Lecture07
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The document discusses static single assignment (SSA) form, a program representation where each variable is assigned only once. SSA form makes data flow analysis simpler by explicitly representing value definitions and merges. The transformation to SSA form involves inserting φ-functions to merge multiple reaching definitions at control flow merges, and renaming variables. Different approaches to placing φ-functions are minimal, Briggs-minimal, and pruned SSA. Briggs-minimal may add φ-functions that pruned SSA omits if they are dead.
![Another Example
1 v := 1 1 v0 := 1
2 v := v+1 v1 := φ(v0,v2)
2
v2 := v1+1
CIS570 Lecture 7 Static Single Assignment Form 12
SSA vs. ud/du Chains
SSA form is more constrained
Advantages of SSA
– More compact
– Some analyses become simpler when each use has only one def
– Value merging is explicit
– Easier to update and manipulate?
Furthermore
– Eliminates false dependences (simplifying context)
for (i=0; i<n; i++)
A[i] = i; Unrelated uses of i are given
for (i=0; i<n; i++) different variable names
print(foo(i));
CIS570 Lecture 7 Static Single Assignment Form 13
6](https://image.slidesharecdn.com/lecture07-110824035133-phpapp02/85/Lecture07-6-320.jpg)
![Dominance Tree Example
The dominance tree shows the dominance relation
1
1
2 5 9
2 4 5 13 9 12
3 6 7 10 11
3 6 8 7 10 11
4 8 12 Dominance Tree
13
CFG
CIS570 Lecture 7 Static Single Assignment Form 28
Variable Renaming (cont)
Data Structures
– Stacks[v] ∀v
Holds the subscript of most recent definition of variable v, initially empty
– Counters[v] ∀v
Holds the current number of assignments to variable v; initially 0
p
Auxiliary Routine po
sh 1
procedure GenName(variable v) pu
i := Counters[v] 2 4 5 13 9 12
push i onto Stacks[v]
Counters[v] := i + 1
3 6 8 7 10 11
Use the Dominance Tree to remember the most
recent definition of each variable
CIS570 Lecture 7 Static Single Assignment Form 29
14](https://image.slidesharecdn.com/lecture07-110824035133-phpapp02/85/Lecture07-14-320.jpg)
![Variable Renaming Algorithm
procedure Rename(block b) Call Rename(entry-node)
for each φ-function p in b
GenName(LHS(p)) and replace v with vi, where i=Top(Stack[v])
for each statement s in b (in order)
for each variable v ∈ RHS(s)
replace v by vi, where i = Top(Stacks[v])
for each variable v ∈ LHS(s)
GenName(v) and replace v with vi, where i=Top(Stack[v])
for each s ∈ succ(b) (in CFG)
j ← position in s’s φ-function corresponding to block b
for each φ-function p in s Φ( , , )
replace the j th operand of RHS(p) by v , where i = Top(Stack[v])
i
for each s ∈ child(b) (in DT)
Rename(s) Recurse using Depth First Search
for each φ-function or statement t in b
for each vi ∈ LHS(t) Unwind stack when done with this node
Pop(Stack[v])
CIS570 Lecture 7 Static Single Assignment Form 30
Transformation from SSA Form
Proposal
– Restore original variable names (i.e., drop subscripts)
– Delete all φ-functions
Complications
x0 =
− What if versions get out of order?
x1 =
(simultaneously live ranges)
= x0
= x1
Alternative
− Perform dead code elimination (to prune φ-functions)
− Replace φ-functions with copies in predecessors
− Rely on register allocation coalescing to remove unnecessary copies
CIS570 Lecture 7 Static Single Assignment Form 31
15](https://image.slidesharecdn.com/lecture07-110824035133-phpapp02/85/Lecture07-15-320.jpg)
Lecture07
- 1. Static Single Assignment Form Last Time – Lattice theoretic frameworks for data-flow analysis Today – Program representations – Static single assignment (SSA) form – Program representation for sparse data-flow – Conversion to and from SSA Next Time – Reuse optimizations CIS570 Lecture 7 Static Single Assignment Form 2 Data Dependence Definition – Data dependences are constraints on the order in which statements may be executed Types of dependences – Flow (true) dependence: s1 writes memory that s2 later reads (RAW) – Anti-dependence: s1 reads memory that s2 later writes (WAR) – Output dependences: s1 writes memory that s2 later writes (WAW) – Input dependences: s1 reads memory that s2 later reads (RAR) True dependences – Flow dependences represent actual flow of data False dependences – Anti- and output dependences reflect reuse of memory, not actual data flow; can often be eliminated CIS570 Lecture 7 Static Single Assignment Form 3 1
- 2. Example s1 a = b; flow s2 b = c + d; anti s3 e = a + d; output s4 b = 3; input s5 f = b * 2; CIS570 Lecture 7 Static Single Assignment Form 4 Representing Data Dependences Implicitly – Use variable defs and uses – Pros: simple – Cons: hides data dependence (analyses must find this info) Def-use chains (du chains) – Link each def to its uses – Pros: explicit; therefore fast – Cons: must be computed and updated, consumes space Alternate representations – e.g., Static single assignment form (SSA), dependence flow graphs (DFG), value dependence graphs (VDG) CIS570 Lecture 7 Static Single Assignment Form 5 2
- 3. DU Chains Definition – du chains link each def to its uses Example s1 a = b; s2 b = c + d; du chain s3 e = a + d; s4 b = 3; s5 f = b * 2; CIS570 Lecture 7 Static Single Assignment Form 6 UD Chains Definition – ud chains link each use to its defs Example s1 a = b; s2 b = c + d; ud chain s3 e = a + d; s4 b = 3; s5 f = b * 2; CIS570 Lecture 7 Static Single Assignment Form 7 3
- 4. Role of Alternate Program Representations Process Original Code (RTL) Optimized Code (RTL) T1 T2 SSA Code1 SSA Code2 SSA Code3 Advantage – Allow analyses and transformations to be simpler & more efficient/effective Disadvantage – May not be “executable” (requires extra translations to and from) – May be expensive (in terms of time or space) CIS570 Lecture 7 Static Single Assignment Form 8 Static Single Assignment (SSA) Form Idea – Each variable has only one static definition – Makes it easier to reason about values instead of variables – Similar to the notion of functional programming Transformation to SSA – Rename each definition – Rename all uses reached by that assignment Example v := ... v0 := ... ... := ... v ... ... := ... v0 ... v := ... v1 := ... ... := ... v ... ... := ... v1 ... What do we do when there’s control flow? CIS570 Lecture 7 Static Single Assignment Form 9 4
- 5. SSA and Control Flow Problem – A use may be reached by several definitions 1 2 v := ... 3 v := ... 1 4 ...v... 2 v0 :=... 3 v1 :=... 4 ...v?... CIS570 Lecture 7 Static Single Assignment Form 10 SSA and Control Flow (cont) Merging Definitions – φ-functions merge multiple reaching definitions Example 1 2 v0 :=... 3 v1 :=... 4 v2 := φ(v0,v1) ...v2... CIS570 Lecture 7 Static Single Assignment Form 11 5
- 6. Another Example 1 v := 1 1 v0 := 1 2 v := v+1 v1 := φ(v0,v2) 2 v2 := v1+1 CIS570 Lecture 7 Static Single Assignment Form 12 SSA vs. ud/du Chains SSA form is more constrained Advantages of SSA – More compact – Some analyses become simpler when each use has only one def – Value merging is explicit – Easier to update and manipulate? Furthermore – Eliminates false dependences (simplifying context) for (i=0; i<n; i++) A[i] = i; Unrelated uses of i are given for (i=0; i<n; i++) different variable names print(foo(i)); CIS570 Lecture 7 Static Single Assignment Form 13 6
- 7. SSA vs. ud/du Chains (cont) Worst case du-chains? switch (c1) { case 1: x = 1; break; case 2: x = 2; break; case 3: x = 3; break; } x4 = φ(x1, x2, x3) switch (c2) { case 1: y1 = x; break; case 2: y2 = x; break; case 3: y3 = x; break; case 4: y4 = x; break; } m defs and n uses leads to m×n du chains CIS570 Lecture 7 Static Single Assignment Form 14 Transformation to SSA Form Two steps – Insert φ-functions – Rename variables CIS570 Lecture 7 Static Single Assignment Form 15 7
- 8. Where Do We Place φ-Functions? Basic Rule – If two distinct (non-null) paths x→z and y→z converge at node z, and nodes x and y contain definitions of variable v, then a φ-function for v is inserted at z x v1 :=... y v2 :=... z v3 := φ(v1,v2) ...v3... CIS570 Lecture 7 Static Single Assignment Form 16 Approaches to Placing φ-Functions Minimal – As few as possible subject to the basic rule Briggs-Minimal – Same as minimal, except v must be live across some edge of the CFG v = v = Briggs Minimal will not place a = v = v φ function in this case because v is not live across any CFG edge no uses of v CIS570 Lecture 7 Static Single Assignment Form 17 8
- 9. Approaches to Placing φ-Functions (cont) Briggs-Minimal – Same as minimal, except v must be live across some edge of the CFG Pruned – Same as minimal, except dead φ-functions are not inserted v = v = Will Pruned place a φ function at = v = v this merge? no uses of v What’s the difference between Briggs Minimal and Pruned SSA? CIS570 Lecture 7 Static Single Assignment Form 18 Briggs Minimal vs. Pruned v = v = = v = v v2= φ(v0,v1) = v2 Briggs Minimal will add a φ function because v is live across the blue edge, but Pruned = φ(v2,v3) SSA will not because the φ function is dead Why would we ever use Briggs Minimal instead of Pruned SSA? CIS570 Lecture 7 Static Single Assignment Form 19 9
- 10. Machinery for Placing φ-Functions entry Recall Dominators – d dom i if all paths from entry to node i include d d d dom i – d sdom i if d dom i and d≠i Dominance Frontiers i – The dominance frontier of a node d is the set of nodes that are “just barely” not dominated by d; i.e., the set of nodes n, such that – d dominates a predecessor p of n, and – d does not strictly dominate n – DF(d) = {n | ∃p∈pred(n), d dom p and d !sdom n} Notational Convenience – DF(S) = ∪s∈S DF(s) CIS570 Lecture 7 Static Single Assignment Form 20 Dominance Frontier Example DF(d) = {n | ∃p∈pred(n), d dom p and d !sdom n} Nodes in Dom(5) Dom(5) = {5, 6, 7, 8} 1 DF(5) = {4, 5, 12, 13} 2 5 9 3 6 7 10 11 4 8 12 13 What’s significant about the Dominance Frontier? In SSA form, definitions must dominate uses CIS570 Lecture 7 Static Single Assignment Form 21 10
- 11. Dominance Frontier Example II DF(d) = {n | ∃p∈pred(n), d dom p and d !sdom n} Nodes in Dom(5) Dom(5) = {5, 6, 7, 8} 1 DF(5) = {4, 5, 13} 2 5 3 6 7 4 8 13 In this new graph, node 4 is the first point of convergence between the entry and node 5, so do we need a φ- function at node 13? CIS570 Lecture 7 Static Single Assignment Form 22 SSA Exercise 1 2 v3 :=... 7 3 4 8 v1 := ... 9 v2 := ... 5 10 v4:=φ(v1,v2) 6 v5:=φ(v3,v4) DF(8) = {10} DF(9) = {10} DF(2) = {6} DF(d) = {n | ∃p∈pred(n), d dom p and d !sdom n} DF({8,9}) = {10} DF(10) = {6} DF({2,6,8,9,10}) = {6,10} CIS570 Lecture 7 Static Single Assignment Form 23 11
- 12. Dominance Frontiers Revisited Suppose that node 3 defines variable x x ∈ Def(3) DF(3) = {5} 1 2 3 4 5 6 Do we need to insert a φ- function for x anywhere else? Yes. At node 6. Why? CIS570 Lecture 7 Static Single Assignment Form 24 Dominance Frontiers and SSA Let – DF1(S) = DF(S) – DFi+1(S) = DF(S ∪ DFi(S)) Iterated Dominance Frontier – DF∞(S) Theorem – If S is the set of CFG nodes that define variable v, then DF∞(S) is the set of nodes that require φ-functions for v CIS570 Lecture 7 Static Single Assignment Form 25 12
- 13. Algorithm for Inserting φ-Functions for each variable v WorkList ← ∅ EverOnWorkList ← ∅ AlreadyHasPhiFunc ← ∅ for each node n containing an assignment to v Put all defs of v on the worklist WorkList ← WorkList ∪ {n} EverOnWorkList ← WorkList while WorkList ≠ ∅ Remove some node n from WorkList for each d ∈ DF(n) if d ∉ AlreadyHasPhiFunc Insert at most one φ function per node Insert a φ-function for v at d AlreadyHasPhiFunc ← AlreadyHasPhiFunc ∪ {d} if d ∉ EverOnWorkList Process each node at most once WorkList ← WorkList ∪ {d} EverOnWorkList ← EverOnWorkList ∪ {d} CIS570 Lecture 7 Static Single Assignment Form 26 Variable Renaming Basic idea – When we see a variable on the LHS, create a new name for it – When we see a variable on the RHS, use appropriate subscript Easy for straightline code x= x0 = =x = x0 x= x1 = =x = x1 Use a stack when there’s control flow – For each use of x, find the definition of x that dominates it x= x0 = Traverse the dominance tree =x = x0 CIS570 Lecture 7 Static Single Assignment Form 27 13
- 14. Dominance Tree Example The dominance tree shows the dominance relation 1 1 2 5 9 2 4 5 13 9 12 3 6 7 10 11 3 6 8 7 10 11 4 8 12 Dominance Tree 13 CFG CIS570 Lecture 7 Static Single Assignment Form 28 Variable Renaming (cont) Data Structures – Stacks[v] ∀v Holds the subscript of most recent definition of variable v, initially empty – Counters[v] ∀v Holds the current number of assignments to variable v; initially 0 p Auxiliary Routine po sh 1 procedure GenName(variable v) pu i := Counters[v] 2 4 5 13 9 12 push i onto Stacks[v] Counters[v] := i + 1 3 6 8 7 10 11 Use the Dominance Tree to remember the most recent definition of each variable CIS570 Lecture 7 Static Single Assignment Form 29 14
- 15. Variable Renaming Algorithm procedure Rename(block b) Call Rename(entry-node) for each φ-function p in b GenName(LHS(p)) and replace v with vi, where i=Top(Stack[v]) for each statement s in b (in order) for each variable v ∈ RHS(s) replace v by vi, where i = Top(Stacks[v]) for each variable v ∈ LHS(s) GenName(v) and replace v with vi, where i=Top(Stack[v]) for each s ∈ succ(b) (in CFG) j ← position in s’s φ-function corresponding to block b for each φ-function p in s Φ( , , ) replace the j th operand of RHS(p) by v , where i = Top(Stack[v]) i for each s ∈ child(b) (in DT) Rename(s) Recurse using Depth First Search for each φ-function or statement t in b for each vi ∈ LHS(t) Unwind stack when done with this node Pop(Stack[v]) CIS570 Lecture 7 Static Single Assignment Form 30 Transformation from SSA Form Proposal – Restore original variable names (i.e., drop subscripts) – Delete all φ-functions Complications x0 = − What if versions get out of order? x1 = (simultaneously live ranges) = x0 = x1 Alternative − Perform dead code elimination (to prune φ-functions) − Replace φ-functions with copies in predecessors − Rely on register allocation coalescing to remove unnecessary copies CIS570 Lecture 7 Static Single Assignment Form 31 15
- 16. Concepts Data dependences – Three kinds of data dependences – du-chains Alternate representations SSA form Conversion to SSA form – φ-function placement – Dominance frontiers – Variable renaming – Dominance trees Conversion from SSA form CIS570 Lecture 7 Static Single Assignment Form 32 Next Time Assignments – Project proposals due Lecture – Reuse optimizations CIS570 Lecture 7 Static Single Assignment Form 33 16