PR tree-optimization/49960

	* gcc.dg/autopar/pr49960.c: New test.
	* gcc.dg/autopar/pr49960-1.c: New test.

git-svn-id: svn+ssh://gcc.gnu.org/svn/gcc/trunk@181409 138bc75d-0d04-0410-961f-82ee72b054a4

diff --git a/gcc/testsuite/gcc.dg/autopar/pr49960-1.c b/gcc/testsuite/gcc.dg/autopar/pr49960-1.c
new file mode 100644 (file)
index 0000000..ac9d567
--- /dev/null
+++ b/gcc/testsuite/gcc.dg/autopar/pr49960-1.c
@@ -0,0 +1,36 @@
+/* { dg-do compile } */
+/* { dg-options "-O2 -ftree-parallelize-loops=4 -fdump-tree-parloops-details -fdump-tree-optimized" } */
+
+#include <stdlib.h>
+#include <stdio.h>
+
+int main() 
+{
+  unsigned int x, y, idx, H = 1024, W = 1024;
+  
+  int * tmps = (int *)malloc(H*W*sizeof(int));
+  
+  /* This loop gets parallelized even though output dependences exist 
+     between writes to 'tmps' that prevent parallelization. 
+     For example: tmps[1] = 1, ..., tmps[1] = 17.  */
+  
+  for(x = 1; x < H; x++) 
+    {
+      for(y = 1; y < W; y++) 
+       {
+         idx = x*W+y;
+         tmps[idx % 4096] = idx;         
+       }
+    }
+  
+  for(x = 1; x < 8; x++)
+    printf("tmps[%d]=%d\n", x, tmps[x]);
+  
+  return 0;
+}
+/* Check that no loop gets parallelized.  */
+
+/* { dg-final { scan-tree-dump-times "SUCCESS: may be parallelized" 0 "parloops" } } */
+/* { dg-final { scan-tree-dump-times "loopfn" 0 "optimized" } } */
+/* { dg-final { cleanup-tree-dump "parloops" } } */
+/* { dg-final { cleanup-tree-dump "optimized" } } */

diff --git a/gcc/testsuite/gcc.dg/autopar/pr49960.c b/gcc/testsuite/gcc.dg/autopar/pr49960.c
new file mode 100644 (file)
index 0000000..1f3482f
--- /dev/null
+++ b/gcc/testsuite/gcc.dg/autopar/pr49960.c
@@ -0,0 +1,49 @@
+/* { dg-do compile } */
+/* { dg-options "-O2 -ftree-parallelize-loops=4 -fdump-tree-parloops-details -fdump-tree-optimized" } */
+
+#include <stdio.h>
+#define MB 100
+#define NA 450
+#define MA 400
+
+int T[MA][MB],A[MA][NA],B[MB][NA];
+void MRTRBR(int MA_1, int NA_1, int MB_1)
+{
+  int i,j, t,k;
+
+  /* The outer most loop is not parallel because for different k's there
+     is write-write dependency for T[i][j].  */
+  
+  /* The two inner loops don't get parallelized due to low number of 
+     iterations.  */
+
+  for (k = 3; k < NA_1; k++)
+    for (i = 3; i < MA_1; i++)
+      for (j = 3; j < MB_1; j++)
+       {
+         t = T[i][j];
+         T[i][j] = t+2+A[i][k]*B[j][k];
+       }
+}
+void main ()
+{
+  int j,i;
+  
+  for (i = 3; i < MA; i++)
+    for (j = 3; j < MB; j++)
+      T[i][j] = (i>j?i:j);
+  
+  MRTRBR (MA,NA,MB);
+  
+  for (i = MA-1; i < MA; i++)
+    for (j = MB-10; j < MB; j++)
+      printf ("i %d j %d T[i][j] = %d\n",i,j,T[i][j]);
+}
+
+
+/* Check that the outer most loop doesn't get parallelized (thus no loop gets parallelized)  */
+
+/* { dg-final { scan-tree-dump-times "SUCCESS: may be parallelized" 0 "parloops" } } */
+/* { dg-final { scan-tree-dump-times "loopfn" 0 "optimized" } } */
+/* { dg-final { cleanup-tree-dump "parloops" } } */
+/* { dg-final { cleanup-tree-dump "optimized" } } */