On Wed, Mar 4, 2015 at 10:17 AM, Bjarne Stroustrup
<bjarne@stroustrup.com> wrote:
>
>
> On 3/3/2015 10:45 PM, Gabriel Dos Reis wrote:
>>
>> Should we declare "auto f(auto) -> auto" an excess in sophism?
>
>
>
> and those two "auto"s can denote different and unrelated types?
>

yes - the first 'auto' is a trailing return type signifier, the second
'auto' is an abbreviated template signifier (adds a template parameter
to the function template's template parameter list) and is deduced
from the supplied argument, the third auto is the placeholder for
return type deduction and is deduced from a return statement within
the body of the function - it  does not add a template parameter to
the function template's template parameter list - and is quite
different from the second auto in that way.

>
>>
>> -- Gaby
>>
>>
>>> On Mar 3, 2015, at 6:39 PM, Faisal Vali <faisalv@gmail.com> wrote:
>>>
>>> In the hopes of soliciting constructive feedback, if anyone has the
>>> time or the interest to play with a patched up Clang that implements
>>> enhanced auto deduction & abbreviated template syntax from the
>>> concepts ts, here it is:
>>>
>>> https://github.com/faisalv/clang/tree/enhanced-auto-c%2B%2B1z .
>>>
>>> The last line of the README file tells you the subversion revision of
>>> LLVM to use.
>>>
>>> For examples of test cases that successfully compile, please refer to:
>>> --
>>> https://github.com/faisalv/clang/blob/enhanced-auto-c%2B%2B1z/test/CXX/auto/cxx1z-auto-vars-generalized.cpp
>>> --
>>> https://github.com/faisalv/clang/blob/enhanced-auto-c%2B%2B1z/test/CXX/auto/cxx1z-abbreviated-templates.cpp
>>>
>>> I would certainly appreciate the feedback!
>>>
>>> 1) Enhanced-Auto Deduction:
>>>
>>> pair<auto...> f() { return make_pair([] { }, [] { }); }
>>> vector<auto> v = vector<int>();
>>>
>>> 2) Abbreviated Template Syntax:
>>> void f(auto) <=> template<class T> void f(T);
>>>
>>>
>>> A few, perhaps not so obvious or interesting, technicalities:
>>>
>>> a)  The equivalence of templates when it comes to trailing return
>>> types in function pointer parameters is based on the order of the
>>> 'auto' that signifies a placeholder, not just the appearance of an
>>> auto in the declaration-specifier of a parameter:
>>>
>>>     template<class R, class P> void f( R(P) );  // #1
>>>     template<class P, class R> void f(  R(P) );  // #2 (order of
>>> templ-params flipped)
>>>
>>>     template<class R, class P> void f( auto (P)->R); // equivalent to
>>> #1, not abbreviated.
>>>     void f(auto(auto));  // equivalent to #1
>>>     void f(auto (auto) -> auto); // equivalent to #2
>>>                                                // the trailing return
>>> type auto
>>>                                                // identifies a
>>> template parameter
>>>     template<class R> void f(R(auto); // equivalent to #1
>>>     template<class P> void f(auto(P)); // equivalent to #2
>>>
>>> b) variadic auto
>>>     Once an ellipsis is seen as part of the declarator, all contained
>>> auto placeholders get transformed into parameter packs.
>>>     void f(auto (*...)(auto) -> std::tuple<auto, std::pair<auto,
>>> auto>...>);
>>> Note there are 4 deducible template parameter packs above.
>>>
>>> c) multi abbreviated template declarations
>>>    void f(auto), g(auto);
>>> are allowed - the above declares two function templates.
>>>
>>> d) template aliases desugar their types *after* each auto has been
>>> identified as a placeholder (and so must be deduced)
>>>
>>>     template<class T> using Int = int;
>>>     Int<auto> i = 3; // Is an error, auto must be deducible.
>>>
>>> e) generic lambda's as default arguments work with abbreviated template
>>> syntax:
>>>   void f(int (*)(int) = [](auto a) { return a; }, auto b =
>>> decltype(b){});
>>>   f<float*>(); // OK
>>>
>>> f) variable templates and enhanced auto deduction interact as expected:
>>>     template<class T> pair<auto...> (*vfp)(auto...) = [](T t1, T* t2)
>>> { return make_pair(t1, t2); };
>>>
>>> Thanks!
>>> Faisal Vali
>>>
>>> --
>>>
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