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Re: [std-proposals] Named auto

From: Oleksandr Koval <oleksandr.koval.dev_at_[hidden]>
Date: Sat, 1 Oct 2022 13:14:35 +0300
That version of Edward looks 100% in line with current syntax but it's too
verbose for me. It requires too much typing even for simple cases.
In the realm of universal template parameters I'd like to have:

auto:[V<T>] v = std::vector<int>{}; // V = std::vector, T = int

void f(const auto:[V]& v);

auto:[T1, T2] [key, value] = get_pair();

Why should we type those `template, class, int` inside [] if it's clear
that it is a placeholder for a template parameter? Taking into account that
this feature is supposed just to provide access to deduced type, I see no
reason to allow it to constrain a kind of template parameters
(type/non-type/template) or parameters themselves with concepts.

Compiler can transform the above into:

template<template auto V, template auto T>
void deduce_v(V<T>);

On Fri, Sep 30, 2022 at 8:31 PM Edward Catmur via Std-Proposals <
std-proposals_at_[hidden]> wrote:

> On Fri, 30 Sept 2022 at 17:38, Lénárd Szolnoki via Std-Proposals <
> std-proposals_at_[hidden]> wrote:
>
>> Hi,
>>
>> One more syntax proposal that extends P1985's unkiversal template
>> parameter: `template auto`
>>
>> 1. allow `template auto` in any deduced context, for example
>> `vector<template auto> vec = get_vec();`
>> 2. use the syntax `template<class T> auto` to introduce the name `T`.
>>
>
> Again, syntax. How on earth do you parse this? `template auto` in P1985's
> usage is fine, since it's just alternate for *template-parameter* (or
> *type-parameter*), with an alternation on the token `template` with
> *template-head* resolved immediately at the next token (`auto` vs. `<`).
> `vector<auto>` is similarly fine, since `auto` is already a
> *simple-type-specifier*, so that's just a semantic relaxation.
>
> template<class T> auto x = foo()
>> vector<template<class V> auto> v = get_vec()
>>
>> Some extra stuff:
>> * Introducing a name is optional, you can use `template<class> auto` for
>> a type placeholder, instead of a universal one
>> * you can have nttp or template placeholders too
>>
>> std::array<string, template<size_t arr_size> auto> arr = get_arr();
>>
>> A bit verbose compared to my previously proposed auto<class T>. Honestly,
>> I like that more, but I could live with this.
>>
>> I definitely wouldn't want a syntax where you are required to immediately
>> repeat the introduced name.
>>
>
> Repeating the name means that you can specify its kind (type, class
> template, value, etc.) and easily see which names are introduced into scope.
>
> I really don't see the necessity to golf everything; C++ is already
> notoriously hard to parse and you want to make it even harder?
>
>
>> Cheers,
>> Lénárd
>>
>> On 30 September 2022 15:01:58 BST, Edward Catmur via Std-Proposals <
>> std-proposals_at_[hidden]> wrote:
>>>
>>>
>>>
>>> On Fri, 30 Sept 2022 at 14:19, Oleksandr Koval <
>>> oleksandr.koval.dev_at_[hidden]> wrote:
>>>
>>>> Sorry if this was already discussed, I like `auto<class T>` syntax but
>>>> the problem I see is this:
>>>>
>>>> std::vector<int> f();
>>>> auto<class T> v = f(); // should T be `std::vector<int>` or just `int`?
>>>>
>>>
>>> That syntax would work as a 2-step process: first you declare entities,
>>> then deduce them. The declaration and deduction should be pretty much the
>>> same as for (non-terse) function templates.
>>>
>>> So T is repeated:
>>>
>>> auto<class T> T v = f();
>>>
>>> or
>>>
>>> auto<class T> std::vector<T> v = f();
>>>
>>> or
>>>
>>> auto<template<class> class TT, class T> TT<T> v = f();
>>>
>>>
>>> It’s a bit confusing when you see them side by side. Because of that I
>>>> don’t think that using angle brackets is a good idea (at least I can’t find
>>>> any form which is not confusing). We need a new syntax which will clearly
>>>> show that typename(s) is closely related to `auto` itself. For example:
>>>>
>>>> auto:[class T] v = f();
>>>> auto:[class T1, classT2] [key, value] = get_pair();
>>>>
>>>> On Fri, Sep 30, 2022 at 4:07 AM Edward Catmur via Std-Proposals <
>>>> std-proposals_at_[hidden]> wrote:
>>>>
>>>>>
>>>>>
>>>>> On Thu, 29 Sept 2022 at 21:44, Arthur O'Dwyer <
>>>>> arthur.j.odwyer_at_[hidden]> wrote:
>>>>>
>>>>>> On Thu, Sep 29, 2022 at 3:57 PM Lénárd Szolnoki via Std-Proposals <
>>>>>> std-proposals_at_[hidden]> wrote:
>>>>>>
>>>>>>> On 29 September 2022 20:25:06 BST, Edward Catmur <
>>>>>>> ecatmur_at_[hidden]> wrote:
>>>>>>> >On Thu, 29 Sept 2022 at 18:15, Lénárd Szolnoki via Std-Proposals <
>>>>>>> std-proposals_at_[hidden]> wrote:
>>>>>>> >
>>>>>>> >> I think this could be addressed by two distinct proposals.
>>>>>>> >>
>>>>>>> >> 1. allow placeholders to appear in any deduced context
>>>>>>> (std::vector<auto>)
>>>>>>> >> 2. allow a placeholder to introduce a name (auto<class T>,
>>>>>>> auto<int i>
>>>>>>> >> might appear as a deduced nttp)
>>>>>>> >>
>>>>>>> >> Then you can have your vector<auto<class T>> to deduce the value
>>>>>>> type and
>>>>>>> >> introduce the name T for that.
>>>>>>> >
>>>>>>> >std::vector<auto> is problematic, because elsewhere auto means a
>>>>>>> value of
>>>>>>> >unconstrained type (e.g. in template<auto>).
>>>>>>>
>>>>>>> I disagree. auto is a placeholder for the type of the non-type
>>>>>>> parameter. If you don't omit the name, then that name refers to the value.
>>>>>>>
>>>>>>
>>>>>> FWIW, I tend to agree with Lénárd here: `auto *p,
>>>>>> std::unique_ptr<auto> q` seems quite reasonable to me.
>>>>>>
>>>>>
>>>>> Ye-esss; looking at the grammar; `auto` is always a
>>>>> placeholder-type-specifier, which is a simple-type-specifier, so it makes
>>>>> sense that it takes the place of a concrete (or inferred) type.
>>>>>
>>>>> So Lénárd, I apologise; you're correct that in these contexts `auto`
>>>>> designates a type, not a value. I'm still trying to get my head round
>>>>> "template<auto> int f();", but the syntax is clear.
>>>>>
>>>>> *However*, I foresee practical problems with allowing auto *anywhere*
>>>>>> in a declaration. Consider
>>>>>> template<class T> void f(T t); // since C++98
>>>>>> template<class T> void g(decltype(T(1)) t); // since C++11
>>>>>> void f(auto t); // since C++20, equivalent to f #1
>>>>>> void g(decltype(auto(1)) t); // since C++23, *not* equivalent
>>>>>> to g #1
>>>>>>
>>>>>> Now, the `T` parameter to `g` is not deducible, so "obviously" the
>>>>>> `auto` in g #2 doesn't mean the same thing as the `auto` in f #2. But are
>>>>>> you sure we can teach that to the computer?
>>>>>>
>>>>>> (Background: I teach that `auto` since C++14 has (like most C++
>>>>>> keywords) had two meanings: concrete type *inference*, as in auto x
>>>>>> = 1, and templatey type *deduction*, as in [](auto x){}. The
>>>>>> physical mechanisms behind, and consequences of, these two usages of `auto`
>>>>>> are vastly different, although their human-level meaning is similar: "I
>>>>>> don't want to bother with types; compiler, please figure it out." So if you
>>>>>> see me talking about "inference" versus "deduction," or "the first meaning
>>>>>> of auto" versus "the second meaning of auto," that's what I'm talking
>>>>>> about.)
>>>>>>
>>>>>
>>>>> The deduction process is ultimately the same; both end up in
>>>>> temp.deduct.call. In an abbreviated function template or generic lambda,
>>>>> the deduction can be overridden by explicit template argument, but that's a
>>>>> relatively minor effect. And syntactically, they're the same; a
>>>>> decl-specifier of a decl-specifier-seq (of a function parameter or variable
>>>>> declaration); `auto` in a function return type or trailing return type has
>>>>> less in common, though I'd suppose you'd class that as inference. Still, I
>>>>> guess it's OK to teach it that way.
>>>>>
>>>>> It would certainly *not* be reasonable to have a rule like "Try to
>>>>>> interpret every `auto` as deduction, but if that would result in a
>>>>>> non-deducible template parameter, then backtrack and assume it's inference
>>>>>> instead." It would be reasonable to have a rule that boils down to "Inside
>>>>>> the operand of a decltype or sizeof or array bound, the `auto` always means
>>>>>> inference not deduction." But is `g` above the *only* problem case?
>>>>>> Are there other corner cases where `auto` is already legal today, and/or we
>>>>>> wouldn't want it to mean a template parameter?
>>>>>> void f(A<auto(int())>); // is this concrete A<0> or templatey
>>>>>> A<T(int())>? I guess the type of A will disambiguate...
>>>>>>
>>>>>
>>>>> Currently a placeholder can appear as the decl-specifier of a function
>>>>> parameter, variable declaration, or template parameter or, as exactly one
>>>>> simple-type-specifier of a return or trailing return type,
>>>>> conversion function id, new expression's type id, or (`auto` only) as the
>>>>> type specifier of a functional cast. If these were relaxed, I think we'd
>>>>> probably be OK; you just go through inventing extra type template
>>>>> parameters and perform deduction as usual.
>>>>>
>>>>> Also I don't think ambiguity *can* be an issue, or it would be
>>>>> already; there must be disambiguators in the syntax. Indeed, I have a
>>>>> strong suspicion that your `A<auto(int())>` is a most vexing parse.
>>>>>
>>>>> Also consider that while
>>>>>> void h1(std::vector<auto, auto> v);
>>>>>> would work fine with Lénárd's proposed syntax,
>>>>>> void h2(std::array<auto, ??> a);
>>>>>> would not. That does seem mildly problematic. However, maybe it's
>>>>>> consistent with C++20, which permits
>>>>>> template<class T> concept Integral = true;
>>>>>> template<Integral T> void ij(); // since C++20
>>>>>> but not
>>>>>> template<int N> concept Odd = true;
>>>>>> template<Odd N> void ij(); // error: Odd does not constrain a
>>>>>> type
>>>>>>
>>>>>
>>>>> Yes, fair enough; this does conform to the grammar.
>>>>> --
>>>>> Std-Proposals mailing list
>>>>> Std-Proposals_at_[hidden]
>>>>> https://lists.isocpp.org/mailman/listinfo.cgi/std-proposals
>>>>>
>>>>
>>>>
>>>> --
>>>> Regards,
>>>> Oleksandr Koval.
>>>>
>>> --
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>>
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>


-- 
Regards,
Oleksandr Koval.

Received on 2022-10-01 10:14:49