CS293S Redundancy Removal. Yufei Ding

Transcription

1 CS293S Redundancy Removal Yufei Ding

2 Review of Last Class Consideration of optimization Sources of inefficiency Components of optimization Paradigms of optimization Redundancy Elimination Types of intermediate representations of a program Removing redundant expressions 2

3 Topics of This Class Removing redundant expressions DAG: version tracking Linear representation: value numbering Scope of optimization Basic block Extended basic block Region (dominator) 3

4 1. IR: Syntax Tree m <-- 2 x y x z n <-- 3 x y x z o <-- 2 x y - z 4

5 DAG: Eliminating Identical Subtrees m <-- 2 x y x z n <-- 3 x y x z o <-- 2 x y - z Using hash-based constructor: key is the textual notion of operators and operands; value is the node number or address. 5

6 Version Tracking m <-- 2 x y x z y <-- 3 x y x z o <-- 2 x y - z Associate a counter with each variable for version tracking m1 <-- 2 x y1 x z1 y2 <-- 3 x y1 x z1 o1 <-- 2 x y2 - z1 6

7 Missing Any Opportunities? Original Code a x + y z y d 17 c x + z Is there any redundant expression? Can the DAG-based approach recognize it? The hash-based constructor uses a textual notion of equality, - so y equals y, independent of its value. - z does not equal y, independent of its value. 7

8 Linear Representation: Linear IR Example: Three Address Code Statement form: x y op z With 1 operator (op ) and, at most, 3 names (x, y, z) Example: z x - 2 * y becomes t 2 * y z x - t Statements are executed sequentially. 8

9 Local Value Numbering Basic Algorithm For each expression e (assuming in the form result e <-- o 1 op o 2 ) Get value numbers for operands from hash lookup table, represented by VN(o 1 ), VN(o 2 ). If no such key exists, insert these keys to the hash table with new value numbers; Search with hash key <op,vn(o 1 ),VN(o 2 )>, If no such key exists, 1. insert the key to the hash table with a new value number; 2. use the same value number for result e and insert them to hash table. else, get the value of this key, use the same value number for result e, and insert them to hash table. 9

10 Local Value Numbering to Find Redundancy An example Original Code a x + y z y d 17 c x + z With VNs a 3 x 1 + y 2 z 2 y 2 d 4 17 c 3 x 1 + z 2 Hash Table for VN {<x,1>, <y,2>, <<+,1,2>,3>, <a,3>} {..., <z,2>} {..., <z,2>, <17,4>, <d,4>} {..., <z,2>, <17,4>, <d,4>, <c,3>} Compare Value Num with Version Num. (The way they get increase; why the latter is insufficient for the example. 10

11 Rewritten for Redundancy Elimination An example Original Code a x + y z y d 17 c x + z With VNs a 3 x 1 + y 2 z 2 y 2 d 4 17 c 3 x 1 + z 2 Hash Table for VN {<x,1>, <y,2>, <<+,1,2>,3>, <a,3>} {..., <z,2>} {..., <z,2>, <17,4>, <d,4>} {..., <z,2>, <17,4>, <d,4>, <c,3>} Rewritten a x + y z y d 17 c a Hash Table for Rewritten {<1,x>, <2,y>, <3,a>} {<1,x>, <2,y>, <3,a>} {<1,x>, <2,y>, <3,a>, <4,17>} {<1,x>, <2,y>, <3,a>, <4,17>}

12 Resolving Overwritten Issue An example Original Code a x + y z y * b x + y a 17 * c x + y Two redundancies marked by * With VNs a 3 x 1 + y 2 z 2 y 2 * b 3 x 1 + y 2 a 4 17 * c 3 x 1 + y 2 Optional Solutions: Use c 3 b 3 Save a 3 in t 3 Rename around it (best) Rewritten a x + y z y * b a a 17 * c a Hash Table for Rewritten {<1,x>, <2,y>, <3,a>} {<1,x>, <2,y>, <3,a>} {<1,x>, <2,y>, <3,a>, <4,17>} {<1,x>, <2,y>, <3,a>, <4,17>} 12

13 Renaming in Value Numbering Example (continued) Renaming: Give each value a unique name Original Code a 0 x 0 + y 0 z 0 y 0 * b 0 x 0 + y 0 a 1 17 * c 0 x 0 + y 0 With VNs a 0 3 x y 0 2 z 2 0 y 0 2 * b 0 3 x y 0 2 a * c 0 3 x y 0 2 Rewritten a 0 x 0 + y 0 z 0 y 0 * b 0 a 0 a 1 17 * c 0 a 0 Hash Table for Rewritten {<1,x 0 >, <2,y 0 >, <3,a 0 >} {<1,x 0 >, <2,y 0 >, <3,a 0 >} {<1,x 0 >, <2,y 0 >, <3,a 0 >, <4,17>} {<1,x 0 >, <2,y 0 >, <3,a 0 >, <4,17>} Result: a 0 is available Rewriting just works 13

14 Reordering based on associativity The order in which expressions are written matters Example: either 2 x y or y x z will be missed in leftor right-associative treatment. m <-- 2 x y x z n <-- 3 x y x z o <-- 2 x y - z Example: 3 x a x 5 14

15 A Hard Problem: Pointer Assignments *p = 17 could force every variable in a program to increase their version number. A major motivation for pointer analysis Original Code a x + y b w + v *p 17 c x + y d w + v 15

16 So far Removing redundant expressions DAG: version tracking Linear representation: value numbering 16

17 Local Value Numbering <-> Linear IR A m a + b n a + b Local Value Numbering 1 block at a time Strong local results B p c + d r c + d C q a + b r c + d No cross-block effects D e b + 18 E e a + 17 s a + b t c + d u e + f u e + f F v a + b w c + d x e + f G y a + b z c + d Missed opportunities (need stronger methods) 17 *

18 Basic blocks A basic block is a maximal-length segment of straight-line, unpredicated code. In another word, it has one entry point (i.e., no code within it is the destination of a jump instruction), one exit point and no jump instructions contained within it. Example L2: L1: m = 2; c = m + n; if(c>0) goto L1; d = 4; goto L2; c = 5; 18

19 CFG B A p c + d r c + d m a + b n a + b C q a + b r c + d D e b + 18 E e a + 17 s a + b t c + d u e + f u e + f Control-flow graph (CFG) Nodes for basic blocks Edges for branches Basis for many program analysis & transformation G F y a + b z c + d v a + b w c + d x e + f This CFG, G = (N,E) N = {A,B,C,D,E,F,G} E = {(A,B),(A,C),(B,G),(C,D), (C,E),(D,F),(E,F),(F,E)} N = 7, E = 8 19

20 Extended basic block (EBB) A m a + b n a + b B p c + d r c + d C q a + b r c + d D e b + 18 E e a + 17 s a + b t c + d u e + f u e + f F v a + b w c + d x e + f G y a + b z c + d An EBB is a set of blocks B1, B2,..., Bn, where Bi, 2<= i <= n has a unique predecessor, which is in the EBB. May have multiple exits A tree structure