[cfe-dev] Help getting started

[John Bytheway]

On 02/01/12 17:57, Dave Abrahams wrote:
> 
> on Mon Jan 02 2012, David Blaikie <dblaikie-Re5JQEeQqe8AvxtiuMwx3w-AT-public.gmane.org> wrote:
> 
>> Sorry, was a bit tired I guess. I meant polymorphic lambdas (template
>> type parameters) - I wasn't intending to suggest some other construct
>> you hadn't suggested.
>>
>> I'm still not sure I follow the ambiguity resolution. Are you saying
>> this named function template syntax would diverge from lambdas? (in the
>> sense that a single identifier specifies an unnamed parameter in a
>> lambda, but an untyped on in this syntax you're proposing)
> 
> It's not surprising that you didn't follow, because what I said didn't
> make any sense (sorry).  I guess I see the problem you're referencing, and I
> don't know what the solution should be yet.

I think you're conflating two new language features that ought to be
considered in isolation:

1. Automatic polymorphization of parameter types.

2. Automatic deduction of return types for non-lambda functions.

Thinking about how to add feature 1 in isolation, in a way that works
for all functions, function templates, and lambda functions, without the
above-mentioned ambiguity between parameters without names and those
without types, the solution that first occurs to me is the one suggested
by Robert Frunzke in a comment on your C++Next article: use "auto" in
place of the type name rather than omitting it.

So we would write

auto min(auto x, auto y)->decltype(x < y ? x : y)
{ return x < y ? x : y; }

which is equivalent to

template <class T, class U>
auto min(T x, U y)->decltype(x < y ? x : y)
{ return x < y ? x : y; }

And then, independently of that, we also allow the "->decltype..." to be
omitted by following the same rules as for lambdas.

Indeed, one might even want to omit both the type and the name of the
parameter, e.g.

void ignore(auto&&...) {}

This also expands more easily to more general pattern-matching, such as:

void f(std::vector<auto> const&);

which can fill in the type in the same way that function templates do.

Though of course it's less clear what to do with:

void f(std::pair<auto, auto> const&);

(my instinct is that the two autos should spawn independent template
parameters, but one might argue that they should be the same type)

But this is now getting rather off-topic...

John Bytheway