CODEWITHSUNDEEP
verificationspinpromela
Jiew Meng
Jiew Meng

Reputation: 88197

Whats causing timeout in Promela/SPIN?

I have the following promela code: 

chan level = [0] of {int};

proctype Sensor (chan levelChan) {
    int x;
    do
    :: true ->
            levelChan ? x;
            if 
            :: (x < 2) -> printf("low %d", x);
            :: (x > 8) -> printf("high %d", x);
            :: else -> printf("normal %d", x);
            fi
    od
}

init {
    run Sensor(level);  
    int lvl = 5;
    level ! lvl;
    lvl = 0;
    do 
    :: true ->
        level ! lvl; 
        lvl++;
        (lvl > 9) -> break;
    od
}

I am expecting to send level (0-9) information into a channel and have the sensor output low|normal|high depending on this level. Its quite simple. But why SPIN saying its timeout all the time? 

  0:    proc  - (:root:) creates proc  0 (:init:)
Starting Sensor with pid 1
  1:    proc  0 (:init:) creates proc  1 (Sensor)
  1:    proc  0 (:init:) 1.pml:18 (state 1) [(run Sensor(level))]
  2:    proc  1 (Sensor) 1.pml:6 (state 11) [(1)]
  3:    proc  0 (:init:) 1.pml:20 (state 2) [lvl = 5]
  4:    proc  0 (:init:) 1.pml:20 (state 3) [level!lvl]
  4:    proc  1 (Sensor) 1.pml:8 (state 2)  [levelChan?x]
  5:    proc  1 (Sensor) 1.pml:9 (state 9)  [else]
  6:    proc  0 (:init:) 1.pml:21 (state 4) [lvl = 0]
normal 5  8:    proc  1 (Sensor) 1.pml:12 (state 8) [printf('normal %d',x)]
  9:    proc  0 (:init:) 1.pml:22 (state 10)    [(1)]
 12:    proc  1 (Sensor) 1.pml:6 (state 11) [(1)]
 13:    proc  0 (:init:) 1.pml:24 (state 6) [level!lvl]
 13:    proc  1 (Sensor) 1.pml:8 (state 2)  [levelChan?x]
 14:    proc  1 (Sensor) 1.pml:9 (state 9)  [((x<2))]
low 0 15:   proc  1 (Sensor) 1.pml:10 (state 4) [printf('low %d',x)]
 17:    proc  0 (:init:) 1.pml:25 (state 7) [lvl = (lvl+1)]
 19:    proc  1 (Sensor) 1.pml:6 (state 11) [(1)]
timeout
#processes: 2
 19:    proc  1 (Sensor) 1.pml:8 (state 2)
 19:    proc  0 (:init:) 1.pml:26 (state 8)
2 processes created

It seem to only do 1 iteration of the do loop why?

Upvotes: 3

Views: 1769

Answers (1)

Crux
Crux

Reputation: 58

The statements in the init process's do loop are run in sequence.

do 
:: true ->
    level ! lvl; 
    lvl++;
    (lvl > 9) -> break;
od

The very first time the do loop is run it will send lvl over the level channel, increase the lvl (so it is now 1) and then test (lvl > 9). This is false so will block and causes the timeout.

To have multiple options in a do loop you need :: to define the start of each option:

do
 :: true ->
    level ! lvl; 
    lvl++;
 :: (lvl > 9) -> break;
od

However this will still not perform as expected. When lvl is greater than 9 both options of the do loop are valid and either one can be chosen, so the loop will not necessarily break when lvl > 9 it could choose to continue to send the lvl over the level channel and increase lvl.

You need to have a condition on the first do option as well as the second:

do 
:: (lvl <= 9) ->
    level ! lvl; 
    lvl++;
:: (lvl > 9) -> break;
od

It might be nicer to use the for loop syntax for this example:

init {
  run Sensor(level);  
  int lvl = 5;
  level ! lvl;

  for (lvl : 0..9){
    level ! lvl; 
  }
}

Note that this example will still have a timeout error caused by Sensor's do loop, when the init process has finished Sensor will still try to read from the level channel and timeout.

Upvotes: 3

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