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Subject: Re: [boost] [Fit] Formal review / Kris
From: Louis Dionne (ldionne.2_at_[hidden])
Date: 2016-03-21 17:00:42


Paul Fultz II wrote
> On Monday, March 21, 2016 at 8:34:16 AM UTC-5, Louis Dionne wrote:
>>
>> [...]
>>
>> I'm not sure I understand what you mean. I mean, it is expected that
>> the function is never called when the first one could be called, since
>> it's overload _linearly_, right?
>>
>
> Yes, but another way to achieve that is to use a ranked tag-dispatching.
> Like this:
>
> template
> <int N>
> struct rank : rank
> <N-1>
> {};
>
> template<>
> struct rank<0>
> {};
>
> // Call F
> template
> <class... Ts>
> auto foo_impl(Ts&&... xs, rank<1>)
> -> decltype(std::declval
> <F>
> ()(std::forward
> <Ts>
> (xs)...));
>
> // Call G
> template
> <class... Ts>
> auto foo_impl(Ts&&... xs, rank<0>)
> -> decltype(std::declval
> <G>
> ()(std::forward
> <Ts>
> (xs)...));
>
> template
> <class... Ts>
> auto foo_impl(Ts&&... xs)
> -> decltype(foo_impl(std::forward
> <Ts>
> (xs)..., rank<1>{}));
>
> So `G` will not be called if `F` is callable, however, the compiler will
> instantiate both `F` and `G` with this implementation. I used to use an
> implementation like this, but it is more costly at compile time.
> Furthermore,
> lazy instantiation is nice feature as well, especially when using
> constexpr,
> as the body of a constexpr function is always instantiated during
> substitution.
>

I'm very surprised to learn that the compiler will instantiate both F and G
in the code you posted above. Can't the compiler determine that the first
`foo_impl` should be chosen if `decltype(std::declval<F>()(...))`does not
SFINAE out because rank<1> is a better match than rank<0>, and this
without ever looking at `std::declval<G>()(...)`?

Anyway, IIUC, `hana::overload_linearly` has this problem, so I'll change it.
Thanks for the heads up.

Louis

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