reinterpret_cast memory not modified - c++

Question

I tried to modify an object of A indirectly by allocating 8 bytes using std::allocator. I used <int> specifically, hoping it will return aligned memory for 2 integers. Then I modified an object of A.
Why do I only see A.x=9 in the bitset (towards the end 1001), what happened to A.y=10? Do I need to align the raw pointer returned by std::allocator?
I was expecting the output below:
0000000000000000000000000000100100000000000000000000000000001010
before=0000000000000000000000000000000000000000000000000000000000000000
---------------------
sizeof(A)=8
alignof(A)=4
x: 9 y: 10
---------------------
x: 9 y: 10
after=0000000000000000000000000000000000000000000000000000000000001001
#include <bitset>
#include <iostream>
struct A {
int x;
int y;
A() = delete;
friend std::ostream &operator<<(std::ostream &os, const A &a) {
os << "x: " << a.x << " y: " << a.y;
return os;
}
};
int main(int argc, char **argv) {
std::allocator<int> alloc;
auto raw = alloc.allocate(2);
std::bitset<64> before(*raw);
before.reset();
std::cout << "before=" << before << "\n";
std::cout << "---------------------\n";
std::cout << "sizeof(A)=" << sizeof(A) << "\n";
std::cout << "alignof(A)=" << alignof(A) << "\n";
auto a = reinterpret_cast<A *>(raw);
a->x = 9;
a->y = 10;
std::cout << *a << "\n";
std::cout << "---------------------\n";
std::bitset<64> after(*raw);
std::cout << *a << "\n";
std::cout << "after=" << after << "\n";
return 0;
}

Answer 1

I actually don't see any undefined behviour here.
The actual problem is that you seem to assume that before(*raw) creates a bitset with a snapshot of that memory region. But that is not what happens.
See: https://en.cppreference.com/w/cpp/utility/bitset/bitset
(you call the unsigned long long constructor)
If you simply write (before after):
unsigned long long raw_converted_to_long = *raw;
You see the value is 9 - this is the reason why the bitset is filled with 9.