Time out when using Spin/Promela

Question

If someone could explain to me why I am getting a timeout with the following code that would be great. I understand, or at least I think I do, the idea of a timeout, but with the do loops I thought this would stop this. Any advice is appreciated.

mtype wantp = false;
mtype wantq = false;
mtype turn = 1;

active proctype p()
{   
  do
  :: printf ("non critical section for p\n");
     wantp = true;
     (wantq ==true);

    if
    :: (turn == 2)-> 
       wantp = false;
      /* wait for turn ==1*/
      (turn ==1);
      wantp = true;
    fi;

    printf("CRITICAL SECTION for P\n");
    turn = 2;
    wantp = false;
  od
}

active proctype q()
{
  do
  :: printf("non critical section for q\n");
     wantq = true;
    (wantp ==true);

    if
    :: (turn == 1)-> 
       wantq = false;
       (turn ==2);
       wantq = true;
    fi;

    printf("CRITICAL SECTION for Q\n");
    turn = 1;
    wantq = false;
  od
}

GoZoner GoZoner · Accepted Answer · 2013-03-19T06:54:03

When you perform a SPIN verification and a problem arises, such as a 'timeout', you have what is known as a 'trail' file. The trail file shows you the exact path through the program execution that leads to the problem.

Assuming the above file is called test.pml. You perform the following:

$ spin -a test.pml
$ gcc -o pan pan.c
$ ./pan
# info about verification, shows timeout
# view the detailed trail file
$ spin -p -t test.pml

Then, look at the details printed out to figure out how the timeout happened.

Time out when using Spin/Promela

1 Answers