Hi
2024. március 13., szerda 12:43 keltezéssel, Jonathan Wakely
írta:
> On Mon, 11 Mar 2024 at 23:36, Barnabás Pőcze wrote:
> >
> > Previously, calling erase(key) on both std::map and std::set
> > would execute that same code that std::multi{map,set} would.
> > However, doing that is
Hi
2024. március 13., szerda 12:43 keltezéssel, Jonathan Wakely
írta:
> On Mon, 11 Mar 2024 at 23:36, Barnabás Pőcze wrote:
> >
> > Previously, calling erase(key) on both std::map and std::set
> > would execute that same code that std::multi{map,set} would.
> > However, doing that is
On Mon, 11 Mar 2024 at 23:36, Barnabás Pőcze wrote:
>
> Previously, calling erase(key) on both std::map and std::set
> would execute that same code that std::multi{map,set} would.
> However, doing that is unnecessary because std::{map,set}
> guarantee that all elements are unique.
>
> It is
Previously, calling erase(key) on both std::map and std::set
would execute that same code that std::multi{map,set} would.
However, doing that is unnecessary because std::{map,set}
guarantee that all elements are unique.
It is reasonable to expect that erase(key) is equivalent
or better than:
