Why does std::get_if(std::variant) take pointer as an argument instead of reference?
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Victor Dyachenko
Jul 28, 2016, 4:54:53 AM7/28/16
to ISO C++ Standard - Discussion
From the current draft:
Why the arguments are pointers? It is not efficient. The implementation has every time to check if the argument is not
This functions can be class members or take a reference as an argument. What are reasons for suboptimal design like this? Just mimic
template <class T, class... Types>
std::add_pointer_t<T> get_if(variant<Types...> *pv);
template <class T, class... Types>
std::add_pointer_t<const T> get_if(const variant<Types...> *pv);
Why the arguments are pointers? It is not efficient. The implementation has every time to check if the argument is not
nullptr. Does nullptr argument makes any sense at all?This functions can be class members or take a reference as an argument. What are reasons for suboptimal design like this? Just mimic
dynamic_castinterface?Victor Dyachenko
Oct 6, 2016, 3:54:53 AM10/6/16
to ISO C++ Standard - Discussion
From: Axel Naumann <Axel.Naumann (at) cern.ch>
Sent: Thursday, October 06, 2016 10:35 AM
To: Victor Dyachenko <victor.dyachenko (at) gmail.com>
Subject: Re: Fwd: Why does std::get_if(std::variant) take pointer as an
argument instead of reference?
Hi Victor,
I think the reason was pointer-in, pointer-out symmetry, which you could indeed
describe as "mimic dynamic_cast". I find that a strong reason.
The reduction in efficiency from the nullptr check is minimal given that the
function needs a runtime check for the index anyway. We could have made it a
pre-condition but then get_if couldn't be noexcept anymore.
The most performant and recommended interface is visitation.
Cheers, Axel
On 10/06/2016 09:25 AM, Victor Dyachenko wrote:
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