openmp/runtime/test/worksharing/for/kmp_doacross_check.c

   1 // RUN: %libomp-compile-and-run
   2 // UNSUPPORTED: gcc
   3 // This test is incompatible with gcc because of the explicit call to
   4 // __kmpc_doacross_fini().  gcc relies on an implicit call to this function
   5 // when the last iteration is executed inside the GOMP_loop_*_next() functions.
   6 // Hence, in gcc, having the explicit call leads to __kmpc_doacross_fini()
   7 // being called twice.
   8 #include <stdio.h>
   9
  10 #define N   1000
  11
  12 struct dim {
  13   long long lo; // lower
  14   long long up; // upper
  15   long long st; // stride
  16 };
  17 extern void __kmpc_doacross_init(void*, int, int, struct dim *);
  18 extern void __kmpc_doacross_wait(void*, int, long long*);
  19 extern void __kmpc_doacross_post(void*, int, long long*);
  20 extern void __kmpc_doacross_fini(void*, int);
  21 extern int __kmpc_global_thread_num(void*);
  22
  23 int main()
  24 {
  25   int i;
  26   int iter[N];
  27   struct dim dims;
  28   for( i = 0; i < N; ++i )
  29     iter[i] = 1;
  30   dims.lo = 1;
  31   dims.up = N-1;
  32   dims.st = 1;
  33   #pragma omp parallel num_threads(4)
  34   {
  35     int i, gtid;
  36     long long vec;
  37     gtid = __kmpc_global_thread_num(NULL);
  38     __kmpc_doacross_init(NULL,gtid,1,&dims); // thread starts the loop
  39     #pragma omp for nowait schedule(dynamic)
  40     for( i = 1; i < N; ++i )
  41     {
  42       // runtime call corresponding to #pragma omp ordered depend(sink:i-1)
  43       vec=i-1;
  44       __kmpc_doacross_wait(NULL,gtid,&vec);
  45       // user's code
  46       iter[i] = iter[i-1] + 1;
  47       // runtime call corresponding to #pragma omp ordered depend(source)
  48       vec=i;
  49       __kmpc_doacross_post(NULL,gtid,&vec);
  50     }
  51     // thread finishes the loop (should be before the loop barrier)
  52     __kmpc_doacross_fini(NULL,gtid);
  53   }
  54   if( iter[N-1] == N ) {
  55     printf("passed\n");
  56   } else {
  57     printf("failed %d != %d\n", iter[N-1], N);
  58     return 1;
  59   }
  60   return 0;
  61 }
  62