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Compiler Baojian Hua bjhua@ustc.edu.cn Value Numbering Compiler Baojian Hua bjhua@ustc.edu.cn.

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Presentation on theme: "Compiler Baojian Hua bjhua@ustc.edu.cn Value Numbering Compiler Baojian Hua bjhua@ustc.edu.cn."— Presentation transcript:

1 Compiler Baojian Hua bjhua@ustc.edu.cn
Value Numbering Compiler Baojian Hua

2 Motivation a = x + y b = x + y x = m + n a = x + y b = x + y c = x + y
d = x + y e = x f = y g = e + f a = x + y b = x + y x = m + n c = x + y d = x + y e = x f = y g = e + f a x has been re-assigned! c c

3 Motivation a = x + y b = x + y x = m + n c = x + y d = x + y e = x
2 1 a = x + y b = x + y x = m + n c = x + y d = x + y e = x f = y g = e + f 2 1 x 1 y 5 3 4 2 x+y 2 a 6 5 1 2 b 3 6 5 1 m 4 n 5 5 5 m+n 5 x 1 1 What’s a good data structure for this? 6 5 1

4 Value Numbering (VN) Algorithm
The algorithm is in Tiger 17.4 This is a deep semantic property of expressions In deciding whether “x+y” has been calculated and thus can be avoided, one can NOT just search the expression (by syntax), but some semantic property (here, the value number of “x+y”) x+y == a+b <==> V(x)=V(a) /\ V(y)=V(b) A form of abstract interpretation

5 VN vs CSE The goals of both of the two algorithms is to perform redundancy elimination how to detect redundancy: VN: value number CSE: available expression analysis The key difference is the viewpoint to define “redundancy” VN: semantics CSE: syntax

6 VN vs CSE VN: Yes! CSE: No! x = a+b t = a s = b y = t+s
But research has revealed that each of them come with pros and cons. So most compilers contain both of them. z = a+b h = a+b Is this a redundancy computation? VN CSE

7 What about control flow?
VN can be extended to extended basic block (EBB) easily (recall that EBBs form a local tree): Do a preorder walk of the EBB tree; Maintain a scoped table along the way (just as we did with the scoped declarations of variables when we discuss elaboration) a = x + y b = x + y x = 2 a = 3 a? d = x + y a? b? Available expression + Reaching expression

8 Example a = x + y x = 2 b = x + y a = 3 a d = x + y x 1 y 2 x+y 2 a 2
x 1 y 2 x+y 2 a = x + y a 2 b b = x + y x = 2 a = 3 a d = x + y

9 Example a = x + y x = 2 b = x + y a = 3 d = x + y x 1 y 2 x+y 2 a 3 x
x 1 y 2 x+y 2 a = x + y a 3 x 4 a b = x + y x = 2 a = 3 d = x + y

10 GVN: Global Value Numbering
It’s hard to do global value numbering for general CFG But for SSA, there are good algorithms: 1988: partition-based algorithm Alpern ‘88 1997: dominator-based algorithm Kooper ‘97 2004: GVN-PRE implemented in GCC

11 Partition-based GVN

12 Partition-based GVN L1: a = x + y L2: L3: x = x + y x = x + y L4:
d = x + y

13 Partition-based GVN L1: Key observations:
Variables defined with different operators can not be equal (arity), thus we group them in different congruent classes; for variables with same definition operator, they are equal iff the corresponding operands are from the same congruent classes. a = x + y L2: L3: x1 = x + y x2 = x + y L4: x3=ϕ(x1, x2) d = x3 + y

14 Partition-based GVN L1: Initial partitions:
{a, x1, x2, d} {x3} {x} {y} this partition is based on the expression operators. a = x + y L2: L3: x1 = x + y x2 = x + y a = x+y x1 = x+y x2 = x+y d = x3+y x3 = ϕ(x1, x2) x = ? y = ? L4: x3=ϕ(x1, x2) d = x3 + y In the same set <==> take good chance of equality.

15 Partition-based GVN L1: {a, x1, x2, d} {x3} {x} {y} a = x + y
Now, consider {x}: There are 3 uses of x, but not x3! L2: L3: x1 = x + y x2 = x + y a = x+y x1 = x+y x2 = x+y d = x3+y x3 = ϕ(x1, x2) x = ? y = ? L4: x3=ϕ(x1, x2) d = x3 + y

16 Partition-based GVN L1: {a, x1, x2, d} {x3} {x} {y} a = x + y
Final partitions: {a, x1, x2} {d} {x3} {x} {y} L2: L3: x1 = x + y x2 = x + y a = x+y x1 = x+y x2 = x+y d = x3+y x3 = ϕ(x1, x2) x = ? y = ? L4: x3=ϕ(x1, x2) d = x3 + y a a a As “a” dominates both x1 and x2, so all uses of x1 and x2 can be replaced by “a”.

17 Partition-based GVN Algorithm
P = all initial partitions (a queue of sets) while (!empty(P)) S = deQueue(P) // suppose S is an array n = arity(S[0]) // example: + ==> n == 2 for (i=0 to n-1) use_v[i] = S[0][i]; // calculate use sets T = {S[0]} foreach (x \in s[1]…s[k-1]) // k=|s| if (use[x_i]!=use_v[i]) break; T \/= {x} if (T!=S) enQueue (T); enQueue (S-T);

18 Example L1: i = 1 j = 1 L1: i1 = 1 j1 = 1 L2: L2: i3 = ϕ (i1, i2)
j3 = ϕ (j1, j2) L4: L4: L3: L3: i = i+1 j = j+1 i = i+3 j = j+3 i4 = i3+1 j4 = j3+1 i5 = i3+3 j5 = j3+3 L5: L5: i2 = ϕ (i4, i5) j2 = ϕ (j4, j5) CFG SSA

19 Example L1: i1 = 1 j1 = 1 {i1, j1} {i3, j3} {i4, j4, i5, j5} {i2, j2}
// Be partitioned into: {i1, j1} {i3, j3} {i4, j4} {i5, j5} {i2, j2} L2: i3 = ϕ (i1, i2) j3 = ϕ (j1, j2) L4: L3: i4 = i3+1 j4 = j3+1 i5 = i3+3 j5 = j3+3 L5: i2 = ϕ (i4, i5) j2 = ϕ (j4, j5) SSA

20 Example L1: i = 1 j = 1 L1: i1 = 1 j1 = 1 L2: L2: i3 = ϕ (i1, i2)
j3 = ϕ (j1, j2) L4: L4: L3: L3: i = i+1 j = j+1 i = i+3 j = j+3 i4 = i3+1 j4 = j3+1 i5 = i3+3 j5 = j3+3 L5: L5: i2 = ϕ (i4, i5) j2 = ϕ (j4, j5) CFG SSA

21 Dominator-based GVN

22 Dominator-based GVN VN can be extended to GVN on dominator trees:
Do a (what order?) walk of the dominator tree; Maintain a scoped table along the way (just as we did with the scope declarations of variables) for each block B, do VN as before, for a successor S of B, modify the ϕ arguments L1: 2 1 a = x + y L2: L3: x1 = x + y x2 = x + y 2 2 1 1 L4: 2 2 L1 2 x3=ϕ(x1, x2) d = x3 + y 3 2 1 L2 L4 L3

23 Another Example VN can be extended to GVN on dominator trees:
Do a (what order?) walk of the dominator tree; Maintain a scoped table along the way (just as we did with the scope declarations of variables) for each block B, do VN as before, for a successor S of B, modify the ϕ arguments. L1: 2 1 a = x + y L2: L3: b = x + y x1 = 2 a1 = 3 3 3 2 1 4 4 L4: L1 5 3 x2=ϕ(x, x1) d = x2 + y 6 5 1 L2 L4 L3


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