Understanding happens-before and synchronization [duplicate]

Answer 1

Here we have hb(w, r) but that doesn't mean that c will contain value 3 after assignment. How do I enforce that c is assigned with 3? Does synchronization order provide such guarantees?

And your example

public volatile int v;
public int c;
Actions:

Thread A
v = 3;  //w

Thread B
c = v;  //r

You don't need volatile for v in your example. Let's take a look at a similar example

int v = 0;
int c = 0;
volatile boolean assigned = false;

Actions:

Thread A

v = 3;
assigned = true;

Thread B

while(!assigned);
c = v;

1. assigned field is volatile.
2. We will have c = v statement in Thread B only after assigned will be true (while(!assigned) is responsible for that).
3. if we have volatile — we have happens before.
4. happens before means that, if we see assigned == true — we will see all that happened before a statement assigned = true: we will see v = 3.
5. So when we have assigned == true -> we have v = 3.
6. We have c = 3 as a result.

What will happen without volatile

int v = 0;
int c = 0;
boolean assigned = false;

Actions:

Thread A

v = 3;
assigned = true;

Thread B

while(!assigned);
c = v;

We have assigned without volatile for now.

The value of c in the Thread B can be equal 0 or 3 in such situation. So there is not any guaranties that c == 3.