How std::memory_order_seq_cst works

Question

I took the example about std::memory_order_seq_cst from: http://en.cppreference.com/w/cpp/atomic/memory_order

#include <thread>
#include <atomic>
#include <cassert>

std::atomic<bool> x = {false};
std::atomic<bool> y = {false};
std::atomic<int> z = {0};

void write_x()
{
    x.store(true, std::memory_order_seq_cst);
}

void write_y()
{
    y.store(true, std::memory_order_seq_cst);
}

void read_x_then_y()
{
    while (!x.load(std::memory_order_seq_cst))
        ;
    if (y.load(std::memory_order_seq_cst)) {
        ++z;
    }
}

void read_y_then_x()
{
    while (!y.load(std::memory_order_seq_cst))
        ;
    if (x.load(std::memory_order_seq_cst)) {
        ++z;
    }
}

int main()
{
    std::thread a(write_x);
    std::thread b(write_y);
    std::thread c(read_x_then_y);
    std::thread d(read_y_then_x);
    a.join(); b.join(); c.join(); d.join();
    assert(z.load() != 0);  // will never happen
}

This example is also mentioned in the question of Acquire/Release versus Sequentially Consistent memory order.

My question is how it is possible that thread c and thread d see different things? If it is possible, why this simple example below always yields to z=3? For instance, thread b could say "okay I see 0 even though thread a is already done so z becomes 0+1 again"

#include <atomic>
#include <iostream>

std::atomic<int> z = {0};

void increment()
{
    z.fetch_add(1, std::memory_order_relaxed);
}
int main()
{
    std::thread a(increment);
    std::thread b(increment);
    std::thread c(increment);
    a.join(); b.join(); c.join();
    std::cout << z.load() << '\n';
}

Answer 1

So by seeing different things in your comment you mean that Thread C see x==1,y==0 and Thread D see x==0 and y==1. Is that possible with sequential consistency?

Let's suppose this total order (the modification is the transition between this symbolized memory states):

{x==0,y==0} : S0
{x==1,y==0} : S1
{x==1,y==1} : S2

When we say "see" we mean that a thread potentialy performs a load. Two loads can not be performed simultaneously in one thread. So how is it possible that thread C see x==1 then see y==0 and Thread D see x==0 then see y==1? Thread C performs the two loads while the memory is in the state S1, and Thread D see x at state S0, then see y at state S2.

In your example code, what happens is that Thread C load x then load y, and Thread D load y repeatedly until it is true then load x. So after y==1, it is guarenteed that x==1 in this total order.

As said by Minee in its comment, nothing could be expected if in place of sequential consistency memory order were used acquire/release memory order: acquire/release semantic does not imply any total ordering,moreover there are no happens before relation between the store to x and the store to y. So the assertion z.load()!=0 could fire.

Answer 2

Because read-modify-write operations have special guarantees.

According to the standard [atomics.order] paragraph 11:

Atomic read-modify-write operations shall always read the last value (in the modification order) written before the write associated with the read-modify-write operation.