https://gcc.gnu.org/bugzilla/show_bug.cgi?id=98835
Bug ID: 98835
Summary: False positive -Wclass-memaccess with class with
ref-qualified copy-assignment operator
Product: gcc
Version: unknown
Status: UNCONFIRMED
Severity: normal
Priority: P3
Component: c++
Assignee: unassigned at gcc dot gnu.org
Reporter: jchl at arista dot com
Target Milestone: ---
Consider the following code:
#include <type_traits>
#include <cstring>
class Good {
public:
Good & operator=(Good const &) = default;
};
class Bad {
public:
Bad & operator=(Bad const &) & = default;
};
template<typename T>
void test() {
static_assert(std::is_trivially_copyable_v<T>);
T value1;
T value2;
std::memcpy(&value1, &value2, sizeof(T));
}
int main() {
test<Good>();
test<Bad>();
}
[See: https://godbolt.org/z/4vj9GT ]
GCC trunk on x86_64 incorrectly gives the following warning:
<source>: In instantiation of 'void test() [with T = Bad]':
<source>:24:15: required from here
<source>:19:16: warning: 'void* memcpy(void*, const void*, size_t)' writing
to an object of type 'class Bad' with no trivial copy-assignment; use
copy-assignment or copy-initialization instead [-Wclass-memaccess]
19 | std::memcpy(&value1, &value2, sizeof(T));
| ~~~~~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Since both Good and Bad have trivial copy-assignment operators and are
trivially copyable, both types are eligible to be memcpy'd; the
ref-qualification of the assignment operator shouldn't be relevant.
