'undefined' or 'end of list'. The name of each node is given by the dfs
number of the corresponding basic block. Please note, that we include the
artificial ENTRY_BLOCK (or EXIT_BLOCK in the post-dom case) in our lists to
- support multiple entry points. As it has no real basic block index we use
- 'last_basic_block' for that. Its dfs number is of course 1. */
+ support multiple entry points. Its dfs number is of course 1. */
/* Type of Basic Block aka. TBB */
typedef unsigned int TBB;
{ \
unsigned int i = 1; /* Catch content == i. */ \
if (! (content)) \
- (var) = xcalloc ((num), sizeof (type)); \
+ (var) = XCNEWVEC (type, num); \
else \
{ \
- (var) = xmalloc ((num) * sizeof (type)); \
+ (var) = XNEWVEC (type, (num)); \
for (i = 0; i < num; i++) \
(var)[i] = (content); \
} \
static void
init_dom_info (struct dom_info *di, enum cdi_direction dir)
{
- /* We need memory for n_basic_blocks nodes and the ENTRY_BLOCK or
- EXIT_BLOCK. */
- unsigned int num = n_basic_blocks + 1 + 1;
+ unsigned int num = n_basic_blocks;
init_ar (di->dfs_parent, TBB, num, 0);
init_ar (di->path_min, TBB, num, i);
init_ar (di->key, TBB, num, i);
/* Ending block. */
basic_block ex_block;
- stack = xmalloc ((n_basic_blocks + 3) * sizeof (edge_iterator));
+ stack = XNEWVEC (edge_iterator, n_basic_blocks + 1);
sp = 0;
/* Initialize our border blocks, and the first edge. */
di->nodes = di->dfsnum - 1;
- /* Make sure there is a path from ENTRY to EXIT at all. */
- gcc_assert (di->nodes == (unsigned int) n_basic_blocks + 1);
+ /* This aborts e.g. when there is _no_ path from ENTRY to EXIT at all. */
+ gcc_assert (di->nodes == (unsigned int) n_basic_blocks - 1);
}
/* Compress the path from V to the root of its set and update path_min at the
{
b->dom[dir] = et_new_tree (b);
}
- n_bbs_in_dom_tree[dir] = n_basic_blocks + 2;
+ n_bbs_in_dom_tree[dir] = n_basic_blocks;
init_dom_info (&di, dir);
calc_dfs_tree (&di, dir);
for (ason = son->right, n = 1; ason != son; ason = ason->right)
n++;
- *bbs = xmalloc (n * sizeof (basic_block));
+ *bbs = XNEWVEC (basic_block, n);
(*bbs)[0] = son->data;
for (ason = son->right, n = 1; ason != son; ason = ason->right)
(*bbs)[n++] = ason->data;
return dom;
}
+/* Given a dominator tree, we can determine whether one thing
+ dominates another in constant time by using two DFS numbers:
+
+ 1. The number for when we visit a node on the way down the tree
+ 2. The number for when we visit a node on the way back up the tree
+
+ You can view these as bounds for the range of dfs numbers the
+ nodes in the subtree of the dominator tree rooted at that node
+ will contain.
+
+ The dominator tree is always a simple acyclic tree, so there are
+ only three possible relations two nodes in the dominator tree have
+ to each other:
+
+ 1. Node A is above Node B (and thus, Node A dominates node B)
+
+ A
+ |
+ C
+ / \
+ B D
+
+
+ In the above case, DFS_Number_In of A will be <= DFS_Number_In of
+ B, and DFS_Number_Out of A will be >= DFS_Number_Out of B. This is
+ because we must hit A in the dominator tree *before* B on the walk
+ down, and we will hit A *after* B on the walk back up
+
+ 2. Node A is below node B (and thus, node B dominates node A)
+
+
+ B
+ |
+ A
+ / \
+ C D
+
+ In the above case, DFS_Number_In of A will be >= DFS_Number_In of
+ B, and DFS_Number_Out of A will be <= DFS_Number_Out of B.
+
+ This is because we must hit A in the dominator tree *after* B on
+ the walk down, and we will hit A *before* B on the walk back up
+
+ 3. Node A and B are siblings (and thus, neither dominates the other)
+
+ C
+ |
+ D
+ / \
+ A B
+
+ In the above case, DFS_Number_In of A will *always* be <=
+ DFS_Number_In of B, and DFS_Number_Out of A will *always* be <=
+ DFS_Number_Out of B. This is because we will always finish the dfs
+ walk of one of the subtrees before the other, and thus, the dfs
+ numbers for one subtree can't intersect with the range of dfs
+ numbers for the other subtree. If you swap A and B's position in
+ the dominator tree, the comparison changes direction, but the point
+ is that both comparisons will always go the same way if there is no
+ dominance relationship.
+
+ Thus, it is sufficient to write
+
+ A_Dominates_B (node A, node B)
+ {
+ return DFS_Number_In(A) <= DFS_Number_In(B)
+ && DFS_Number_Out (A) >= DFS_Number_Out(B);
+ }
+
+ A_Dominated_by_B (node A, node B)
+ {
+ return DFS_Number_In(A) >= DFS_Number_In(A)
+ && DFS_Number_Out (A) <= DFS_Number_Out(B);
+ } */
/* Return TRUE in case BB1 is dominated by BB2. */
bool
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