Can the operations on a volatile variable be interleaved?

Question

volatile int a;
//thread A
int b = a;
b++;
a=b;
//thread B
int b = a;
b--;
a=b;
Does declaring 'a' as volatile permit interleaving of operations of threads A & B ?
int b = a;//thread A
int b = a;//thread B
b++; //thread A
b--; //thread B
a = b; //thread A
a = b; //thread B
I know that a synchronized block or a method does not allow this. I'm not sure of what volatile does... Please help!!!

Answer 1

Yes. The sequence you show absolutely can happen.
Volatile means exactly the following, nothing more, nothing less:
1. Every read and write of that variable will be against the master copy, no thread-local copies.
2. double and long will be read and written atomically. (All other types are always read and written atomically, regardless of whether they're volatile or not.)
In addition, note that plus-plus and minus-minus are not atomic, and declaring the variable volatile doesn't change that.
If thread T1 does b++; that consists of the following operations:
1. Read current value of b.
2. Add 1 to that value.
3. Store the new value back into b.
So even if b is volatile, if T2 is also executing b++; concurrently, you can have the following (say b starts at 1):
1. T1 read 1 from b.
2. T2 read 2 from b.
3. T1 add 1, result is 2.
4. T2 add 1, result is 2.
5. T1 store 2 in b.
6. T2 store 2 in b.
So even though we incremented twice, the value only increased by 1. This can happen whether b is volatile or not.