Question

Consider the following Java code:

volatile boolean v1 = false;
volatile boolean v2 = false;

//Thread A
v1 = true;
if (v2)
    System.out.println("v2 was true");

//Thread B
v2 = true;
if (v1)
    System.out.println("v1 was true");

If there was a globally visible total order for volatile accesses then at least one println would always be reached.

Is that actually guaranteed by the Java Standard? Or is an execution like this possible:

A: v1 = true;
B: v2 = true;
A: read v2 = false;
B: read v1 = false;
A: v2 = true becomes visible (after the if)
B: v1 = true becomes visible (after the if)

I could only find statements about accesses to the same volatile variable in the Standard (but I might be missing something).

“A write to a volatile variable (§8.3.1.4) v synchronizes-with all subsequent reads of v by any thread (where subsequent is defined according to the synchronization order).”

http://java.sun.com/docs/books/jls/third_edition/html/memory.html#17.4.4

Thanks!

Answer 1

I think the quote answers your question more than you realize it does.

The write to v2 by thread B in line #2 is “seen” by thread A when it attempts to read the variable.

I’ve bolded the key part of the quote:

“A write to a volatile variable (§8.3.1.4) v synchronizes with all subsequent reads of v by any thread …”