On Friday, 29 June 2012 at 20:13:58 UTC, Jonathan M Davis wrote:
On Friday, June 29, 2012 21:54:42 Zhenya wrote:
I see, I just thought that opCall @ property equivalent
opAssign
and wanted to check it out, and now I would be interested to
understand why it is not
I don't see how you could think that it _would_ be. The _only_
way that opCall
can be invoked is by using the variable as if it were a
function.
struct X
{
bool _x;
A opCall(bool x) {_x = x;return this;}
}
x(false);
Without those parens, the compiler has no idea that you're
trying to use
opCall. opCall is specifically for being able to call a
variable as if it were
a function. By using =, you're making the compiler look for
opAssign
x = false;
because that's the function for overloading =. You're only
going to be able to
make a function a property when it would be used as a function
if it wasn't
declared as a property, and neither opCall or opAssign is used
as a function
(e.g x.opCall(), x.opAssign()). They're both overloading
operators. @property
is specifically for making a function act as if it were a
variable, and
overloaded operators aren't used as either functions or
variables. They
overload _operators_.
Off the top of my head, the _only_ overloaded operator that I
can think of
where it would make any sense to declare it @property would be
opDispatch,
because it's replacing function calls, but there are problems
with that,
because you can't have two opDispatches, so you can't use it
for both
properties and normal functions.
- Jonathan M Davis
Thank you,I understood.
