https://gcc.gnu.org/bugzilla/show_bug.cgi?id=84855
--- Comment #3 from Jonathan Wakely <redi at gcc dot gnu.org> ---
Because 'a' and 'b' are not local variables, they refer directly to subobjects
of the tuple-like object, not separate variables. So if you didn't make a copy
of the foobar then 'a' would refer to f.s and b would refer to f.v and you
would get very surprising results if you modify them:
#include <vector>
#include <string>
#include <cassert>
struct foobar
{
std::string s{"original"};
std::vector<int> v;
};
int main()
{
foobar f;
assert(f.s == "original");
auto [ s, v ] = f;
s = "changed";
assert(f.s == "original"); //fails if &s == &f.s
}
You have to make a copy of f, and then &s == &__copy_of_f.s
