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### GeorgiaTech04-byu-live

Course: CS 486, Fall 2009
School: BYU
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Word Count: 1300

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Analysis Important Dependence for Optimization Instruction scheduling Data-cache optimization Software engineering Program understanding Reverse engineering Debugging Various kinds of dependences Data related Control related Data-Dependence Graph A data-dependence graph has one node for every variable (basic block) and one edge representing the flow of data between the two nodes Different types of data...

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Analysis Important Dependence for Optimization Instruction scheduling Data-cache optimization Software engineering Program understanding Reverse engineering Debugging Various kinds of dependences Data related Control related Data-Dependence Graph A data-dependence graph has one node for every variable (basic block) and one edge representing the flow of data between the two nodes Different types of data dependence Flow: def to use Anti: use to def Output: def to def Input: use to use B2 X = 1 Z > 2 entry Z > 1 B1 X = 2 B4 Further classifiable as Loop-carried: requires iterations Loop-independent: occurs anyway B3 Y = X + 1 B5 Z = X 3 X = 4 B6 exit Z = X + 7 1 Data-Dependence Graph entry entry Z > 1 B1 X = 1 Z > 2 X = 2 B4 X = 1 Z > 2 Z > 1 B1 X = 2 B4 B2 B2 Z = X 3 X = 4 B3 Y = X + 1 B5 B3 Y = X + 1 B5 Z = X 3 X = 4 B6 exit Z = X + 7 B6 exit Z = X + 7 Data-Dependence Graph entry entry Z > 1 B1 X = 1 Z > 2 X = 2 B4 X = 1 Z > 2 Z > 1 B1 X = 2 B4 B2 X X B5 Z = X 3 X = 4 B2 X B3 Y = X + 1 B3 Y = X + 1 B5 Z = X 3 X = 4 X B6 exit Z = X + 7 B6 exit Z = X + 7 2 Data-Dependence Graph entry entry Z Z Z > 1 B1 Z > 1 B1 X = 2 B4 X = 1 Z > 2 X = 2 B4 B2 X = 1 Z > 2 Z X Y X B5 Z = X 3 X = 4 B2 X B3 X Y = X + 1 B3 Y = X + 1 B5 Z = X 3 X = 4 X Y B6 exit X B6 exit Z = X + 7 Z = X + 7 Z Control-Dependence Analysis Intuition A statement S1 is control dependent on a statement S2 if the outcome of S2 directly determines whether S1 may not be reached What are the control dependences for each statement in the CFG on the right? entry Z > 1 B1 T X = 1 F X = 2 B4 B2 Z > 2 T B3 Y = X + 1 F Z = X 3 B5 X = 4 B6 Z = X + 7 exit 3 Control-Dependence Analysis Intuition A statement S1 is control dependent on a statement S2 if the outcome of S2 directly determines whether S1 may not be reached What are the control dependences for each statement in the CFG on the right? entry, B1, exit entering code B2 B1T B3 B2T B4 B1F B5 B2F, B1F B6 B2F, B1F entry Z > 1 B1 T X = 1 F X = 2 B4 B2 Z > 2 T B3 Y = X + 1 F Z = X 3 B5 X = 4 B6 Z = X + 7 exit Control-Dependence Analysis Definition Let G be a CFG, with X and Y nodes in G. Y is controldependent on X iff 1. There exists a path P from X to Y with any Z in P (excluding X and Y) postdominated by Y and 2. X is not postdominated by Y (there are two edges out of X; traversing one edges always leads to Y, the other may not lead to Y) entry Z > 1 B1 T X = 1 F X = 2 B4 B2 Z > 2 T B3 Y = X + 1 F Z = X 3 B5 X = 4 B6 Z = X + 7 exit 4 Computing CD Using FOW 1. Construct basic CDG 2. Add region nodes Computing CD Using FOW Augment the CFG by adding a node Start with edge (Start, entry) labeled "T" and edge (Start, exit) labeled "F"; call this AugCFG AugCFG T En T 2 1 F 4 Start T 3 F F 5 6 Ex 5 Computing CD Using FOW Construct the postdominator tree for AugCFG AugCFG T En T 2 1 F 4 Start T 3 F F 5 6 Ex Computing CD Using FOW Construct the postdominator tree for AugCFG Pdom Tree Start 2 Ex 3 6 1 5 En 4 6 Computing CD Using FOW Consider set S of edges (m, n) in AugCFG such that n does not postdominate m For AugCFG S consists of: AugCFG T En T 2 1 F 4 Pdom Tree Start 2 Ex Start 6 1 T 3 F 3 F 5 5 En Ex 6 4 Computing CD Using FOW Consider set S of edges (m, n) in AugCFG such that n does not postdominate m For AugCFG S consists of: (Start, En) (1,2) (1,4) (2,3) (2,5) AugCFG T En T 2 1 F 4 Start T 3 F F 5 6 Ex 7 Computing CD Using FOW Consider, for each edge (A,B) in S, those nodes in the Pdom tree from to B least common ancestor L of A and B Including L if L is A Excluding L if L is not A Pdom Tree Start 2 Ex 3 6 1 5 En 4 Computing CD Using FOW Consider, for each edge (A,B) in S, those nodes in the Pdom tree from B to least common ancestor L of A and B Including L if L is A Excluding L if L is not A Pdom Tree Start 2 Ex 3 6 1 5 En Edge Start, En 1, 2 1, 4 2, 3 2, 5 L Nodes 4 8 Computing CD Using FOW Consider, for each edge (A,B) in S, those nodes in the Pdom tree from B to least common ancestor L of A and B Including L if L is A Excluding L if L is not A Pdom Tree Start 2 Ex 3 6 1 5 En Edge Start, En 1, 2 1, 4 2, 3 2, 5 L Ex Ex Ex Ex Ex Nodes En, 1 2 4, 5, 6 3 5, 6 4 All identified nodes are control dependent on A Computing CD Using FOW Consider, for each edge (A,B) in S, those nodes in the Pdom tree from B to least common ancestor L of A and B Including L if L is A Excluding L if L is not A Pdom Tree Start 2 Ex 3 6 1 5 En Edge Start, En 1, 2 1, 4 2, 3 2, 5 L Ex Ex Ex Ex Ex Nodes En, 1 2 4, 5, 6 3 5, 6 CD on Start, T 1, T 1, F 2, T 2, F 4 All identified nodes are control dependent on A 9 Computing CD Using FOW Given (A,B), how can you find the set of nodes control dependent on A easily? Why is the set we obtain correct? Pdom Tree Start 2 Ex 3 6 1 5 En Edge Start, En 1, 2 1, 4 2, 3 2, 5 L Ex Ex Ex Ex Ex Nodes En, 1 2 4, 5, 6 3 5, 6 CD on Start, T 1, T 1, F 2, T 2, F 4 Computing CD Using FOW Edge Start, En 1, 2 1, 4 2, 3 2, 5 L Ex Ex Ex Ex Ex Nodes En, 1 2 4, 5, 6 3 5, 6 CD on Start, T Create CDG T Start T T T 1, T 1, F 2, T 2, F En 2 1 F F F F F 3 6 5 4 10 Computing CD Using FOW 1. Construct basic CDG 2. Add region nodes Computing CD Using FOW Why adding regions? Group nodes with same set of control conditions Only two successors for predicate nodes (=> hierarchical organization) 11 Computing CD Using FOW Method for adding regions: Consider set CD of control dependence predecessors of each non-region node with more than one unlabeled controldependence predecessor Create region node R for each CD and move edges accordingly Nodes with control dependence CD connected to R R connected to CD Computing CD Using FOW ...

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