Why are opCall's not implicitely assignable?

[Karen Lanrap]

Jarrett Billingsley wrote:
> keep in mind that you're not_really_ assigning anything to f,
> it's just sugar. 

Yes, and that is the problem with this sugar. It dilutes what an 
assignment _really_ is---if there is a definition for a _real_ 
assignment at all.

Assume there is a definition for what an assignment in D _really_ is.

If you then claim that "f=2" is not a _real_ assignment, you have to 
define what it is other than a _real_ assignment.

If you then say it is not an overloading of the assignment operator 
you have to give a reason, why it differs from overloading the 
assignment operator.

I do not see any difference. So please explain. I am posting in the 
learn group with the aim to understand the language. If I would have 
a clear opinion on this matter I would post in the main D group.