Question

public static MyClass operator++(MyClass op)
{
    MyClass result = new MyClass(); // MyClass() x=y=z=0; 

    result.x = op.x + 1;
    result.y = op.y + 1;
    result.z = op.z + 1;

    return result:
}

//...

public void Main()
{
    MyClass c = new MyClass();
    MyClass b = new MyClass(1,2,3); //ctor x = 1, ...
    c = b++;
}

The question is why variable b going to change?
because result.x = op.x + 1; shouldn’t change op.x
result actually is c is (1,2,3) b is (2,3,4)
I don’t understand why not c is (2,3,4) and b is (1,2,3)

Answer 1

You say, “because result.x = op.x + 1 shouldn’t change op.x.” But ++ is not the same as + 1. It is the standard postfix increment operator which increments the operand b by using your custom overload — it assigns the new value to b immediately after passing the previous value to c. If it were the prefix operator (++b), it would have incremented before passing the value to c.

Without any overloads at all, the code:

b++

Is equivalent to:

b = b + 1

And

c = b++ 

Is equivalent to:

c = b;
b = b + 1;

Now using your overload, it’s more equivalent to:

c = b;
b = call_your_++_overload(b);