Re: [boost] [range] treating N-ary function with N-1 default args as unary

Subject: Re: [boost] [range] treating N-ary function with N-1 default args as unary
From: Stewart, Robert (Robert.Stewart_at_[hidden])
Date: 2012-07-03 10:28:58

Mathias Gaunard wrote:
> On 01/07/2012 01:18, Vicente J. Botet Escriba wrote:
>
> > Could a lambda help on this context?
> >
> > int main()
> > {
> >      int A[5];
> >      A | boost::adaptors::transformed(
> >         [](int a) { return foo(a);});
> > }
>
> The OP already pointed this one out. It's not satisfying
> because it requires specifying the type of the first argument.
>
> Something you could do in C++11
>
> int main()
> {
>       struct foo_type
>       {
>            template<class Sig>
>            struct result;
>
>            template<class This, class A0>
>            struct result<This(A0)>
>            {
>                 typedef decltype(foo(declval<A0>())) type;
>            };
>
>            template<class A0>
>            typename result<foo_(A0&&)>::type operator()(
                             ^^^^
That clearly is a typo and I'm not certain what you actually meant.

>               A0&& a0) const
>            {
>                 return foo(forward<A0>(a0));
>            }
>       };
>       foo_type foo_;
>       int A[5];
>       A | boost::adaptors::transformed(foo_);
> }
>
> The functor could be generated by a macro that would just
> take the function name and the number of arguments.

Wouldn't there need to be a different macro for each number of actual arguments?  That is, one for unary, one for binary, etc. or did you mean something else?

_____
Rob Stewart                           robert.stewart_at_[hidden]
Software Engineer                     using std::disclaimer;
Dev Tools & Components
Susquehanna International Group, LLP  http://www.sig.com

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