Re: [boost] Variadic append for std::string

Subject: Re: [boost] Variadic append for std::string
From: Richard Hodges (hodges.r_at_[hidden])
Date: 2017-01-18 03:53:08

That's pretty straightforward with another overload:

auto& s = join(to(y), separator(", "), "A", "b", 42);

where to(y) is something like

template<String>
struct to_existing_type<String> {
    String& get() { return s_; }
    String s_;
};

template<class String>
auto to(String& s)
{
  return to_existing_type<S>(s);
}

With a bit of template unwrapping, we could imagine something like this:

    join(to(x), 2, 3, to(y), "foo", "bar", create(), "baz", 42);

which would return a tuple:

std::tuple<std::string&, std::string&, std::string>

in c++17 this would allow:

auto&& [x, y, z] = join(to(x), 2, 3, to(y), "foo", "bar", create(), "baz",
42);

But this maybe taking it a bit far... What do you think?

On 18 January 2017 at 09:06, Olaf van der Spek <ml_at_[hidden]> wrote:

> On Mon, Jan 16, 2017 at 11:41 AM, Richard Hodges <hodges.r_at_[hidden]>
> wrote:
>
> > Sorry to chime in so late in the discussion.
> >
> > What about a syntax similar to this?
> >
> > int main()
> > {
> >     auto s = join("Hello ", ", World.", " The hex for ", 58, " is ",
> > std::hex, 58);
> >     std::cout << s << std::endl;
> >
> >     s = join(separator(" : "), "a", "b", std::hex, 200 ,
> > std::quoted("banana"));
> >     std::cout << s << std::endl;
> >
> > }
> >
> > Which would produce the following output:
> >
> > Hello , World. The hex for 58 is 3a
> > a : b : c8 : “banana"
> >
>
> The syntax is fine but it's missing an appending variant, like append(s,
> "A", "B", 42);
> This variant is important as it (also) allows you to reuse existing
> storage.
>
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