Question

我使用Spin

创建了一个面包店锁

1   int n=3;
2       int choosing[4] ; // initially 0 
3       int number[4];  // initially 0 
4        
5   active [3] proctype p()
6   {
7        
8        choosing[_pid] = 1; 
9        int max = 0; 
10       int i=0;
11  
12      do      
13      ::(number[i] > max) -> max=number[i];
14      ::i++;
15      :: (i == n) -> break;
16      od;
17      
18      number[_pid] = max + 1;
19      choosing[_pid] = 0;
20  
21      int j=0;
22      
23      do
24      ::(j==n) -> break; 
25      ::  do
26          ::(choosing[j] == 0)-> break;
27          od;
28      ::  if
29          ::(number[j] ==0) -> j++;
30          ::(number[j] > number[_pid]) -> j++;
31          ::((number[j] == number[_pid]) && ( j> _pid)) -> j++;
32          fi;
33      od;
34  
35      number[_pid]=0
36      
37  }

当我测试它时出现错误：pan：1：断言违反 - 无效的数组索引（深度为5）

当我跑步时我得到了回来

  1:    proc  2 (p) Bakery_lock.pml:8 (state 1) [choosing[_pid] = 1]
  2:    proc  2 (p) Bakery_lock.pml:10 (state 2)    [max = 0]
  2:    proc  2 (p) Bakery_lock.pml:12 (state 3)    [i = 0]
  3:    proc  2 (p) Bakery_lock.pml:14 (state 6)    [i = (i+1)]
  4:    proc  2 (p) Bakery_lock.pml:14 (state 6)    [i = (i+1)]
  5:    proc  2 (p) Bakery_lock.pml:14 (state 6)    [i = (i+1)]
spin: indexing number[3] - size is 3
spin: Bakery_lock.pml:13, Error: indexing array 'number'
  6:    proc  2 (p) Bakery_lock.pml:13 (state 4)    [((number[i]>max))]

任何人都可以告诉我它跳过这一行的原因（i == n） - ＆gt;打破; ？

Answer 1

它没有“跳过”该行。 Spin执行可执行的每一行。在do行中i++ 始终可执行，因此，因为Spin会探索所有可能的执行，即使i++行也会执行(i == n)行}。修复是：

 do      
 :: (number[i] > max) -> max=number[i];
 :: (i  < n) -> i++
 :: (i == n) -> break;
 od;

旋转随机错误，面包店锁定

1 个答案: