Firstly let me say that I am aware of this being a fairly common topic here but searching for it I couldn't quite find another question that clarifies the following situation. I am very sorry if this is a possible duplicate but here you go:
I am new to concurrency and have been given the following code in order to answer questions:
- a) Why any other output aside from "00" would be possible?
- b) How to amend the code so that "00" will ALWAYS print.
boolean flag = false;
void changeVal(int val) {
if(this.flag){
return;
}
this.initialInt = val;
this.flag = true;
}
int initialInt = 1;
class MyThread extends Thread {
public void run(){
changeVal(0);
System.out.print(initialInt);
}
}
void execute() throws Exception{
MyThread t1 = new MyThread();
MyThread t2 = new MyThread();
t1.start(); t2.start(); t1.join(); t2.join();
System.out.println();
}
For a) my answer would be the following: In the absence of any volatile / synchronization construct the compiler could reorder some of the instructions. In particular, "this.initialInt = val;" and "this.flag = true;" could be switched so that this situation could occur: The threads are both started and t1 charges ahead. Given the reordered instructions it first sets flag = true. Now before it reaches the now last statement of "this.initialInt = val;" the other thread jumps in, checks the if-condition and immediately returns thus printing the unchanged initialInt value of 1. Besides this, I believe that without any volatile / synchronization it is not for certain whether t2 might see the assignment performed to initialInt in t1 so it may also print "1" as the default value.
For b) I think that flag could be made volatile. I have learned that when t1 writes to a volatile variable setting flag = true then t2, upon reading out this volatile variable in the if-statement will see any write operations performed before the volatile write, hence initialInt = val, too. Therefore, t2 will already have seen its initialInt value changed to 0 and must always print 0. This will only work, however, if the use of volatile successfully prevents any reordering as I described in a). I have read about volatile accomplishing such things but I am not sure whether this always works here in the absence of any further synchronized blocks or any such locks. From this answer I have gathered that nothing happening before a volatile store (so this.flag = true) can be reordered as to appear beyond it. In that case initialInt = val could not be moved down and I should be correct, right? Or not ? :)
Thank you so much for your help. I am looking forward to your replies.
This example will alway print 00 , because you do
changeVal(0)
before the printing .to mimic the case where 00 might not be printed , you need to move
initialInt = 1;
to the context of a thread like so :now you might have a race condition , that sets initialInt back to 1 in thread1 before it is printed in thread2
another alternative that might results in a race-condition but is harder to understand , is switching the order of setting the flag and setting the value