llvm-6502/include/llvm/Analysis/ET-Forest.h
Devang Patel 8d3ab25335 Use iterative algorith to assign DFS number. This reduces
call stack depth.


git-svn-id: https://llvm.org/svn/llvm-project/llvm/trunk@30575 91177308-0d34-0410-b5e6-96231b3b80d8
2006-09-22 01:05:33 +00:00

304 lines
8.1 KiB
C++

//===- llvm/Analysis/ET-Forest.h - ET-Forest implementation -----*- C++ -*-===//
//
// The LLVM Compiler Infrastructure
//
// This file was written by Daniel Berlin from code written by Pavel Nejedy, and
// is distributed under the University of Illinois Open Source License. See
// LICENSE.TXT for details.
//
//===----------------------------------------------------------------------===//
//
// This file defines the following classes:
// 1. ETNode: A node in the ET forest.
// 2. ETOccurrence: An occurrence of the node in the splay tree
// storing the DFS path information.
//
// The ET-forest structure is described in:
// D. D. Sleator and R. E. Tarjan. A data structure for dynamic trees.
// J. G'omput. System Sci., 26(3):362 381, 1983.
//
// Basically, the ET-Forest is storing the dominator tree (ETNode),
// and a splay tree containing the depth first path information for
// those nodes (ETOccurrence). This enables us to answer queries
// about domination (DominatedBySlow), and ancestry (NCA) in
// logarithmic time, and perform updates to the information in
// logarithmic time.
//
//===----------------------------------------------------------------------===//
#ifndef LLVM_ANALYSIS_ETFOREST_H
#define LLVM_ANALYSIS_ETFOREST_H
#include <cassert>
#include <cstdlib>
namespace llvm {
class ETNode;
/// ETOccurrence - An occurrence for a node in the et tree
///
/// The et occurrence tree is really storing the sequences you get from
/// doing a DFS over the ETNode's. It is stored as a modified splay
/// tree.
/// ET occurrences can occur at multiple places in the ordering depending
/// on how many ET nodes have it as their father. To handle
/// this, they are separate from the nodes.
///
class ETOccurrence {
public:
ETOccurrence(ETNode *n): OccFor(n), Parent(NULL), Left(NULL), Right(NULL),
Depth(0), Min(0), MinOccurrence(this) {};
void setParent(ETOccurrence *n) {
assert(n != this && "Trying to set parent to ourselves");
Parent = n;
}
// Add D to our current depth
void setDepthAdd(int d) {
Min += d;
Depth += d;
}
// Reset our depth to D
void setDepth(int d) {
Min += d - Depth;
Depth = d;
}
// Set Left to N
void setLeft(ETOccurrence *n) {
assert(n != this && "Trying to set our left to ourselves");
Left = n;
if (n)
n->setParent(this);
}
// Set Right to N
void setRight(ETOccurrence *n) {
assert(n != this && "Trying to set our right to ourselves");
Right = n;
if (n)
n->setParent(this);
}
// Splay us to the root of the tree
void Splay(void);
// Recompute the minimum occurrence for this occurrence.
void recomputeMin(void) {
ETOccurrence *themin = Left;
// The min may be our Right, too.
if (!themin || (Right && themin->Min > Right->Min))
themin = Right;
if (themin && themin->Min < 0) {
Min = themin->Min + Depth;
MinOccurrence = themin->MinOccurrence;
} else {
Min = Depth;
MinOccurrence = this;
}
}
private:
friend class ETNode;
// Node we represent
ETNode *OccFor;
// Parent in the splay tree
ETOccurrence *Parent;
// Left Son in the splay tree
ETOccurrence *Left;
// Right Son in the splay tree
ETOccurrence *Right;
// Depth of the node is the sum of the depth on the path to the
// root.
int Depth;
// Subtree occurrence's minimum depth
int Min;
// Subtree occurrence with minimum depth
ETOccurrence *MinOccurrence;
};
class ETNode {
public:
ETNode(void *d) : data(d), DFSNumIn(-1), DFSNumOut(-1),
Father(NULL), Left(NULL),
Right(NULL), Son(NULL), ParentOcc(NULL) {
RightmostOcc = new ETOccurrence(this);
};
// This does *not* maintain the tree structure.
// If you want to remove a node from the forest structure, use
// removeFromForest()
~ETNode() {
delete RightmostOcc;
}
void removeFromForest() {
// Split us away from all our sons.
while (Son)
Son->Split();
// And then split us away from our father.
if (Father)
Father->Split();
}
// Split us away from our parents and children, so that we can be
// reparented. NB: setFather WILL NOT DO WHAT YOU WANT IF YOU DO NOT
// SPLIT US FIRST.
void Split();
// Set our parent node to the passed in node
void setFather(ETNode *);
// Nearest Common Ancestor of two et nodes.
ETNode *NCA(ETNode *);
// Return true if we are below the passed in node in the forest.
bool Below(ETNode *);
/*
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 B)
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);
}
*/
bool DominatedBy(ETNode *other) const {
return this->DFSNumIn >= other->DFSNumIn &&
this->DFSNumOut <= other->DFSNumOut;
}
// This method is slower, but doesn't require the DFS numbers to
// be up to date.
bool DominatedBySlow(ETNode *other) {
return this->Below(other);
}
void assignDFSNumber (int);
bool hasFather() const {
return Father != NULL;
}
// Do not let people play around with fathers.
const ETNode *getFather() const {
return Father;
}
template <typename T>
T *getData() const {
return static_cast<T*>(data);
}
unsigned getDFSNumIn() const {
return DFSNumIn;
}
unsigned getDFSNumOut() const {
return DFSNumOut;
}
private:
// Data represented by the node
void *data;
// DFS Numbers
int DFSNumIn, DFSNumOut;
// Father
ETNode *Father;
// Brothers. Node, this ends up being a circularly linked list.
// Thus, if you want to get all the brothers, you need to stop when
// you hit node == this again.
ETNode *Left, *Right;
// First Son
ETNode *Son;
// Rightmost occurrence for this node
ETOccurrence *RightmostOcc;
// Parent occurrence for this node
ETOccurrence *ParentOcc;
};
} // end llvm namespace
#endif