从C++内存模型视角看，该Seqlock实现是否正确？

我阅读了这篇论文，发现其中图6的示例从C++内存模型角度来看并不正确。

atomic<unsigned> seq; // seqlock representation
// assume big enough to ignore overflow
atomic<int> data1, data2; // protected by seq

T reader() {
    int r1, r2;
    unsigned seq0, seq1;
    do {
        seq0 = seq.load(m_o_acquire);
        r1 = data1.load(m_o_relaxed);
        r2 = data2.load(m_o_relaxed);
        atomic_thread_fence(m_o_acquire);
        seq1 = seq.load(m_o_relaxed);
    } while (seq0 != seq1 || seq0 & 1);
    // do something with r1 and r2;
}

void writer(...) {
    unsigned seq0 = seq;
    while (seq0 & 1 ||
           !seq.compare_exchange_weak(seq0, seq0+1)) {}
    data1 = ...;
    data2 = ...;
    seq = seq0 + 2;
}

Figure 6. Seqlock reader with acquire fence

该实现假设读取seq0时，reader与“最后一个”writer之间存在synchronize-with关系，但C++内存模型并不能保证这一点。

要正确实现这一点，必须在读取seq1时通过RMW操作（例如seq.fetch_add(0)或seq.compare_exchange_weak(seq0, seq0)）强制建立synchronize-with关系；只有当seq1 == seq0时，才能确保读取seq0时与“最后一个”writer之间存在synchronize-with关系，进而满足happens-before要求。

内容的提问来源于stack exchange，提问作者Alex