Advanced Compilers CMPSCI 710 Spring 2003 Dominators, etc.

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1 Advanced ompilers MPSI 710 Spring 2003 ominators, etc. mery erger University of Massachusetts, Amherst ominators, etc. Last time Live variable analysis backwards problem onstant propagation algorithms def-use chains Today SSA-form dominators 2 ef-use hains: Problem SSA orm switch (j) case x: i à 1; break; case y: i à 2; break; case z: i à 3; break; switch (k) case x: a à i; break; case y: b à i; break; case z: c à i; break; Static single assignment each assignment to variable gets unique name all uses reached by that assignment are renamed exactly one def per use sparse program representation: use-def chain = (variable, [use 1, use 2 ]) worst-case size of graph = O(*U) = O(N 2 ) 3 4 SSA Transformations -unctions New variable for each assignment, rename uses v 1 à 4 à v+5 1 v 2 à 6 à v+7 2 At each join, add special assignment: φ function : operands indicate which assignments reach join j th operand = j th predecessor If control reaches join from j th predecessor, then value of φ(r,s, ) is value of j th operand asy for straight-line code, but what about control flow? 5 6 1

2 SSA Transformation, function SSA xample II if P then v à 4 else v à 6 /* use v */ if P then v 1 à 4 else v 2 à 6 v 3 à φ(v 1, v 2 ) /* use v3 */ v à 1 while (v < 10) v à v + 1 1: v à 1 2: if (v < 10) 3: v à v + 1 4: goto SSA xample III Placing functions switch (j) case x: i à 1; break; case y: i à 2; break; case z: i à 3; break; switch (k) case x: a à i; break; case y: b à i; break; case z: c à i; break; Safe to put φ functions for every variable at every join point ut: inefficient not necessarily sparse! loses information Goal: minimal f nodes, subject to need 9 10 unction Requirement Minimal Placement of functions Node Z needs φ function for V if: Z is convergence point for two paths originating at different nodes both originating nodes contain assignments to V or also need φ functions for V v = 1 v = 2 Z v 1 = 1 v 2 = 2 v 3 =φ(v 1,v 2 ) Naïve computation of need is expensive: must examine all triples in graph an be done in O(N) time Relies on dominance frontier computation [ytron et al., 1991] Also can be used to compute control dependence graph Z

3 ontrol ependence Graph ontrol ependence xample Identifies conditions affecting statement execution Statement is control dependent on branch if: one edge from branch definitely causes statement to execute another edge can cause statement to be skipped if (a == 1) q à 2 else if (a == 2) goto goto A b à 1 A: c à 1 : d à 1 q à 2 b à 1 if (a==1) c à 1 d à 1 if (a==2) else then ominators inding ominators efore we do dominance frontiers, we need to discuss other dominance relationships x dominates y (x dom y) in G, all paths to y go through x om(v) = set of all vertices that dominate v ntry dominates every vertex Reflexive: a dom a Transitive: a dom b, b dom c! a dom c Antisymmetric: a dom b, b dom a! b=a Notice: in SSA form, a definition dominates its use om(v) = {v} Šp 2 PR(v) om(p) Algorithm: OM(ntry) = {ntry} for v 2 V-{ntry} OM(v) = V repeat changed = false for n 2 V-{ntry} olddom = OM(n) OM(n) = {n} [ (Šp 2 PR(n) OM(p)) if OM(n) olddom changed = true ominator Algorithm xample Other ominators om om om om om om OM(ntry) = {ntry} for v 2 V-{ntry} OM(v) = V repeat changed = false for n 2 V-{ntry} olddom = OM(n) OM(n) = {n} [ (Šp 2 PR(n) OM(p)) if OM(n) \neq olddom changed = true om Strict dominators om!(v) = om(v) {v} antisymmetric & transitive Immediate dominator (v) = closest strict dominator of v d v if d om! v and 8w 2 om!(v): w om d antisymmetric induces tree

4 ominator xample ominator Tree om om! om om! om om! om om! om om! om om! om om! Inverse ominators Inverse ominator xample om -1 (v) = set of all vertices dominated by v reflexive, antisymmetric, transitive om(a): om -1 (A): om(): om -1 (): om(): om -1 (): om(): om -1 (): om(): om -1 (): om(): om -1 (): om(g): om -1 (G): {A} {A,} {A,,} {A,,} {A,,} {A,,,} {A,,,G} inally: ominance rontiers! Why ominance rontiers The dominance frontier (X) is set of all nodes Y such that: X dominates a predecessor of Y ut X does not strictly dominate Y (X) = {Y (9 P 2 PR(Y): X om P) Æ q X om! Y) ominance frontier criterion: if node x contains def of a, then any node z in (x) needs a f function for a intuition: at least two non-intersecting paths converge to z, and one path must contain node strictly dominated by x

5 ominance rontier xample A S X node y is in dominance frontier of node x if: x dominates predecessor of y but does not strictly dominate y (X) = {Y ( 9 P 2 PR(Y): X om P) Æ q X om! Y) ominance rontier xample om -1 SU(om -1 ) om -1 SU(om -1 ) om -1 SU(om -1 ) om -1 SU(om -1 ) om -1 SU(om -1 ) om -1 SU(om -1 ) (v) = SU(om -1 (v)) om! -1 (v) where om! -1 (v) = om -1 (v) {v} om -1 SU(om -1 ) Next Time omputing dominance frontiers omputing SSA form 27 5