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Subject: Re: [std-proposals] Unify '.' and '->' operator behavior
From: J Decker (d3ck0r_at_[hidden])
Date: 2020-02-15 22:45:31


On Sat, Feb 15, 2020 at 8:40 PM Arthur O'Dwyer <arthur.j.odwyer_at_[hidden]>
wrote:

>
> On Sat, Feb 15, 2020 at 11:10 PM J Decker <d3ck0r_at_[hidden]> wrote:
>
>> On Sat, Feb 15, 2020 at 7:07 PM Arthur O'Dwyer <arthur.j.odwyer_at_[hidden]>
>> wrote:
>>
>>> Hi Jim,
>>>
>>> Sadly your idea doesn't work in C++, because C++ has *member functions*.
>>> That is, we can write things like this:
>>>
>>> struct Widget {
>>> void reset(); // reset the widget
>>> };
>>> [...]
>>> std::unique_ptr<Widget> p = std::make_unique<Widget>();
>>> p->reset(); // reset the pointed-to widget by calling Widget::reset()
>>> member function
>>> p.reset(); // reset the pointer's own value to nullptr
>>>
>>>
>> But, isn't the .reset() implemented on the unique_ptr<> template as a
>> operator. override? which takes precedence over the default behavior...
>>
>
> No, C++ doesn't permit overloading the "." operator at all.
> It's just that `p.reset()` calls `unique_ptr<Widget>::reset`,
> and `p->reset()` calls `unique_ptr<Widget>::operator->` followed by
> `Widget::reset`.
>
> I think you'll have to learn more about how C++ currently works, before
> working on proposals to change it.
>
A appricate you leveraing my acknowledgment of limitation as a dismissal;
it's more a factor that I need to learn the right way to say it.

It doesn't look like that's either a instance of a class/struct/union or a
pointer to a class/struct/union, but rather is a unique pointer to a
class/struct/union.
Also would have no effect on references... I would bet that the
modification of the compiler to accomplish a compatible, non-breaking
change is either a small addition to 'if left hand is a '*' pointer or a
instance of a struct/class/union.

Because the cases that matter are all error cases now.

>
> –Arthur
>



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