Nothing Orders Memory

Both atomics start at 0.

std::atomic<int> x{0}, y{0};

// thread A
x.store(1, std::memory_order_relaxed);
y.store(1, std::memory_order_relaxed);

// thread B
int a = y.load(std::memory_order_relaxed);   // reads 1
int b = x.load(std::memory_order_relaxed);   // reads ?

Thread B observes a == 1. Is b == 0 a legal outcome under the C++ memory model?

Yes, but only if x and y land on different cache lines; sharing a line would force them to become visible together.

Yes: relaxed orders no memory beyond making each individual operation atomic, so B may see A's store to y without seeing its earlier store to x.

No: the two stores are program-ordered in A, and atomic operations preserve program order as seen by every thread.

No on x86 and yes on ARM, so the answer depends on the target rather than on the standard.