We're in a shared memory concurrency model where all reads and writes to integer variables are atomic.

  • do: $S_1$ in parallel with: $S_2$   means to execute $S_1$ and $S_2$ in separate threads, concurrently.
  • atomically($E$)   means to evaluate $E$ atomically, i.e. all other threads are stopped during the execution of $E$.

Consider the following program:

x = 0; y = 4
do:                 # thread T1
    while x != y:
        x = x + 1; y = y - 1
in parallel with:   # thread T2
    while not atomically (x == y): pass
    x = 0; y = 2

Does the program always terminate? When it does terminate, what are the possible values for x and y?

Acknowledgement: this is a light rephrasing of exercise 2.19 in Foundations of Multithreaded, Parallel, and Distributed Programming by Gregory R. Andrews.

  • $\begingroup$ You can't just atomically wait for something: if other threads are stopped, then you'll feel alone. $\endgroup$
    – jmad
    Mar 17, 2012 at 0:04
  • $\begingroup$ @jmad You're right, I forgot to mention that waiting exits the monitor. I've reformulated the problem to avoid having to define this. Thanks. $\endgroup$ Mar 17, 2012 at 0:16
  • $\begingroup$ Perhaps you want to say, does this program always terminate or is there a run of this program which terminates to make the question more precise, depending on what you want. $\endgroup$ Mar 17, 2012 at 8:56
  • $\begingroup$ @DaveClarke Ok, I've reformulated. But even with the original phrasing, I would have expected an answer to show that there are both terminating and non-terminating traces. $\endgroup$ Mar 17, 2012 at 13:34
  • $\begingroup$ As you have mentioned, this question belongs to Parallel/Distributed Programming, please consider adding at least one of the tags parallel-computing/distributed-computing. $\endgroup$
    – Kaveh
    Mar 26, 2012 at 18:36

1 Answer 1


This trace is possible, in two separate threads T1 and T2. $state$ is $(x,y)$.

  • T1: ... $state=(0, 4)$
  • T1: x = x + 1; y = y - 1 $~~state=(1, 3)$
  • T1: x = x + 1; y = y - 1 $~~state=(2, 2)$
  • T2: x == y evaluates to true, pass and then x = 0; $~~state=(0, 2)$
  • T1: x != y evaluates to true, x = x + 1; y = y - 1 $~~state=(1, 1)$
  • T2: y = 2 $~~state=(1, 2)$
  • T1: x != y evaluates to true, x = x + 1; y = y - 1 $~~state=(2, 1)$
  • $state=(3, 0)$
  • $state=(4, -1)$
  • ...

(Note that it works even if x = expr; is atomic)

There are other possible interleavings. The point $(2,2)$ is common to all of them, where T1 has pending (logically) atomic instructions:

T1: push x; push y; eq ? stop : push(x + 1); pop@x; push(y - 1); pop@y; repeat
T2: (x != y) ? repeat : x = 0; y = 2;

In the first case, T1 proceeds to stop and then T2 can only proceed and the final state is $(0,2)$.

If T2 finally skips the repeat and (T1:push x) is run before (T2:x = 0) then T1 will stop looping and the same final state is reached.

If T2 finally skips the repeat and (T1:push x) is run after (T2:x = 0) then T2 can proceed after the stop independently of (T1:y = 2).

state = (0, 2)
T1: push(x + 1); pop@x; push(y - 1); pop@y; ...
T2: y = 2;

If T2 is run now then it will loop as above, so T1 proceeds:

state = (1, 2); stack = 1
T1: pop@y; ...
T2: y = 2;

If T2 is run now, this will go to the final state $(1,1)$. Otherwise:

state = (1, 1)
T1: push x; push y; eq ? stop : push(x + 1); pop@x; push(y + 1); pop@y; repeat
T2: y = 2;

If T2 does not act before push y, this will stop and go to the state $(1,2)$. If it does, then the state is $(1, 2)$ and this will loop into $(2,1)$, $(3,0)$, ...

To sum up the possible final states are $(0,2)$, $(1,1)$, $(1,2)$. I don't think it was worth the effort though, since I probably made mistakes.

  • $\begingroup$ This is one possible trace, but there are also terminating traces. $\endgroup$ Mar 17, 2012 at 1:29
  • $\begingroup$ @Gilles: yes, quite obviously. Unfortunately I'm used to despise possibly non-terminating programs so I didn't bother answering "if it does". I'm ready to complete my answer, but please specify whether x = expr; is atomic or not. ("not" would imply more answers so I'm trying to minimize my work here.) $\endgroup$
    – jmad
    Mar 17, 2012 at 1:50
  • $\begingroup$ x=$E$ is not atomic; each variable in $E$ is read atomically, then the value is computed, then the assignment is performed atomically, but the thread can be interrupted between two reads or between the reads and the write. Even if you're not going to list all the traces, at least explain how to observe them: describe the possible interleavings. $\endgroup$ Mar 17, 2012 at 1:57

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