Common Sub-expression Elim Want to compute when an expression is available in a var Domain:

Common Sub-expression Elim

• Want to compute when an expression is available in a var

• Domain:

Common Sub-expression Elim

• Want to compute when an expression is available in a var

• Domain:

Flow functions

X := Y op Z

in

out

FX := Y op Z(in) =

X := Y

in

out

FX := Y(in) =

Flow functions

X := Y op Z

in

out

FX := Y op Z(in) = in – { X ! * } – { * ! ... X ... } [{ X ! Y op Z | X Y Æ X Z}

X := Y

in

out

FX := Y(in) = in – { X ! * } – { * ! ... X ... } [{ X ! E | Y ! E 2 in }

Example

x := read()

v := a + b

x := x + 1

w := x + 1

w := x + 1

a = w

v = a + b

z := x + 1

t = a + b

Direction of analysis

• Although constraints are not directional, flow functions are

• All flow functions we have seen so far are in the forward direction

• In some cases, the constraints are of the formin = F(out)

• These are called backward problems.

• Example: live variables– compute the set of variables that may be live

Example: live variables

• Set D =

• Lattice: (D, v, ?, >, t, u) =

Example: live variables

• Set D = 2 Vars

• Lattice: (D, v, ?, >, t, u) = (2Vars, µ, ; ,Vars, [, Å)

x := y op z

in

out

Fx := y op z(out) =

Example: live variables

• Set D = 2 Vars

• Lattice: (D, v, ?, >, t, u) = (2Vars, µ, ; ,Vars, [, Å)

x := y op z

in

out

Fx := y op z(out) = out – { x } [ { y, z}

Example: live variables

x := 5

y := x + 2

x := x + 1 y := x + 10

... y ...

Example: live variables

x := 5

y := x + 2

x := x + 1 y := x + 10

... y ...How can we remove the x := x + 1 stmt?

Revisiting assignment

x := y op z

in

out

Fx := y op z(out) = out – { x } [ { y, z}

Revisiting assignment

x := y op z

in

out

Fx := y op z(out) = out – { x } [ { y, z}

Theory of backward analyses

• Can formalize backward analyses in two ways

• Option 1: reverse flow graph, and then run forward problem

• Option 2: re-develop the theory, but in the backward direction

Precision

• Going back to constant prop, in what cases would we lose precision?

Precision

• Going back to constant prop, in what cases would we lose precision?

if (p) { x := 5;} else x := 4;}...if (p) { y := x + 1} else { y := x + 2}... y ...

if (...) { x := -1;} else x := 1;}y := x * x;... y ...

x := 5if (<expr>) { x := 6}... x ...

where <expr> is equiv to false

Precision

• The first problem: Unreachable code– solution: run unreachable code removal before– the unreachable code removal analysis will do its

best, but may not remove all unreachable code

• The other two problems are path-sensitivity issues– Branch correlations: some paths are infeasible– Path merging: can lead to loss of precision

MOP: meet over all paths

• Information computed at a given point is the meet of the information computed by each path to the program point

if (...) { x := -1;} else x := 1;}y := x * x;... y ...

MOP

• For a path p, which is a sequence of statements [s1, ..., sn] , define: Fp(in) = Fsn

( ...Fs1(in) ... )

• In other words: Fp =

• Given an edge e, let paths-to(e) be the (possibly infinite) set of paths that lead to e

• Given an edge e, MOP(e) =

• For us, should be called JOP (ie: join, not meet)

MOP vs. dataflow

• MOP is the “best” possible answer, given a fixed set of flow functions– This means that MOP v dataflow at edge in the CFG

• In general, MOP is not computable (because there can be infinitely many paths)– vs dataflow which is generally computable (if flow fns

are monotonic and height of lattice is finite)

• And we saw in our example, in general,MOP dataflow

MOP vs. dataflow

• However, it would be great if by imposing some restrictions on the flow functions, we could guarantee that dataflow is the same as MOP. What would this restriction be?

x := -1;y := x * x;... y ...

x := 1;y := x * x;... y ...

Merge

x := -1; x := 1;

Merge

y := x * x;... y ...

Dataflow MOP

MOP vs. dataflow

• However, it would be great if by imposing some restrictions on the flow functions, we could guarantee that dataflow is the same as MOP. What would this restriction be?

• Distributive problems. A problem is distributive if:

8 a, b . F(a t b) = F(a) t F(b)

• If flow function is distributive, then MOP = dataflow

Summary of precision

• Dataflow is the basic algorithm

• To basic dataflow, we can add path-separation– Get MOP, which is same as dataflow for distributive

problems– Variety of research efforts to get closer to MOP for

non-distributive problems

• To basic dataflow, we can add path-pruning– Get branch correlation

• To basic dataflow, can add both: – meet over all feasible paths

Program Representations

Representing programs

• Goals

Representing programs

• Primary goals– analysis is easy and effective

• just a few cases to handle• directly link related things

– transformations are easy to perform– general, across input languages and target machines

• Additional goals– compact in memory– easy to translate to and from– tracks info from source through to binary, for source-level

debugging, profilling, typed binaries– extensible (new opts, targets, language features)– displayable

Option 1: high-level syntax based IR

• Represent source-level structures and expressions directly

• Example: Abstract Syntax Tree

Option 2: low-level IR

• Translate input programs into low-level primitive chunks, often close to the target machine

• Examples: assembly code, virtual machine code (e.g. stack machines), three-address code, register-transfer language (RTL)

• Standard RTL instrs:

Option 2: low-level IR

Comparison

Comparison

• Advantages of high-level rep– analysis can exploit high-level knowledge of

constructs– easy to map to source code (debugging, profiling)

• Advantages of low-level rep– can do low-level, machine specific reasoning– can be language-independent

• Can mix multiple reps in the same compiler

Components of representation

• Control dependencies: sequencing of operations– evaluation of if & then– side-effects of statements occur in right order

• Data dependencies: flow of definitions from defs to uses– operands computed before operations

• Ideal: represent just those dependencies that matter– dependencies constrain transformations– fewest dependences ) flexibility in implementation

Control dependencies

• Option 1: high-level representation– control implicit in semantics of AST nodes

• Option 2: control flow graph (CFG)– nodes are individual instructions– edges represent control flow between instructions

• Options 2b: CFG with basic blocks– basic block: sequence of instructions that don’t have

any branches, and that have a single entry point– BB can make analysis more efficient: compute flow

functions for an entire BB before start of analysis

Control dependencies

• CFG does not capture loops very well

• Some fancier options include:– the Control Dependence Graph– the Program Dependence Graph

• More on this later. Let’s first look at data dependencies

Data dependencies

• Simplest way to represent data dependencies: def/use chains

Def/use chains

• Directly captures dataflow– works well for things like constant prop

• But...

• Ignores control flow– misses some opt opportunities since conservatively considers all

paths– not executable by itself (for example, need to keep CFG around)– not appropriate for code motion transformations

• Must update after each transformation

• Space consuming

SSA

• Static Single Assignment– invariant: each use of a variable has only one def

SSA

• Create a new variable for each def

• Insert pseudo-assignments at merge points

• Adjust uses to refer to appropriate new names

• Question: how can one figure out where to insert nodes using a liveness analysis and a reaching defns analysis.

Converting back from SSA

• Semantics of x3 := (x1, x2)

– set x3 to xi if execution came from ith predecessor

• How to implement nodes?

Converting back from SSA

• Semantics of x3 := (x1, x2)

– set x3 to xi if execution came from ith predecessor

• How to implement nodes?– Insert assignment x3 := x1 along 1st predecessor

– Insert assignment x3 := x2 along 2nd predecessor

• If register allocator assigns x1, x2 and x3 to the same register, these moves can be removed– x1 .. xn usually have non-overlapping lifetimes, so this

kind of register assignment is legal