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From: "redi at gcc dot gnu.org" <gcc-bugzilla@gcc.gnu.org>
To: gcc-bugs@gcc.gnu.org
Subject: [Bug libstdc++/108645] Change in behavior, std::accumulate doesn't
 always work as expected in C++20 builds
Date: Fri, 03 Feb 2023 10:01:33 +0000
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https://gcc.gnu.org/bugzilla/show_bug.cgi?id=3D108645

--- Comment #4 from Jonathan Wakely <redi at gcc dot gnu.org> ---
(In reply to Jonathan Wakely from comment #3)
> 	  const auto __size =3D __lhs.size() + __rhs.size();
> 	  if (__size > __lhs.capacity() && __size <=3D __rhs.capacity())
> 	    return std::move(__rhs.insert(0, __lhs));

It would be possible to change that logic to:

          const auto __size =3D __lhs.size() + __rhs.size();
          if (__size <=3D __rhs.capacity() && __rhs.capacity() >
__lhs.capacity())
            return std::move(__rhs.insert(0, __lhs));

That way if both strings have sufficient capacity, we would return the larg=
er
of the two capacities. In general, the returned string is more likely to be=
 the
one that gets reused, and this way it would be more likely to have addition=
al
spare capacity.

In some usage patterns this would be a pessimization, because it would tend=
 to
coalesce all the larger capacities into one string, leaving smaller, less
useful strings behind. Also, inserting into the beginning of the RHS takes =
more
work than simply appending to the end of the LHS. So I don't think it's cle=
ar
that this would be a definite improvement.

It wouldn't change anything in your example anyway, because there is no
dynamically allocated memory anywhere except the accumulator value, which is
always the LHS.=