Advanced Compiler Techniques

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Advanced Compiler Techniques

LIU XianhuaSchool of EECS, Peking University

Loops

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“Advanced Compiler Techniques”

Content

Concepts:DominatorsDepth-First OrderingBack edgesGraph depthReducibilityNatural LoopsEfficiency of Iterative AlgorithmsDependences & Loop Transformation

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“Advanced Compiler Techniques”

Loops are Important!

Loops dominate program execution timeNeeds special treatment during optimizationLoops also affect the running time of program analysese.g., A dataflow problem can be solved in just a single pass if a program has no loops

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“Advanced Compiler Techniques”

Dominators

Node d dominates node n if every path from the entry to n goes through d.written as: d dom nQuick observations:Every node dominates itself.The entry dominates every node.Common Cases:The test of a while loop dominates all blocks in the loop body.The test of an if-then-else dominates all blocks in either branch.

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Dominator Tree

Immediate dominance: d idom n d dom n, d  n, no m s.t. d dom m and m dom nImmediate dominance relationships form a tree

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Finding Dominators

A dataflow analysis problem: For each node, find all of its dominators.Direction: forwardConfluence: set intersectionBoundary: OUT[Entry] = {Entry}Initialization: OUT[B] = All nodesEquations:OUT[B] = IN[B] U {B}IN[B] = p is a predecessor of B OUT[p]

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Example: Dominators

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{1,5}

{1,4}

{1,2,3}

{1,2}

{1}

{1}

{1}

{1}

{1}

{1,2}

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“Advanced Compiler Techniques”

Depth-First Search

Start at entry.If you can follow an edge to an unvisited node, do so.If not, backtrack to your parent (node from which you were visited).

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Depth-First Spanning Tree

Root = entry.Tree edges are the edges along which we first visit the node at the head.

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Depth-First Node Order

The reverse of the order in which a DFS retreats from the nodes. 1-4-5-2-3Alternatively, reverse of postorder traversal of the tree. 3-2-5-4-1

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Four Kinds of Edges

Tree edges.Advancing edges (node to proper descendant).Retreating edges (node to ancestor, including edges to self).Cross edges (between two nodes, neither of which is an ancestor of the other.

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A Little Magic

Of these edges, only retreating edges go from high to low in DF order.Example of proof: You must retreat from the head of a tree edge before you can retreat from its tail.Also surprising: all cross edges go right to left in the DFST.Assuming we add children of any node from the left.

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Example: Non-Tree Edges

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Retreating

Forward

Cross

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Back Edges

An edge is a back edge if its head dominates its tail.Theorem: Every back edge is a retreating edge in every DFST of every flow graph.Converse almost always true, but not always.

Back

edge

Head reached before

tail in any DFST

Search must reach the

tail before retreating

from the head, so tail is

a descendant of the head

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Example: Back Edges

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{1,5}

{1,4}

{1,2,3}

{1,2}

{1}

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Reducible Flow Graphs

A flow graph is reducible if every retreating edge in any DFST for that flow graph is a back edge.Testing reducibility: Remove all back edges from the flow graph and check that the result is acyclic.Hint why it works: All cycles must include some retreating edge in every DFST.In particular, the edge that enters the first node of the cycle that is visited.

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DFST on a Cycle

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Depth-first search

reaches here first

Search must reach

these nodes before

l

eaving the cycle

So this is a

retreating edge

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Why Reducibility?

Folk theorem: All flow graphs in practice are reducible.Fact: If you use only while-loops, for-loops, repeat-loops, if-then(-else), break, and continue, then your flow graph is reducible.

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Example: Remove Back Edges

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Remaining graph is acyclic.

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Example: Nonreducible Graph

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A

C

B

In any DFST, one

of these edges will

be a retreating edge.

A

B

C

A

B

C

But no heads dominate their tails, so deleting

back edges leaves the cycle.

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Why Care AboutBack/Retreating Edges?

Proper ordering of nodes during iterative algorithm assures number of passes limited by the number of “nested” back edges.Depth of nested loops upper-bounds the number of nested back edges.

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DF Order and Retreating Edges

Suppose that for a RD analysis, we visit nodes during each iteration in DF order.The fact that a definition d reaches a block will propagate in one pass along any increasing sequence of blocks.When d arrives at the tail of a retreating edge, it is too late to propagate d from OUT to IN.The IN at the head has already been computed for that round.

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Example: DF Order

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d

d

d

d

d

d

d

d

d

d

Definition d is

Gen’d

by node 2.

The first pass

The second pass

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Depth of a Flow Graph

The depth of a flow graph with a given DFST and DF-order is the greatest number of retreating edges along any acyclic path.For RD, if we use DF order to visit nodes, we converge in depth+2 passes.Depth+1 passes to follow that number of increasing segments.1 more pass to realize we converged.

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Example: Depth = 2

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1->4->7 ---> 3->10->17 ---> 6->18->20

increasing

retreating

increasing

increasing

retreating

Pass 1

Pass 2

Pass 3

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“Advanced Compiler Techniques”

Similarly . . .

AE also works in depth+2 passes.Unavailability propagates along retreat-free node sequences in one pass.So does LV if we use reverse of DF order.A use propagates backward along paths that do not use a retreating edge in one pass.

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In General . . .

The depth+2 bound works for any monotone framework, as long as information only needs to propagate along acyclic paths.Example: if a definition reaches a point, it does so along an acyclic path.

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However . . .

Constant propagation does not have this property.

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a = b

b = c

c = 1

L: a = b

b = c c = 1 goto L

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Why Depth+2 is Good

Normal control-flow constructs produce reducible flow graphs with the number of back edges at most the nesting depth of loops.Nesting depth tends to be small.A study by Knuth has shown that average depth of typical flow graphs =~2.75.

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Example: Nested Loops

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3 nested while-

loops; depth =

3

3 nested repeat-

loops; depth = 1

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Natural Loops

A natural loop is defined by:A single entry-point called headera header dominates all nodes in the loopA back edge that enters the loop headerOtherwise, it is not possible for the flow of control to return to the header directly from the "loop" ; i.e., there really is no loop.

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Find Natural Loops

The natural loop of a back edge a->b is {b} plus the set of nodes that can reach a without going through bRemove b from the flow graph, find all predecessors of aTheorem: two natural loops are either disjoint, identical, or nested.

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Example: Natural Loops

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Natural loop

of 3 -> 2

Natural loop

of 5 -> 1

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Relationship between Loops

If two loops do not have the same headerthey are either disjoint, orone is entirely contained (nested within) the otherinnermost loop: one that contains no other loop.If two loops share the same headerHard to tell which is the inner loopCombine as one

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Basic Parallelism

Examples:FOR i = 1 to 100 a[i] = b[i] + c[i]FOR i = 11 TO 20 a[i] = a[i-1] + 3FOR i = 11 TO 20 a[i] = a[i-10] + 3Does there exist a data dependence edge between two different iterations?A data dependence edge is loop-carried if it crosses iteration boundariesDoAll loops: loops without loop-carried dependences

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Data Dependence of Variables

True dependence

Anti-dependence

a = = a

= aa =

a = a =

= a = a

Output dependenceInput dependence

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Affine Array Accesses

Common patterns of data accesses: (i, j, k are loop indexes)A[i], A[j], A[i-1], A[0], A[i+j], A[2*i], A[2*i+1], A[i,j], A[i-1, j+1]Array indexes are affine expressions of surrounding loop indexesLoop indexes: in, in-1, ... , i1Integer constants: cn, cn-1, ... , c0Array index: cnin + cn-1in-1+ ... + c1i1+ c0Affine expression: linear expression + a constant term (c0)

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Formulating DataDependence Analysis

FOR i := 2 to 5 do A[i-2] = A[i]+1;

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Between read access A[i] and write access A[i-2] there is a dependence if:there exist two iterations ir and iw within the loop bounds, s.t.iterations ir & iw read & write the same array element, respectively ∃integers iw, ir 2≤iw,ir≤5 ir=iw-2Between write access A[i-2] and write access A[i-2] there is a dependence if:∃integers iw, iv 2≤iw,iv≤5 iw–2=iv–2To rule out the case when the same instance depends on itself: add constraint iw ≠ iv

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Memory Disambiguation

Undecidable at Compile Time read(n) For i = … a[i] = a[n]

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Domain of Data Dependence Analysis

Only use loop bounds and array indexes that are affine functions of loop variablesfor i = 1 to nfor j = 2i to 100a[i + 2j + 3][4i + 2j][i * i] = … … = a[1][2i + 1][j]Assume a data dependence between the read & write operation if there exists:∃integers ir,jr,iw,jw 1 ≤ iw, ir ≤ n 2iw ≤ jw ≤ 100 2ir ≤ jr ≤ 10 iw + 2jw + 3 = 1 4iw + 2jw = 2ir + 1Equate each dimension of array access; ignore non-affine onesNo solution No data dependenceSolution  There may be a dependence

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Iteration Space

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An abstraction for loops Iteration is represented as coordinates in iteration space.

for i= 0, 5 for j = 0, 3 a[i, j] = 3

i

j

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Iteration Space

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An abstraction for loops

for i = 0, 5 for j = i, 3 a[i, j] = 0

i

j

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Iteration Space

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An abstraction for loops

for

i = 0, 5 for j = i, 7 a[i, j] = 0

i

j

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Affine Access

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Affine Transform

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i

j

u

v

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Loop Transformation

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for i = 1, 100 for j = 1, 200 A[i, j] = A[i, j] + 3 end_forend_for

for u = 1, 200 for v = 1, 100 A[v,u] = A[v,u]+ 3 end_forend_for

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Old Iteration Space

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for

i = 1, 100

for

j = 1, 200 A[i, j] = A[i, j] + 3 end_forend_for

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New Iteration Space

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for

u = 1, 200

for

v = 1, 100 A[v,u] = A[v,u]+ 3 end_forend_for

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Old Array Accesses

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for

i = 1, 100

for

j = 1, 200

A[i, j] = A[i, j] + 3

end_forend_for

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New Array Accesses

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for

u = 1, 200

for v = 1, 100 A[v,u] = A[v,u]+ 3 end_forend_for

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Interchange Loops?

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for i = 2, 1000 for j = 1, 1000 A[i, j] = A[i-1, j+1]+3 end_forend_for

e.g. dependence vector dold = (1,-1)

i

j

for

u

=

1, 1000 for v = 2, 1000 A[v, u] = A[v-1, u+1]+3 end_forend_for

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Interchange Loops?

A transformation is legal, if the new dependence is lexicographically positive, i.e. the leading non-zero in the dependence vector is positive.Distance vector (1,-1) = (4,2)-(3,3)Loop interchange is not legal if there exists dependence (+, -)

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GCD Test

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Is there any dependence?Solve a linear Diophantine equation2*iw = 2*ir + 1

for i = 1, 100 a[2*i] = … … = a[2*i+1] + 3

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GCD

The greatest common divisor (GCD) of integers a1, a2, …, an, denoted gcd(a1, a2, …, an), is the largest integer that evenly divides all these integers. Theorem: The linear Diophantine equation has an integer solution x1, x2, …, xn iff gcd(a1, a2, …, an) divides c

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Examples

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Example 1: gcd(2,-2) = 2. No solutionsExample 2: gcd(24,36,54) = 6. Many solutions

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Loop Fusion

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for i = 1, 1000 A[i] = B[i] + 3end_forfor j = 1, 1000 C[j] = A[j] + 5end_for

for i = 1, 1000 A[i] = B[i] + 3 C[i] = A[i] + 5end_for

Better reuse between A[i] and A[i]

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Loop Distribution

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for i = 1, 1000 A[i] = A[i-1] + 3end_forfor i = 1, 1000 C[i] = B[i] + 5end_for

for i = 1, 1000 A[i] = A[i-1] + 3 C[i] = B[i] + 5end_for

2nd loop is parallel

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Register Blocking

for j = 1, 2*m for i = 1, 2*n A[i, j] = A[i-1, j] + A[i-1, j-1] end_forend_for

for j = 1, 2*m, 2 for i = 1, 2*n, 2 A[i, j] = A[i-1,j] + A[i-1,j-1] A[i, j+1] = A[i-1,j+1] + A[i-1,j] A[i+1, j] = A[i, j] + A[i, j-1] A[i+1, j+1] = A[i, j+1] + A[i, j] end_forend_for

Better reuse between A[

i,j] and A[i,j]

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Virtual Register Allocation

for j = 1, 2*M, 2 for i = 1, 2*N, 2 r1 = A[i-1,j] r2 = r1 + A[i-1,j-1] A[i, j] = r2 r3 = A[i-1,j+1] + r1 A[i, j+1] = r3 A[i+1, j] = r2 + A[i, j-1] A[i+1, j+1] = r3 + r2 end_forend_for

Memory operations reduced to register load/store

8MN loads to 4MN loads

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Scalar Replacement

for i = 2, N+1 = A[i-1]+1 A[i] =end_for

t1 = A[1]for i = 2, N+1 = t1 + 1 t1 = A[i] = t1end_for

Eliminate loads and stores for array references

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Unroll-and-Jam

for j = 1, 2*M for i = 1, N A[i, j] = A[i-1, j] + A[i-1, j-1] end_forend_for

for j = 1, 2*M, 2 for i = 1, N A[i, j]=A[i-1,j]+A[i-1,j-1] A[i, j+1]=A[i-1,j+1]+A[i-1,j] end_forend_for

Expose more opportunity for scalar replacement

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Large Arrays

for i = 1, 1000 for j = 1, 1000 A[i, j] = A[i, j] + B[j, i] end_forend_for

Suppose arrays A and B have row-major layout

B has poor cache locality.Loop interchange will not help.

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Loop Blocking

for v = 1, 1000, 20 for u = 1, 1000, 20 for j = v, v+19 for i = u, u+19 A[i, j] = A[i, j] + B[j, i] end_for end_for end_forend_for

Access to small blocks of the arrays has good cache locality.

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Loop Unrolling for ILP

for i = 1, 10 a[i] = b[i]; *p = ... end_for

for I = 1, 10, 2 a[i] = b[i]; *p = … a[i+1] = b[i+1]; *p = …end_for

Large scheduling regions. Fewer dynamic branchesIncreased code size

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Next Time

Homework9.6.2, 9.6.4, 9.6.7Single Static Assignment (SSA)Readings: Cytron'91, Chow'97

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