413CS 701 Fall 2008©Data Flow FrameworksRevisitedRecall that a Data Flow problem ischaracterized as:(a) A Control Flow Graph(b) A Lattice of Data Flow values(c) A Meet operator to join solutions from Predecessors or Successors(d) A Transfer Function Out = fb(In) or In = fb(Out)414CS 701 Fall 2008©Value LatticeThe lattice of values is usually a meetsemilattice defined by:A: a set of valuesT and ⊥ (“top” and “bottom”):distinguished values in the lattice≤: A reflexive partial order relatingvalues in the lattice∧: An associative and commutativemeet operator on lattice values415CS 701 Fall 2008©Lattice AxiomsThe following axioms apply to thelattice defined by A, T, ⊥, ≤ and ∧: a ≤ b ⇔ a ∧ b = a a ∧ a = a (a ∧ b) ≤ a (a ∧ b) ≤ b (a ∧ T) = a (a ∧⊥) = ⊥416CS 701 Fall 2008©Monotone Transfer FunctionTransfer Functions, fb:L → L (where Lis the Data Flow Lattice) are normallyrequired to be monotone.That is x ≤ y ⇒ fb(x) ≤ fb(y).This rule states that a “worse” inputcan’t produce a “better” output.Monotone transfer functions allow usto guarantee that data flow solutionsare stable.If we had fb(T) = ⊥ and fb(⊥)=T,then solutions might oscillatebetween T and ⊥ indefinitely.Since ⊥≤ T, fb(⊥) should be ≤ fb(T).But fb(⊥) = T which is not ≤ fb(T) =⊥. Thus fb isn’t monotone.417CS 701 Fall 2008©Dominators fit the Data FlowFrameworkGiven a set of Basic Blocks, N, wehave:A is 2N (all subsets of Basic Blocks).T is N.⊥ is φ.a ≤ b ≡ a ⊆ b.fZ(in) = In ∪ {Z}∧ is ∩ (set intersection).418CS 701 Fall 2008©The required axioms are satisfied: a ⊆ b ⇔ a ∩ b = a a ∩ a = a (a ∩ b) ⊆ a (a ∩ b) ⊆ b (a ∩ N) = a (a ∩φ) = φAlso fZ is monotone sincea ⊆ b ⇒ a ∪ {Z} ⊆ b ∪ {Z} ⇒fZ(a) ⊆ fZ(b)419CS 701 Fall 2008©Constant PropagationWe can model Constant Propagationas a Data Flow Problem. For eachscalar integer variable, we willdetermine whether it is known tohold a particular constant value at aparticular basic block.The value lattice isT represents a variable holding aconstant, whose value is not yetknown.i represents a variable holding aknown constant value.T⊥..., −2, −1, 0, 1, 2, ...420CS 701 Fall 2008©⊥ represents a variable whose value isnon-constant.This analysis is complicated by thefact that variables interact, so wecan’t just do a series of independentone variable analyses.Instead, the solution lattice willcontain functions (or vectors) thatmap each variable in the program toits constant status (T, ⊥, or someinteger).Let V be the set of all variables in aprogram.421CS 701 Fall 2008©Let t : V → N U {T,⊥}t is the set of all total mappings fromV (the set of variables) to N U {T,⊥}(the lattice of “constant status”values).For example, t1=(T,6,⊥) is a mappingfor three variables (call them A, B andC) into their constant status. t1 saysA is considered a constant, with valueas yet undetermined. B holds thevalue 6, and C is non-constant.We can create a lattice composed of tfunctions:tT(V) = T (∀ V) (tT=(T,T,T, ...)t⊥(V) = ⊥ (∀ V) (t⊥=(⊥,⊥,⊥, ...)422CS 701 Fall 2008©ta≤ tb⇔∀v ta(v) ≤ tb(v)Thus (1,⊥) ≤ (T,3) since 1 ≤ T and ⊥≤ 3.The meet operator ∧ is appliedcomponentwise:ta∧tb = tc where ∀v tc(v) = ta(v) ∧ tb(b)Thus (1,⊥) ∧ (T,3) = (1,⊥) since 1 ∧ T = 1 and ⊥∧ 3 = ⊥.423CS 701 Fall 2008©The lattice axioms hold: ta≤ tb⇔ ta∧ tb = ta (since thisaxiom holds for each component) ta∧ ta = ta (trivially holds) (ta∧ tb) ≤ ta (per variable def of ∧) (ta∧ tb) ≤ tb (per variable def of ∧) (ta∧ tT) = ta (true for allcomponents) (ta∧ t⊥) = t⊥(true for allcomponents)424CS 701 Fall 2008©The Transfer FunctionConstant propagation is a forwardflow problem, so Cout = fb(Cin)Cin is a function, t(v), that mapsvariables to T,⊥, or an integer valuefb(t(v)) is defined as:(1) Initially, let t’(v)=t(v) (∀v)(2) For each assignment statement v = e(w1,w2,...,wn) in b, in order of execution, do: If any t’(wi) = ⊥ ( 1≤i≤n ) Then set t’(v) = ⊥ (strictness) Elsif any t’(wi) = T (1≤i≤n ) Then set t’(v) = T (delay eval of v) Else t’(v) = e(t’(w1),t’(w2),...)(3) Cout = t’(v)425CS 701 Fall 2008©Note that in valid programs, we don’tuse uninitialized variables, sovariables mapped to T should onlyoccur prior to initialization.Initially, all variables are mapped to T,indicating that initially their constantstatus is unknown.426CS 701 Fall 2008©Examplea=1b=2b=a+1b=a+2b=b-1T, T1,21,21,2 1,31,⊥1,⊥1,⊥427CS 701 Fall 2008©Distributive FunctionsFrom the properties of ∧ and f’smonotone property, we can show that f(a∧b) ≤ f(a) ∧ f(b)To see this note that a∧b ≤ a, a∧b ≤ b ⇒f(a∧b) ≤ f(a), f(a∧b) ≤ f(b) (*)Now we can establish that x≤y, x≤z ⇒ x ≤ y∧z (**)To see that (**) holds, note that x≤y ⇒ x∧y = x x≤z ⇒ x∧z = x (y∧z)∧x ≤ y∧z (y∧z)∧x = (y∧z)∧(x∧x) = (y∧x)∧(z∧x) = x∧x = x Thus x ≤ y∧z, establishing (**).428CS 701 Fall 2008©Now substituting f(a∧b) for x, f(a) for y and f(b) for z in (**) andusing (*) we get f(a∧b) ≤ f(a) ∧ f(b).Many Data Flow problems have flowequations that satisfy the distributiveproperty:f(a∧b) = f(a) ∧ f(b)For example, in our formulation ofdominators:Out = fb(In) = In U {b}whereIn =∩ Out(p)p ∈ Pred(b)429CS 701 Fall 2008©In this case, ∧ =∩.Now fb(S1∩S2) = (S1∩S2) U {b}Also, fb(S1)∩fb(S2) = (S1 U {b})∩(S2 U {b}) = (S1∩S2) U {b}So dominators are distributive.430CS 701 Fall 2008©Not all Data Flow Problemsare DistributiveConstant propagation is notdistributive.Consider the following (with variables(x,y,z)):Now f(t)=t’ wheret’(y) = t(y), t’(z) = t(z),t’(x) = if t(y)=⊥ or t(z) = ⊥ then ⊥ elseif t(y)=T or t(z) =T then T else t(y)+t(z)x=y+zt1 = (T,1,3) t2=(T,2,2)431CS 701 Fall 2008©Now f(t1∧t2) = f(T,⊥,⊥) = (⊥,⊥,⊥)f(t1) = (4,1,3)f(t2) = (4,2,2)f(t1)∧f(t2) = (4,⊥,⊥) ≥ (⊥,⊥,⊥)432CS 701 Fall 2008©Why does it Matter if a DataFlow Problem isn’tDistributive?Consider actual program executionpaths from b0 to (say) bk.One path might be b0,bi1,bi2,...,binwhere bin=bk.At bk the Data Flow information wewant isfin(...fi2(fi1(f0(T)))...) ≡ f(b0,b1,...,bin)On a different path to bk,