CMPUT680 - Winter 2006

CMPUT 680 - Compiler Design and Optimization

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CMPUT680 - Winter 2006

Topic G: Static Single-Assignment Form

José Nelson Amaralhttp://www.cs.ualberta.ca/~amaral/courses/680


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Reading Material

Chapter 19 of the “Tiger book” (with a grain of salt!!).

Bilardi, G., Pingali, K., “The Static Single Assignment Form and its Computation,” unpublished? (citeseer).Cytron, R., Ferrante, J., Rosen, B. K., Wegman, M. N., Zadeck, F. K., “An Efficient Method of Computing Static Single Assignment Form,” ACM Symposium on Principles of Programming Languages (PoPL), pp. 25-35, Austin, TX, Jan., 1989.Cytron, R., Ferrante, J., Rosen, B. K., Wegman, M. N., “Efficiently Computing Static Single Assignment Form and the Control Dependence Graph,” ACM Transactions on Programming Languages and Systems (TOPLAS), Vol. 13, No. 4, October, 1991, pp. 451-490.Sreedhar, V. C., Gao, G. R., “A Linear Time Algorithm for Placing -Nodes,” ACM Symposium on Principles of Programming Languages (PoPL), pp. 62-73, 1995.


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Static Single-Assignment Form

Each variable has only one definition in the program text.

This single static definition can be in a loop andmay be executed many times. Thus even in aprogram expressed in SSA, a variable can be

dynamically defined many times.


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Advantages of SSA

Simpler dataflow analysisNo need to use use-def/def-use chains, which requires NM space for N uses and M definitions

SSA form relates in a useful way with dominance structures. SSA simplifies algorithms that construct interference graphs.


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SSA Form in Control-Flow Path Merges

b M[x]a 0

if b<4

a b

c a + b

B1

B2

B3

B4

Is this code in SSA form?

No, two definitions of a appear inthe code (in B1 and B3)

How can we transform this codeinto a code in SSA form?

We can create two versions ofa, one for B1 and another for B3.


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b M[x]a1 0

if b<4

a2 b

c a? + b

B1

B2

B3

B4

But which version should weuse in B4 now?

We define a fictional function that

“knows” which control path was

taken to reach the basic block B4:

( )⎩⎨⎧

=B3 from B4at arrive weif a2

B2 from B4at arrive weif a1a2a1,φ


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b M[x]a1 0

if b<4

a2 b

a3 (a2,a1) c a3 + b

B1

B2

B3

B4

But which version should weuse in B4 now?

We define a fictional function that

“knows” which control path was

taken to reach the basic block B4:

( )⎩⎨⎧

=B3 from B4at arrive weif a2

B2 from B4at arrive weif a1a2a1,φ


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A Loop Examplea 0

b a+1c c+ba b*2if a < N

return

a1 0b0 undefc0 undef

a3 (a1,a2)b1 (b0,b2)c2 (c0,c1)b2 a3+1c1 c2+b2a2 b2*2if a < N

return

(b0,b2) is not necessary because b1 is never used. But the phase that generates

functions does not know it. Unnecessary functions are later

eliminated by dead code elimination.


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The Function

How can we implement a function that “knows”which control path was taken?

Answer 1: We don’t!! The function is used onlyto connect use to definitions during

optimization, but is never implemented.

Answer 2: If we must execute the function, we can implement it by inserting MOVE instructions

in all control paths.


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Criteria for Inserting Functions

We could insert one function for each variableat every join point (a point in the CFG with morethan one predecessor). But that would be wasteful.

What criteria should we use to insert a function for a variable a at node z of the CFG?

Intuitively, we should add a function if there are two definitions of a that can reach

the point z through distinct paths.


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Path Convergence Criterion

(Cytron-Ferrante/89)

Insert a function for a variable a at a node z ifall the following conditions are true:1. There is a block x that defines a2. There is a block y x that defines a3. There is a non-empty paths xz and yz4. Paths xz and yz don’t have any nodes in common other than z5. The node z does not appear within both xz and yz prior to the end, but it may appear in one or the other.

Note: The start node contains an implicit definition of every variable.


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-Candidates are Join Nodes

Notice that according to the path convergencecriterion, the node z that will receive the functionmust be a join node.

z is the first node that joins the paths Pxz and Pyz.


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Iterated Path-Convergence Criterion

The function itself is a definition of a. Therefore the path-convergence criterion

is a set of equations that must be satisfied.

while there are nodes x, y, z satisfying conditions 1-5 and z does not contain a function for ado insert a (a0, a1, …, an) at node z

This algorithm is extremely costly, because itrequires the examination of every triple of

nodes x, y, z and every path from x to z and from y to z.

Can we do better?


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The SSA Conversion Problem

For each variable x defined in a CFG G=(V,E), given the set of nodes S V that contain a definition for x, find the minimal set, J(S) of nodes that requires a (xi,xj) function.

By definition, the START node defines allthe variables, therefore S V, START S.

If we need to compute nodes for severalvariables, it may be efficient to precomputedata structures based on the CFG.


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Processing Time for SSA Conversion

The performance of an SSA conversion algorithmshould be measured by the processing time Tp,the preprocessing space Sp, and the query time Tq.

(Shapiro and Saint 1970): outline an algorithm

(Reif and Tarjan 1981): extend the Lengauer-Tarjan dominator algorithm to compute -nodes.

(Cytron et al. 1991): show that SSA conversion can use the idea of dominance frontiers, resulting on an O(|V|2) algorithm.

(Sreedhar and Gao, 1995): An O(|E|) algorithm, but in private commun. with Pingali in 1996 admits that it is in practice 5 times slower than Cytron et al.


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Processing Time for SSA Conversion

Bilardi, Pingali, 1999: present a generalized framework and a parameterized Augmented Dominator Tree (ADT) algorithm that allows for a space-time tradeoff. They show that Cytron et al. and Gao-Shreedhar are special cases of the ADT algorithm.

Bilardi and Pingali describe three strategies to compute-placement:

•Two-Phase Algorithms•Lock-Step Algorithms•Lazy Algorithms


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Two-Phase Algorithms

First build the entire Dominance Frontier Graph, then find the nodes reachable from S

SimpleDF Graph may be quite large

DF Computation

Reachability

DF Graph

J(S)

S

CFG


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Lock-Step Algorithms

Performs the reachability computation incrementallywhile the DF relation is computed.

DF ComputationReachability

J(S)

• Avoid storing the DF Graph.• Perform computations at all nodes of the graph, even though most are irrelevant• Inneficient when computing the -nodes for many variables.

CFG

S


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Lazy Algorithms

Lazily compute only the portion fo the DF Graph thatis needed. Carefully select a portion of the DF Graph to compute eagerly (before it is needed).

A Two-Phase Algorithmis an extreme case of a lazy algorithm.

DF Computation

Reachability

DF GraphSubGraph

J(S)

S

CFG


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Computing a Dominator Tree

(Lowry and Medlock, 1969): Introduce the problem and give an O(n4) algorithm.

(Lengauer and Tarjan, 1979): Give a complicated O(m(m.n)) algorithm [(m.n) is the inverse Ackermann’s function].

(Harel, 1985): Give a linear time algorithm.

(Alstrup, Harel and Thorup, 1997): Give a simpler version ofHarel’s algorithm.

(n: # of nodes; m: # of edges)


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Dominance Property of the SSA Form

In SSA form definitions dominate uses, i.e.:1. If x is used in a function in block n, then the definition of x dominates every predecessor of n.2. If x is used in a non- statement in block n, then the definition of x dominates n.


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The Dominance Frontier

A node x dominates a node w if every path from the startnode to w must go through x.

A node x strictly dominates a node w if x dominatesw and x w.

The dominance frontier of a node x is the set ofall nodes w such that x dominates a predecessor

of w, but x does not strictly dominates w.


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Example

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13

2

3

412

10 11

9

8

6 7

5

What is the dominance frontier of node 5?


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Example

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10 11

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6 7

5

First we must find all nodes that node 5 dominates.


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Example

A node w is in the dominance frontier of node 5 if 5 dominates a predecessor of w, but 5 does not strictlydominates w itself. What is the dominance frontier of 5?

1

13

2

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412

10 11

9

8

6 7

5


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Example

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412

10 11

9

8

6 7

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Example

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2

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412

10 11

9

8

6 7

5

DF(5) = {4, 5, 12, 13}



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Dominance Frontier Criterion

Dominance Frontier Criterion:If a node x contains a definition of variable a,then any node z in the dominance frontier ofx needs a function for a.

Can you think of an intuitive explanation for why a node in the dominance frontier of another node must be a join node?


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Example

1

13

2

3

412

10 11

9

8

6 7

5

If a node (12) is in the dominance frontier of

another node (5), than there must beat least two paths converging to (12).

These paths must be non-intersecting, andone of them (5,7,12)

must contain a node strictlydominated by (5).


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Dominator Tree

To compute the dominance frontiers, we first computethe dominator tree of the CFG.

There is an edge from node x to node y in the dominator tree if node x immediately dominates node y.

I.e., x dominates yx, and x does notdominate any other dominator of y.

Dominator trees can be computed using the Lengauer-Tarjan algorithm(1979).

See sec. 19.2 of Appel.


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Example: Dominator Tree

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8

6 7

5

1

3

6 7

10 11

4 5 12 92 13

8Control Flow Graph

Dominator Tree


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Local Dominance Frontier

Cytron-Ferrante define the local dominance frontier of a node n as:

DFlocal[n] = successors of n in the CFG that are not strictly dominated by n


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Example: Local Dominance Frontier

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2

3

412

10 11

9

8

6 7

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Control Flow Graph

In the example, what arethe local dominance

frontier of nodes 5, 6 and 7?

DFlocal[5] = DFlocal[6] = {4,8}DFlocal[7] = {8,12}


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Dominance Frontier Inherited From Its

Children

The dominance frontier of a node n is formed by itslocal dominance frontier plus nodes that are passed up by the children of n in the dominator tree.

The contribution of a node c to its parents dominance frontier is defined as [Cytron-Ferrante, 1991]:

DFup[c] = nodes in the dominance frontier of c that are not strictly dominated by the immediate dominator of c


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1

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2

3

412

10 11

9

8

6 7

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Control Flow Graph

In the example, what arethe contributions of nodes6, 7, and 8 to its parent

dominance frontier?

First we computethe DF and the immediatedominator of each node:DF[6] = {4,8}, idom(6)= 5DF[7] = {8,12}, idom(7)= 5DF[8] = {5,13}, idom(8)= 5


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1

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2

3

412

10 11

9

8

6 7

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Control Flow Graph

First we computethe DF and the immediatedominator of each node:DF[6] = {4,8}, idom(6)= 5DF[7] = {8,12}, idom(7)= 5DF[8] = {5,13}, idom(8)= 5

Now we check for the DFup condition: DFup[6] = {4}DFup[7] = {12}DFup[8] = {5,13}


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A note on implementation

We want to represent these sets efficiently:DF[6] = {4,8} DF[7] = {8,12} DF[8] = {5,13}

If we use bitvectors to represent these sets:DF[6] = 0000 0001 0001 0000DF[7] = 0001 0001 0000 0000DF[8] = 0010 0000 0010 0000


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Strictly Dominated Sets

1

3

6 7

10 11

4 5 12 92 13

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Dominator Tree

We can also represent the strictly dominated sets as vectors:SD[1] = 0011 1111 1111 1100SD[2] = 0000 0000 0000 1000SD[5] = 0000 0001 1100 0000SD[9] = 0000 1100 0000 0000


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A note on implementation

DFup[c] = nodes in the dominance frontier of c that are not strictly dominated by the immediate dominator of c

If we use bitvectors to represent these sets:DF[6] = 0000 0001 0001 0000DF[7] = 0001 0001 0000 0000DF[8] = 0010 0000 0010 0000

SD[5] = 0000 0001 1100 0000

DFup[c] = DF[6] ^ ~SD[5]


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Dominance Frontier Inherited From Its

Children

The dominance frontier of a node n is formed by itslocal dominance frontier plus nodes that are passed up by the children of n in the dominator tree. Thus the dominance frontier of a node n is defined as [Cytron-Ferrante, 1991]:

[ ] [ ][ ]

[ ]cDFnDFnDF upnDTchildrenc

local ∈∪∪=


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1

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2

3

412

10 11

9

8

6 7

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Control Flow Graph

What is DF[5]?Remember that:

DFlocal[5] = DFup[6] = {4}DFup[7] = {12}DFup[8] = {5,13}DTchildren[5] = {6,7,8}


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1

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2

3

412

10 11

9

8

6 7

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Control Flow Graph

What is DF[5]?Remember that:

DFlocal[5] = DFup[6] = {4}DFup[7] = {12}DFup[8] = {5,13}DTchildren[5] = {6,7,8}

Thus, DF[5] = {4, 5, 12, 13}


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Join Sets

In order to insert -nodes for a variable x that is defined in a set of nodes S={n1, n2, …, nk} we needto compute the iterated set of join nodes of S.

Given a set of nodes S of a control flow graph G, the set of join nodes of S, J(S), is defined as follows:

J(S) ={z G| two paths Pxz and Pyz in G that have z as its first common node, x S and y S}


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Iterated Join Sets

Because a -node is itself a definition of a variable,once we insert -nodes in the join set of S, we needto find out the join set of S J(S). Thus, Cytron-Ferrante define the iterated join set of a set of nodes S, J+(S), as the limit of the sequence:

( )( )ii JSJJ

SJJ

∪=

=

+1

1


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Iterated Dominance Frontier

We can extend the concept of dominance frontierto define the dominance frontier of a set of nodes as:

)()( XDFSDFSX∈

= U

Now we can define the iterated dominance frontier,DF+(S), of a set of nodes S as the limit of the sequence:

( )ii DFSDFDF

SDFDF

∪=

=

+1

1 )(

Exercise:

Find an example in which

the IDF of a set S is different

from the DF of the set!

Exercise:

Find an example in which

the IDF of a set S is different

from the DF of the set!


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Location of -Nodes

Given a variable x that is defined in a set of nodes S={n1, n2, …, nk} the set of nodes that must receive-nodes for x is J+(S).

An important result proved by Cytron-Ferrante is that:

( ) ( )SDFSJ ++ =

Thus we are mostly interested in computing the iterated dominance frontier of a set of nodes.


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Algorithms to Compute Dominance Frontier

The algorithm to insert -nodes, due to Cytron andFerrante (1991), computes the dominance frontierof each node in the set S before computing theiterated dominance frontier of the set.

In 1994, Shreedar and Gao proposed a simple,linear algorithm for the insertion of -nodes.

In the worst case, the combination of the dominancefrontier of the sets can be quadratic in the number of nodes in the CFG. Thus, Cytron-Ferrante’salgorithm has a complexity O(N2).


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Sreedhar and Gao’s DJ Graph

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1

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8Control Flow Graph

Dominator Tree


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1

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1

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6 7

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4 5 12 92 13

8Control Flow Graph

Dominator Tree

D nodes


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8Control Flow Graph

Dominator Tree

D nodes

J nodes


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Shreedar-Gao’s Dominance Frontier Algorithm

DominanceFrontier(x)0: DF[x] = 1: foreach y SubTree(x) do2: if((y z == J-edge) and3: (z.level x.level))4: then DF[x] = DF[x] z

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What is the DF[5]?


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SubTree(5) = {5, 6, 7, 8}

Initialization: DF[5] =


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SubTree(5) = {5, 6, 7, 8}

There are three edgesoriginating in 5: {56, 57, 58}

but they are all D-edges

Initialization: DF[5] =


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6 7

10 11

4 5 12 92 13

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SubTree(5) = {5, 6, 7, 8}

There are two edgesoriginating in 6:

{64, 68}but 8.level > 5.level

Initialization: DF[5] = After visiting 6: DF = {4}


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1

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6 7

10 11

4 5 12 92 13

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SubTree(5) = {5, 6, 7, 8}


{78, 712}again 8.level > 5.level


After visiting 7: DF = {4,12}


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1

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6 7

10 11

4 5 12 92 13

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SubTree(5) = {5, 6, 7, 8}


{85, 813}both satisfy cond. in steps 2-3


After visiting 7: DF = {4,12}After visiting 8: DF = {4, 12, 5, 13}


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Shreedhar-Gao’s -Node Insertion Algorithm

Using the D-J graph, Shreedhar and Gao proposea linear time algorithm to compute the iterateddominance frontier of a set of nodes.

An important intuition in Shreedhar-Gao’s algorithm is:

If two nodes x and y are in S, and y is an ancestor of x in the dominator tree, then if we compute DF[x] first, we do not need to recompute DF[x] when computing DF[y].


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Shreedhar-Gao’s -Node Insertion Algorithm

Shreedhar-Gao’s algorithm also use a work listof nodes hashed by their level in the dominatortree and a visited flag to avoid visiting thesame node more than once.

The basic operation of the algorithm is similar totheir dominance-frontier algorithm, but it requiresa careful implementation to deliver the linear-timecomplexity.


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Dead-Code Elimination in SSA Form

Because there is only one definition for eachvariable, if the list of uses of the variable

is empty, the definition is dead.

When a statement v x y is eliminated becausev is dead, this statement must be removed fromthe list of uses of x and y. Which might cause

those definitions to become dead.

Thus we need to iterate the dead code elimination algorithm.


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Simple Constant Propagation in SSA

If there is a statement v c, where c is a constant,then all uses of v can be replaced for c.

A function of the form v (c1, c2, …, cn) where allci are identical can be replaced for v c.

Using a work-list algorithm in a program in SSA form,we can perform constant propagation in linear time

In the next slide we assume that x, y, z are variablesand a, b, c are constants.


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Linear Time Optimizations in SSA

formCopy propagation: The statement x (y) or the statementx y can be deleted and y can substitute every use of x.

Constant folding: If we have the statement x a b, we can evaluate c a b at compile time and replace the statement for x c

Constant conditions: The conditional

if a < b goto L1 else L2

can be replaced for goto L1 or goto L2, according to thecompile time evaluation of a < b, and the CFG, use lists,adjust accordingly

Unreachable Code: eliminate unreachable blocks.


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Single Assignment Form

i=1;j=1;k=0; while(k<100) { if(j<20) { j=i; k=k+1; } else { j=k; k=k+2; } } return j;}

i 1j 1k 0

j ik k+1

j kk k+2

return jif j<20

if k<100

B1

B2

B3

B5 B6

B4

B7


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i 1j 1k1 0

j ik3 k+1

j kk5 k+2

return jif j<20

if k<100

B1

B2

B3

B5 B6

B4

B7


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i 1j 1k1 0

j ik3 k+1

j kk5 k+2

return jif j<20

if k<100

k4 (k3,k5)

B1

B2

B3

B5 B6

B4

B7


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i 1j 1k1 0

j ik3 k+1

j kk5 k+2

return jif j<20

k2 (k4,k1)if k<100

k4 (k3,k5)

B1

B2

B3

B5 B6

B4

B7


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i 1j 1k1 0

j ik3 k2+1

j kk5 k2+2

return jif j<20

k2 (k4,k1)if k2<100

k4 (k3,k5)

B1

B2

B3

B5 B6

B4

B7


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i1 1j1 1k1 0

j3 i1k3 k2+1

j5 k2k5 k2+2

return j2if j2<20

j2 (j4,j1)k2 (k4,k1)if k2<100

j4 (j3,j5)k4 (k3,k5)

B1

B2

B3

B5 B6

B4

B7


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Example:Constant Propagation

i1 1j1 1k1 0

j3 i1k3 k2+1

j5 k2k5 k2+2

return j2if j2<20

j2 (j4,j1)k2 (k4,k1)if k2<100

j4 (j3,j5)k4 (k3,k5)

B1

B2

B3

B5 B6

B4

B7

i1 1j1 1k1 0

j3 1k3 k2+1

j5 k2k5 k2+2

return j2if j2<20

j2 (j4,1)k2 (k4,0)if k2<100

j4 (j3,j5)k4 (k3,k5)

B1

B2

B3

B5 B6

B4

B7


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Example:Dead-code Elimination

i1 1j1 1k1 0

j3 1k3 k2+1

j5 k2k5 k2+2

return j2if j2<20

j2 (j4,1)k2 (k4,0)if k2<100

j4 (j3,j5)k4 (k3,k5)

B1

B2

B3

B5 B6

B4

B7

j3 1k3 k2+1

j5 k2k5 k2+2

return j2if j2<20

j2 (j4,1)k2 (k4,0)if k2<100

j4 (j3,j5)k4 (k3,k5)

B2

B3

B5 B6

B4

B7


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Constant Propagation and Dead Code Elimination

j3 1k3 k2+1

j5 k2k5 k2+2

return j2if j2<20

j2 (j4,1)k2 (k4,0)if k2<100

j4 (1,j5)k4 (k3,k5)

B2

B3

B5 B6

B4

B7

j3 1k3 k2+1

j5 k2k5 k2+2

return j2if j2<20

j2 (j4,1)k2 (k4,0)if k2<100

j4 (j3,j5)k4 (k3,k5)

B2

B3

B5 B6

B4

B7


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Example:Is this the end?

But block 6 is neverexecuted! How can we

find this out, and simplifythe program?

SSA conditional constantpropagation finds the

least fixed point for theprogram and allows

further elimination ofdead code.

See algorithm on pg. 454-455 of Appel.

k3 k2+1 j5 k2k5 k2+2

return j2if j2<20

j2 (j4,1)k2 (k4,0)if k2<100

j4 (1,j5)k4 (k3,k5)

B2

B3

B5 B6

B4

B7


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Example:Dead code elimination

k3 k2+1 j5 k2k5 k2+2

return j2if j2<20

j2 (j4,1)k2 (k4,0)if k2<100

j4 (1,j5)k4 (k3,k5)

B2

B3

B5 B6

B4

B7

B4

k3 k2+1

return j2

j2 (j4,1)k2 (k4,0)if k2<100

j4 (1)k4 (k3)

B2

B5

B7


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Example: Single Argument -Function

Elimination

k3 k2+1

return j2

j2 (j4,1)k2 (k4,0)if k2<100

j4 (1)k4 (k3)

B2

B5

B7

B4

k3 k2+1

return j2

j2 (j4,1)k2 (k4,0)if k2<100

j4 1k4 k3

B2

B5

B7

B4


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Example: Constant and Copy Propagation

k3 k2+1

return j2

j2 (j4,1)k2 (k4,0)if k2<100

j4 1k4 k3

B2

B5

B7

k3 k2+1

return j2

j2 (1,1)k2 (k3,0)if k2<100

j4 1k4 k3

B2

B5

B7

B4B4


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Example: Dead Code Elimination

k3 k2+1

return j2

j2 (1,1)k2 (k3,0)if k2<100

j4 1k4 k3

B2

B5

B7

B4

k3 k2+1

return j2

j2 (1,1)k2 (k3,0)if k2<100

B2

B5

B4


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Example: -Function

Simplification

k3 k2+1

return j2

j2 (1,1)k2 (k3,0)if k2<100

B2

B5

B4

k3 k2+1

return j2

j2 1k2 (k3,0)if k2<100

B2

B5

B4


77

Example: Constant Propagation

k3 k2+1

return j2

j2 1k2 (k3,0)if k2<100

B2

B5

B4

k3 k2+1

return 1

j2 1k2 (k3,0)if k2<100

B2

B5

B4


78

Example: Dead Code Elimination

k3 k2+1

return 1

j2 1k2 (k3,0)if k2<100

B2

B5

B4 return 1 B4

CMPUT680 - Winter 2006

Documents

compiler design

bssa form

single static definition

optimizationssa form

definitionsssa form

knowswhich control path

controlflow path mergesb

paths xz