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From: "carlosgalvezp at gmail dot com" <gcc-bugzilla@gcc.gnu.org>
To: gcc-bugs@gcc.gnu.org
Subject: [Bug c++/107361] New: Why does -Wclass-memaccess require trivial
 types, instead of trivially-copyable types?
Date: Sun, 23 Oct 2022 07:22:03 +0000
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X-Bugzilla-Version: 13.0
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https://gcc.gnu.org/bugzilla/show_bug.cgi?id=3D107361

            Bug ID: 107361
           Summary: Why does -Wclass-memaccess require trivial types,
                    instead of trivially-copyable types?
           Product: gcc
           Version: 13.0
            Status: UNCONFIRMED
          Severity: normal
          Priority: P3
         Component: c++
          Assignee: unassigned at gcc dot gnu.org
          Reporter: carlosgalvezp at gmail dot com
  Target Milestone: ---

I find -Wclass-memaccess warns on this kind of code:

#include <cstring>

struct Foo
{
    int x{};
    char y{};
    int z{};
};

int main()
{
    Foo a;
    std::memset(&a, 0, sizeof(a));
}

All because Foo has default member initializers, making it a non-trivial ty=
pe.
It's still trivially-copyable, which is the requirement for std::memset.
Therefore I ask: why is the warning stricter than it needs to be?

Making sure structs are initialized is defensive programming, and consistent
with coding guidelines that require classes to initialize their data member=
s.=20

Also, there are cases where we cannot use the C++ initialization (Foo a =3D=
 {};).
For example, this class has implicit padding, and the C++ initialization wi=
ll
not initialize that. This is a problem for serializer/deserializer type of =
code
(copying/reading uninitialized padding bytes, valgrind will complain).=