;
}

public static IntWrapper operator ++(IntWrapper instance)
{
    // CORRECT! Only a will be incremented
    IntWrapper result = new IntWrapper();
    result.IntValue = instance.IntValue + 1;
    return result;
}