topic 16: static single assignment -...
TRANSCRIPT
Topic 16: Static Single Assignment
Compiler Design
Prof. Hanjun Kim
CoreLab (Compiler Research Lab)
POSTECH
1
Def-Use Chains, Use-Def Chains
• Many optimization need to find all use-sites for each definition, and all definition-sites for each use
• Constant propagation: refer to the definition-site of the unique reaching definition
• Copy propagation, common sub-expression elimination, …
• 𝑑𝑒𝑓 − 𝑢𝑠𝑒 𝑐ℎ𝑎𝑖𝑛• For each definition 𝑑 of 𝑟, list of pointers to all uses of 𝑟
that 𝑑 reaches
• 𝑢𝑠𝑒 − 𝑑𝑒𝑓 𝑐ℎ𝑎𝑖𝑛• For each use 𝑢 of 𝑟, list of pointers to all definitions of 𝑟
that reach 𝑢
2
Node OUT IN
1 1
2 1,2 1
3 1,2 1,2
4 1,4 1,2,4
5 1,4 1,4
6 1,4,6 1,4
7 4,6,7 1,4,6
8 4,6,7 4,6,7
r1 = 5
r3 = 1
Br r3>r1, 6
r3 = r3 + 1
1:
2:
3:
4:
goto 35:
r4 = 10
r1 = r1 + r4
M[r3] = r1
6:
7:
8:
Def-Use Chains, Use-Def Chains
Reaching Definition
Static Single Assignment
• Static Single Assignment (SSA)• Improvement on def-use chains
• Each register has only one definition in program
• For each use 𝑢 of 𝑟, only one definition of 𝑟 reaches 𝑢
4
r1 = 5
r1 = r1 + 1
r2 = r1 + 1 r3 = r1 - 1
r1 = 5
r4 = r1 + 1
r2 = r4 + 1 r3 = r4 - 1
Why SSA?
• Static Single Assignment Advantages• Dataflow analysis and code optimization made simpler
• Variables have only one definition – no ambiguity
• Dominator information is encoded in the assignment
• Less space required to represent def-use chain
• For each variable, space is proportional to uses * defs
• Defs = 1 in SSA
• Eliminates unnecessary relationshipsfor i = 1 to N do A[i] = 0
for i = 1 to N do A[i] = 0
• No reason why both loops should be forced to use same register to hold index register
• SSA renames second i to a new register which may lead to better register allocation / optimization
5
Phi Function
6
r1 = 5
r2 = r1 + 1
r3 = r1 + 1 r3 = r1 - 1
r4 = r3 * 4
r1 = 5
r2 = r1 + 1
r3 = r1 + 1 r3’ = r1 - 1
r4 = ? * 4
Phi Function
• 𝑟 = ∅(𝑟′, 𝑟")• Set of move operations on each incoming edge
• It enables the use of 𝑟 to be reached by exactly one definition of 𝑟
• 𝑟 = 𝑟′ if control enters from left
• 𝑟 = 𝑟" if control enters from right
7
Phi Function
8
r1 = 5
r2 = r1 + 1
r3’ = r1 + 1 r3” = r1 - 1
r3=𝜙(r3’,r3”)r4 = r3 * 4
Conversion to SSA Form
• Where to insert a 𝜙-function?• Insert a 𝜙-function for a register 𝑟 at node 𝑧 of the flow graph if ALL
of the following are true:
1. There is a block 𝑥 contacting a definition of 𝑟
2. There is another block 𝑦 (𝑦 ≠ 𝑥) contacting a definition of 𝑟
3. There is a non-empty path 𝑃𝑥𝑧 of edges from 𝑥 to 𝑧
4. There is a non-empty path 𝑃𝑦𝑧 of edges from 𝑦 to 𝑧
5. Paths 𝑃𝑥𝑧 and 𝑃𝑦𝑧 do not share any node other than 𝑧
6. The node 𝑧 does not appear within both 𝑃𝑥𝑧 and 𝑃𝑦𝑧 prior to the end, though it may appear in one or the other
• Assume CFG entry node contains implicit definition of every variable
• The variable is either a formal parameter or uninitialized
• 𝜙-function is considered as a definition
9
Conversion to SSA Form
• Iterated path-convergence algorithm
while(there are nodes 𝑥, 𝑦, 𝑧 satisfying conditions 1-6)
&& (𝑧 does not contain a 𝜙-function for 𝑟)do
insert 𝑟 = 𝜙(𝑟, 𝑟, … , 𝑟) (one per predecessor) at node 𝑧
• Costly to compute every triple of nodes 𝑥, 𝑦, 𝑧 and every path from 𝑥 and 𝑦
• Any solution?• Definitions dominate uses – it will simplify computation!!
10
Dominance Frontier
• Dominance Frontier of node 𝑥• Set of all nodes 𝑤 such that 𝑥 dominates a predecessor
of 𝑤, but does not strictly dominate 𝑤
• Strict Domination• 𝑥 strictly dominates 𝑤 if 𝑥 dominates 𝑤 and 𝑥 ≠ 𝑤
• Dominance Frontier Criterion• Whenever node 𝑥 contains definition of some register 𝑟,
then any node 𝑧 in the dominance frontier of 𝑥 needs 𝜙-function for 𝑟
• Iterated Dominance Frontier• Since a 𝜙-function itself is a definition, we must
repeatedly apply the dominance frontier
11
Dominance Frontier Example
12
1
2
3
4
5
6
8
7
9
11
10
Which nodes are in the dominance frontier of node 5?
Strictly Dominated
Dominance Frontier
Dominance Frontier Computation
𝐷𝐹 𝑛 = 𝐷𝐹𝑙𝑜𝑐𝑎𝑙 𝑛 ∪ ∪𝑐∈𝑐ℎ𝑖𝑙𝑑𝑟𝑒𝑛 𝑛 𝐷𝐹𝑢𝑝[𝑐]
• 𝐷𝐹[𝑛]: dominance frontier of 𝑛
• 𝐷𝐹𝑙𝑜𝑐𝑎𝑙[𝑛]: successors of 𝑛 in CFG that are not strictly dominated by 𝑛
• 𝐷𝐹𝑢𝑝[𝑐]: nodes in dominance frontier of 𝑐 that are not strictly dominated by 𝑐’s immediate dominator
• 𝑐ℎ𝑖𝑙𝑑𝑟𝑒𝑛 𝑛 : the nodes whose idom is n
• Use dominator tree• Work bottom up in dominator tree
13
SSA Example
14
Node 𝐷𝑜𝑚[𝑛] 𝐼𝐷𝑜𝑚[𝑛]
1 1
2 1,2 1
3 1,2,3 2
4 1,2,3,4 3
5 1,2,3,4,5 4
6 1,2,3,4,6 4
7 1,2,3,4,5,7 5
8 1,2,3,4,5,7,8 7
9 1,2,3,4,5,9 5
10 1,2,3,4,5,9,10 9
11 1,2,3,4,5,11 5
r1 = 1
r2 = 1
r3 = 0
1:
2:
3:
br r2<205:
r2 = r1
r3 = r3 + 1
r2 = r37:
8:
9:
r3 = r3 + 210:
br r3<100
4:
11:
return r26:
Dominator Analysis
• If 𝑑 dominates each of the 𝑝𝑖, then 𝑑 dominates 𝑛
• If 𝑑 dominates 𝑛, then 𝑑 dominates each of the 𝑝𝑖• 𝐷𝑜𝑚[𝑛] = set of nodes that dominate node n
• 𝑁 = set of all nodes
• Computation:1. 𝐷𝑜𝑚 𝑠0 = 𝑠02. for 𝑛 ∈ 𝑁 − {𝑠0} do 𝐷𝑜𝑚 𝑛 = 𝑁
3. While (changes to any 𝐷𝑜𝑚 𝑛 occur) dofor 𝑛 ∈ 𝑁 − 𝑠0 do𝐷𝑜𝑚 𝑛 = 𝑛 ∪ ( ∩ 𝑝∈𝑝𝑟𝑒𝑑 𝑛 𝐷𝑜𝑚 𝑝 )
15
SSA Example
16
r1 = 1
r2 = 1
r3 = 0
1:
2:
3:
br r2<205:
r2 = r1
r3 = r3 + 1
r2 = r37:
8:
9:
r3 = r3 + 210:
br r3<100
4:
11:
return r26:
Dominator Tree:1
2
6
7
3
10
4
8
5
119
SSA Example
17
Dominance Frontier:
1
2
6
7
3
10
4
8
5
119
Node 𝐷𝐹𝑙𝑜𝑐𝑎𝑙[𝑛] 𝐷𝐹𝑢𝑝 [𝑛] 𝐷𝐹 [𝑛]
1 - - -
2 - - -
3 - - -
4 - - 4
5 - 4 4
6 - - -
7 - - 11
8 11 11 11
9 - - 11
10 11 11 11
11 4 4 4
SSA Example
18
r1 = 1
r2 = 1
r3 = 0
1:
2:
3:
br r2<205:
r2 = r1
r3 = r3 + 1
r2 = r37:
8:
9:
r3 = r3 + 210:
br r3<100
4:
r2 = 𝜙(r2,r2)r3 = 𝜙(r3,r3)
11:
return r26:
Insert 𝜙-functions :
Node 𝐷𝐹 [𝑛]
1 -
2 -
3 -
4 4
5 4
6 -
7 11
8 11
9 11
10 11
11 4
Dominance Frontier CriterionWhenever node 𝑥 contains definition of some register 𝑟, any node 𝑧 in the dominance frontier of 𝑥 needs 𝜙-function for 𝑟
SSA Example
19
r1 = 1
r2 = 1
r3 = 0
1:
2:
3:
br r2<205:
r2 = r1
r3 = r3 + 1
r2 = r37:
8:
9:
r3 = r3 + 210:
r2 = 𝜙(r2,r2)r3 = 𝜙(r3,r3)br r3<100
4:
r2 = 𝜙(r2,r2)r3 = 𝜙(r3,r3)
11:
return r26:
Insert 𝜙-functions :
Node 𝐷𝐹 [𝑛]
1 -
2 -
3 -
4 4
5 4
6 -
7 11
8 11
9 11
10 11
11 4
Dominance Frontier CriterionWhenever node 𝑥 contains definition of some register 𝑟, any node 𝑧 in the dominance frontier of 𝑥 needs 𝜙-function for 𝑟
SSA Example
• Rename Variables1. Traverse dominator tree, renaming different definitions
of 𝑟 to 𝑟1, 𝑟2, 𝑟3, …
2. Rename each regular use of 𝑟 to most recent definition of 𝑟
3. Rename 𝜙-function arguments with each incoming edge’s unique definition
20
SSA Example
21
r1 = 1
r2 = 1
r3 = 0
1:
2:
3:
br r21<205:
r22 = r1
r32 = r31+1
r23 = r317:
8:
9:
r33 = r31+210:
r21=𝜙(r2,r24)r31=𝜙(r3,r34)br r31<100
4:
r24= 𝜙 (r22,r23)
r34= 𝜙 (r32,r33)
11:
return r216:
Rename Variables:
SSA Dominance Property
• Dominance property of SSA form:• Definitions dominate uses
• If 𝑥 is used in non-𝜙-statement in node 𝑛, then definition of 𝑥 dominates 𝑛
• If 𝑥 is 𝑖𝑡ℎ argument of 𝜙-function in node 𝑛, then definition of 𝑥 dominates 𝑖𝑡ℎ predecessor of 𝑛
22
SSA Dead Code Elimination
Given d: t = x op y
• t is live at end of node d if there exists a path from the end of d to use of t that does not go through definition of t
• If program is not in SSA form• Liveness analysis is required to determine if t live at the
end of d
• If program is in SSA form• There is only one definition of t• If there exists use of t, there is a path from the end of d
to use• Every use has a unique definition
• t is live at end of node d if t is used at least once
23
SSA Dead Code Elimination
• Algorithmwhile (for each temporary t with no uses &&
statement defining t has no side-effect)
do
delete statement definition t
24
r1 = 5
r2 = 10
1:
2:
br r3 > r23:
r21 = r2+15
r41 = r3+1
r42 = r34:
5:
6:
r22=𝜙(r21,r2)r4=𝜙(r41,r42)
8:
7:
M[r4]=r22
SSA Simple Constant Propagation
Given d: t = c, c is constant Given u: x = t op b
• If program is not in SSA form• Reaching definition analysis is required
• Use of t in u may be replaced by c if d reaches u and no other definition of t reaches u
• If program is in SSA form• d is only one definition of t and definitions dominate
uses, so d reaches u without any other definition • Any use of t can be replaced by c
• Any 𝜙-function of form v = 𝜙(𝑐1, 𝑐2, … , 𝑐𝑛) where 𝑐𝑖 = 𝑐 can be replaced by v = c
25
SSA Simple Constant Propagation
26
r2 = 102:
br r3 > r23:
r21 = r2+15
r41 = r3+1
r42 = r34:
5:
6:
r22=𝜙(r21,r2)r4=𝜙(r41,r42)
8:
7:
M[r4]=r22
SSA Conditional Constant Propagation
• r2 always has value of 1
• Nodes 9, 10 never executed
• “simple” constant propagation algorithm
• Assumes nodes 9, 10 may be executed (through reaching definitions analysis)
• Cannot optimize use of r2 in node 5 since definitions 7 and 9 both reach 5
27
r1 = 1
r2 = 1
r3 = 0
1:
2:
3:
br r21<205:
r22 = r1
r32 = r31+1
r23 = r317:
8:
9:
r33 = r31+210:
r21=𝜙(r2,r24)r31=𝜙(r3,r34)br r31<100
4:
r24= 𝜙 (r22,r23)
r34= 𝜙 (r32,r33)
11:
return r216:
SSA Conditional Constant Propagation
Much smarter than “simple” constant propagation required
• Does not assume a node can execute until evidence exists
• Does not assume a register is non-constant unless evidence exists
Track runtime value of each register 𝑟 using lattice of values
• 𝑉 𝑟 =⊥ (bottom): compiler has seen no evidence that any assignment to 𝑟is ever executed
• 𝑉 𝑟 = 4: compiler has seen evidence that an assignment 𝑟 = 4 is executed, but has seen no evidence that 𝑟 is ever assigned to another value
• 𝑉 𝑟 = ⊤ (top): compiler has seen evidence that 𝑟 will have, at various time, two different values or some value that is not predictable at compile-time
Also
• All registers start at bottom of lattice
• New information can only move registers up in lattice
28
SSA Conditional Constant Propagation
Track executability of each node in 𝑁• 𝐸 𝑁 = 𝑓𝑎𝑙𝑠𝑒: compiler has seen no evidence that node 𝑁 can be executed
• 𝐸 𝑁 = 𝑡𝑟𝑢𝑒: compiler has seen evidence that node 𝑁can be executed
Initially,• 𝑉 𝑟 =⊥, for all register 𝑟
• 𝐸 𝑠0 = 𝑡𝑟𝑢𝑒, 𝑠0 is CFG start node
• 𝐸 𝑁 = 𝑓𝑎𝑙𝑠𝑒, for all CFG nodes 𝑁 ≠ 𝑠0
29
SSA Conditional Constant Propagation
Algorithm: apply following conditions until no more changes occur to 𝐸 or 𝑉 values
1. Given: register 𝑟 with no definition (formal parameter, uninitialized)Action: 𝑉 𝑟 =⊥
2. Given: executable node 𝐵 with only one successor 𝐶Action: 𝐸 𝐶 = 𝑡𝑟𝑢𝑒
3. Given: executable assignment 𝑟 = 𝑥 𝑜𝑝 𝑦, 𝑉 𝑥 = 𝑐1 and 𝑉 𝑦 = 𝑐2Action: 𝑉 𝑟 = 𝑐1 𝑜𝑝 𝑐2
4. Given: executable assignment 𝑟 = 𝑥 𝑜𝑝 𝑦, 𝑉 𝑥 = ⊤ or 𝑉 𝑦 = ⊤Action: 𝑉 𝑟 = ⊤
5. Given: executable assignment 𝑟 = 𝑀[. . ] or r = f(. . )Action: 𝑉 𝑟 = ⊤
6. Given: executable assignment 𝑟 = 𝜙(𝑥1, 𝑥2, … , 𝑥𝑛), 𝑉 𝑥𝑖 = 𝑐1, 𝑉 𝑥𝑗 = 𝑐2, and predecessors 𝑖 and 𝑗 are executableAction: 𝑉 𝑟 = ⊤
7. Given: executable assignment 𝑟 = 𝜙(𝑥1, 𝑥2, … , 𝑥𝑛), 𝑉 𝑥𝑖 = ⊤, and predecessor 𝑖 is executableAction: 𝑉 𝑟 = ⊤
30
SSA Conditional Constant Propagation
8. Given: executable assignment 𝑟 = 𝜙(𝑥1, 𝑥2, … , 𝑥𝑛), 𝑉 𝑥𝑖 = 𝑐𝑖, and predecessor 𝑖 is executable; and for all 𝑗 ≠ 𝑖, predecessor 𝑗 is not executable, or 𝑉 𝑥𝑗 =⊥,
or 𝑉 𝑥𝑗 = 𝑐𝑖Action: 𝑉 𝑟 = 𝑐𝑖
9. Given: executable branch 𝑏𝑟 𝑥 𝑏𝑜𝑝 𝑦, 𝐿1 (𝑒𝑙𝑠𝑒 𝐿2), 𝑉 𝑥 = ⊤ or 𝑉 𝑦 = ⊤Action: 𝐸 𝐿1 = 𝑡𝑟𝑢𝑒, 𝐸 𝐿2 = 𝑡𝑟𝑢𝑒
10.Given: executable branch 𝑏𝑟 𝑥 𝑏𝑜𝑝 𝑦, 𝐿1 (𝑒𝑙𝑠𝑒 𝐿2), 𝑉 𝑥 = 𝑐1 and 𝑉 𝑦 = 𝑐2Action: 𝐸 𝐿1 = 𝑡𝑟𝑢𝑒 or 𝐸 𝐿2 = 𝑡𝑟𝑢𝑒 depending on 𝑐1 𝑏𝑜𝑝 𝑐2
31
SSA Conditional Constant Propagation
Given 𝑉, 𝐸 values, program can be optimized as follows• If 𝐸 𝐵 = 𝑓𝑎𝑙𝑠𝑒, delete node 𝐵 from CFG
• If 𝑉 𝑟 = 𝑐, replace each use of 𝑟 by 𝑐, delete assignment to 𝑟
32
SSA Conditional Constant Propagation
33
r1 = 1
r2 = 1
r3 = 0
1:
2:
3:
br r21<205:
r22 = r1
r32 = r31+1
r23 = r317:
8:
9:
r33 = r31+210:
r21=𝜙(r2,r24)r31=𝜙(r3,r34)br r31<100
4:
r24= 𝜙 (r22,r23)
r34= 𝜙 (r32,r33)
11:
return r216:
𝑁 𝐸[𝑁]
1 T
2 F
3 F
4 F
5 F
6 F
7 F
8 F
9 F
10 F
11 F
𝑟 V[𝑟]
r1 ⊥
r2 ⊥
r21 ⊥
r22 ⊥
r23 ⊥
r24 ⊥
r3 ⊥
r31 ⊥
r32 ⊥
r33 ⊥
r34 ⊥
SSA Conditional Constant Propagation
34
r1 = 1
r2 = 1
r3 = 0
1:
2:
3:
br r21<205:
r22 = r1
r32 = r31+1
r23 = r317:
8:
9:
r33 = r31+210:
r21=𝜙(r2,r24)r31=𝜙(r3,r34)br r31<100
4:
r24= 𝜙 (r22,r23)
r34= 𝜙 (r32,r33)
11:
return r216:
𝑁 𝐸[𝑁]
1 T
2 T
3 T
4 T
5 F
6 F
7 F
8 F
9 F
10 F
11 F
𝑟 V[𝑟]
r1 1
r2 1
r21 ⊥
r22 ⊥
r23 ⊥
r24 ⊥
r3 0
r31 ⊥
r32 ⊥
r33 ⊥
r34 ⊥
SSA Conditional Constant Propagation
35
r1 = 1
r2 = 1
r3 = 0
1:
2:
3:
br r21<205:
r22 = r1
r32 = r31+1
r23 = r317:
8:
9:
r33 = r31+210:
r21=𝜙(r2,r24)r31=𝜙(r3,r34)br r31<100
4:
r24= 𝜙 (r22,r23)
r34= 𝜙 (r32,r33)
11:
return r216:
𝑁 𝐸[𝑁]
1 T
2 T
3 T
4 T
5 T
6 F
7 F
8 F
9 F
10 F
11 F
𝑟 V[𝑟]
r1 1
r2 1
r21 1
r22 ⊥
r23 ⊥
r24 ⊥
r3 0
r31 0
r32 ⊥
r33 ⊥
r34 ⊥
SSA Conditional Constant Propagation
36
r1 = 1
r2 = 1
r3 = 0
1:
2:
3:
br r21<205:
r22 = r1
r32 = r31+1
r23 = r317:
8:
9:
r33 = r31+210:
r21=𝜙(r2,r24)r31=𝜙(r3,r34)br r31<100
4:
r24= 𝜙 (r22,r23)
r34= 𝜙 (r32,r33)
11:
return r216:
𝑁 𝐸[𝑁]
1 T
2 T
3 T
4 T
5 T
6 F
7 T
8 T
9 F
10 F
11 T
𝑟 V[𝑟]
r1 1
r2 1
r21 1
r22 1
r23 ⊥
r24 1
r3 0
r31 0
r32 1
r33 ⊥
r34 1
SSA Conditional Constant Propagation
37
r1 = 1
r2 = 1
r3 = 0
1:
2:
3:
br r21<205:
r22 = r1
r32 = r31+1
r23 = r317:
8:
9:
r33 = r31+210:
r21=𝜙(r2,r24)r31=𝜙(r3,r34)br r31<100
4:
r24= 𝜙 (r22,r23)
r34= 𝜙 (r32,r33)
11:
return r216:
𝑁 𝐸[𝑁]
1 T
2 T
3 T
4 T
5 T
6 T
7 T
8 T
9 F
10 F
11 T
𝑟 V[𝑟]
r1 1
r2 1
r21 1
r22 1
r23 ⊥
r24 1
r3 0
r31 ⊤
r32 1
r33 ⊥
r34 1
SSA Conditional Constant Propagation
38
r1 = 1
r2 = 1
r3 = 0
1:
2:
3:
br r21<205:
r22 = r1
r32 = r31+1
r23 = r317:
8:
9:
r33 = r31+210:
r21=𝜙(r2,r24)r31=𝜙(r3,r34)br r31<100
4:
r24= 𝜙 (r22,r23)
r34= 𝜙 (r32,r33)
11:
return r216:
𝑁 𝐸[𝑁]
1 T
2 T
3 T
4 T
5 T
6 T
7 T
8 T
9 F
10 F
11 T
𝑟 V[𝑟]
r1 1
r2 1
r21 1
r22 1
r23 ⊥
r24 1
r3 0
r31 ⊤
r32 ⊤
r33 ⊥
r34 ⊤
SSA Conditional Constant Propagation
39
r1 = 1
r2 = 1
r3 = 0
1:
2:
3:
r22 = r1
r32 = r31+1
7:
8:
r21=𝜙(r2,r24)r31=𝜙(r3,r34)br r31<100
4:
r24= 𝜙 (r22,r23)
r34= 𝜙 (r32,r33)
11:
return r216:
𝑁 𝐸[𝑁]
1 T
2 T
3 T
4 T
5 T
6 T
7 T
8 T
9 F
10 F
11 T
𝑟 V[𝑟]
r1 1
r2 1
r21 1
r22 1
r23 ⊥
r24 1
r3 0
r31 ⊤
r32 ⊤
r33 ⊥
r34 ⊤
SSA Conditional Constant Propagation
40
r1 = 1
r2 = 1
r3 = 0
1:
2:
3:
r22 = 1
r32 = r31+1
7:
8:
r21 = 1
r31=𝜙(r3,r34)br r31<100
4:
r24 = 1
r34= 𝜙 (r32,r33)
11:
return 16:
𝑁 𝐸[𝑁]
1 T
2 T
3 T
4 T
5 T
6 T
7 T
8 T
9 F
10 F
11 T
𝑟 V[𝑟]
r1 1
r2 1
r21 1
r22 1
r23 ⊥
r24 1
r3 0
r31 ⊤
r32 ⊤
r33 ⊥
r34 ⊤
SSA Conditional Constant Propagation
41
r3 = 03:
r32 = r31+18:
r31=𝜙(r3,r34)br r31<100
4:
r34= 𝜙 (r32,r33)
11:
return 16:
𝑁 𝐸[𝑁]
1 T
2 T
3 T
4 T
5 T
6 T
7 T
8 T
9 F
10 F
11 T
𝑟 V[𝑟]
r1 1
r2 1
r21 1
r22 1
r23 ⊥
r24 1
r3 0
r31 ⊤
r32 ⊤
r33 ⊥
r34 ⊤
SSA Conditional Constant Propagation
42
r3 = 03:
r32 = r31+18:
r31=𝜙(r3,r32)br r31<100
4:
return 16:
𝑁 𝐸[𝑁]
1 T
2 T
3 T
4 T
5 T
6 T
7 T
8 T
9 F
10 F
11 T
𝑟 V[𝑟]
r1 1
r2 1
r21 1
r22 1
r23 ⊥
r24 1
r3 0
r31 ⊤
r32 ⊤
r33 ⊥
r34 ⊤